The true statements about motion are A, B, and D. While changing direction or speed often involves a force, speed can also change due to factors like gravity. An object in a vacuum will continue to move indefinitely in a straight line, if no other forces act on it.
Explanation:The statements about motion that are true are A, B, and D. Statement A is true because according to Newton's first law of motion, an object will continue to move in a straight line unless acted upon by an external force. Therefore, changing direction implies a force.
B is true because both acceleration and deceleration (changing speed) can be caused by forces; however, an object can also change its speed due to factors like gravity or friction, particularly in fluid mechanics. Statement D is true in the context of an ideal vacuum, where an object would indeed continue to move indefinitely if no other forces were present. However, C is not necessarily true because an object can also move due to inertia or if it was already in motion.
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Final answer:
The true statements about motion, based on Newton's Laws, are that an object changing direction requires a force, and an object in outer space will maintain its motion in a straight line indefinitely in the absence of external forces. Statements asserting the necessity of a force to maintain motion or to move to the right are false due to Newton's first law of inertia.
Explanation:
The discussion of motion and force is grounded in the principles of Newton's Laws of Motion, specifically the first law which is also known as the law of inertia. Here's an assessment of the provided statements:
A. True, according to Newton's first law, an object will not change its state of motion unless a net force is acting on it. Changing direction is a form of acceleration, and hence a force must be present.
B. False. If an object is changing speed, a force must be acting on it. This is based on Newton's second law, which states that the acceleration (change in speed) of an object is due to a net force acting on it.
C. False. An object can continue to move to the right without any forces pushing or pulling it, provided that it was already moving to the right and no forces are acting to stop it or change its direction. This is also an implication of Newton's first law.
D. True. This statement is a direct consequence of Newton's first law. An object in motion will continue to move in a straight line at constant velocity if no external force acts on it.
Can someone please help me with these 2 questions
1) A car takes 76 metres to stop once a driver has noticed a hazard. If the thinking distance of the driver was 14m and the speed of the car was 8m/s. how long did it take the brakes to slow the car down to a stop?
2) A lorry travels 28m when stopping from a speed of 4m/s. If its braking distance was 18m, what was the driver's reaction time?
A car traveling at a constant speed travels 175 miles in 4 hours. How many feet will the car travel in 10 minutes?
Answer: 0.26 minutes
Explanation:
Answer: 38,500 feet
Explanation:
look at the attachment where the work is done out
A 866 kg car is traveling at 16.7 m/s around a banked curve with a radius
of 50 m. At what angle would the car have to be banked so that no
friction is required?
Answer: [tex]29.64 \°[/tex]
Explanation:
The expression to calculate the angle [tex]\theta[/tex] of an ideally banked curve in which friciton is not needed is:
[tex]\theta=tan^{-1}(\frac{V^{2}}{r.g})[/tex]
Where:
[tex]V=16.7 m/s[/tex] is the car's velocity
[tex]r.=50 m[/tex] is the radius of the curve
[tex]g=9.8 m/s^{2}[/tex] is the acceleration due gravity
As you may see, this angle does not depend on the mass of the car.
Solving with the given values:
[tex]\theta=tan^{-1}(\frac{(16.7 m/s)^{2}}{(50 m)(9.8 m/s^{2})})[/tex]
Finally:
[tex]\theta=29.64 \°[/tex]
a. A child is twirling a 1.52 kg object in a vertical circle with a radius of 67.6
cm at constant speed. The tension in the string is 54.7 N at the bottom of
the circle. (a.) What is the speed of the object?
b. (b) What is the tension in the string at the top of the circle?
Answer:
(a) 4.21 m/s
(b) 24.9 N
Explanation:
(a) Draw a free body diagram of the object when it is at the bottom of the circle. There are two forces on the object: tension force T pulling up and weight force mg pulling down.
Sum the forces in the radial (+y) direction:
∑F = ma
T − mg = m v² / r
v = √(r (T − mg) / m)
v = √(0.676 m (54.7 N − 1.52 kg × 9.8 m/s²) / 1.52 kg)
v = 4.21 m/s
(b) Draw a free body diagram of the object when it is at the top of the circle. There are two forces on the object: tension force T pulling down and weight force mg pulling down.
Sum the forces in the radial (-y) direction:
∑F = ma
T + mg = m v² / r
T = m v² / r − mg
T = (1.52 kg) (4.21 m/s)² / (0.676 m) − (1.52 kg) (9.8 m/s²)
T = 24.9 N
Whích best explains parallel forces?
Answer:
Parallel forces lie on the same plane and they have lines of action that never intersect each other.
Explanation:
Parallel forces definition is - forces acting in parallel lines.
In physics, parallel and antiparallel forces are forces that are oriented in the same or opposite directions, as described by Newton's third law of motion. An example is the force applied to push a filing cabinet, along with the force of friction opposing it.
Explanation:Parallel forces refer to forces that are oriented along the same direction, and antiparallel forces are forces that are oriented in opposite directions. An example of parallel or antiparallel forces can be observed in everyday situations such as pushing a filing cabinet. The force applied by a person to push the cabinet and the frictional force that opposes the motion are an example of antiparallel forces. Additionally, Newton's third law of motion illustrates this concept by stating that for every action, there is an equal and opposite reaction, forming a pair of parallel or antiparallel forces.
Choosing a system of interest is crucial when analyzing forces because it determines which forces are considered internal and which are external. For instance, if the system of interest includes both the Earth and a falling object, the gravitational force between them does not cause acceleration of the system as a whole because it is an internal force. However, if the system is only the falling object, then the gravitational force is external, and does not have an equal and opposite force within the system, leading to acceleration.
In physics, forces can sometimes be unified at extremely short distances and high energies, making the distinctions between them less noticeable. While forces like electromagnetic, weak nuclear, and strong nuclear forces appear distinct, they may be different manifestations of a more fundamental force, as suggested by modern physics. This unification is part of the quest for simplicity and understanding the underlying principles governing the forces of nature.
If a wave has a wavelength of 4.5 m and A. Of 0.007 seconds what is the velocity of the wave? be sure to show the steps of your work.
Answer:The wave period is the time it takes to complete one cycle. The standard unit of a wave period is in seconds, and it is inversely proportional to the frequency of a wave, which is the number of cycles of waves that occur in one second.
Explanation:
Answer:
v = 640m/s
Explanation:
An airplane has a mass of 2×10^6 kg and air flows past the power surface of the wings at 100ms¯¹. If the wings have a surface area of 1200m², how fast must the air flow over the upper surface of the wing if the plane is to stay in the air? Consider only the Bernoulli's effect.
Answer:
190 m/s
Explanation:
For the plane to stay in the air, the lift force must equal the weight.
The lift force is also equal to the pressure difference across the wings (pressure at the bottom minus pressure at the top) times the area of the wings.
Therefore:
mg = (P₂ − P₁) A
P₂ − P₁ = mg / A
Using Bernoulli equation:
P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂
P₁ + ½ ρ v₁² = P₂ + ½ ρ v₂²
½ ρ (v₁² − v₂²) = P₂ − P₁
½ ρ (v₁² − v₂²) = mg / A
v₁² − v₂² = 2mg / (Aρ)
v₁² = v₂² + 2mg / (Aρ)
Substituting values (assuming air density of 1.225 kg/m³):
v₁² = (100 m/s)² + 2 (2×10⁶ kg) (9.8 m/s²) / (1200 m² × 1.225 kg/m³)
v₁² = 36,666.67 m²/s²
v₁ = 191 m/s
Rounding to two significant figures, the air must move at 190 m/s over the top of the wing.
To keep the plane in the air, the air flowing over the upper surface of the wings must be faster than the air flowing past the lower surface. This is due to Bernoulli's principle, which states that as the speed of a fluid increases, its pressure decreases. We can use the equation v₂ = sqrt(v₁² + 2(P₁ - P₂)/ρ) to calculate the speed of the air over the upper surface of the wing.
Explanation:To keep the plane in the air, the air flowing over the upper surface of the wings must be faster than the air flowing past the lower surface. This is due to Bernoulli's principle, which states that as the speed of a fluid increases, its pressure decreases. The difference in pressure between the upper and lower surfaces of the wing creates lift.
To calculate the speed of the air over the upper surface of the wing, we can use the equation:
P₁ + ½ρv₁² = P₂ + ½ρv₂²
P₁ is the pressure below the wing, P₂ is the pressure above the wing, ρ is the density of the air, v₁ is the speed of the air below the wing, and v₂ is the speed of the air above the wing.
We can rearrange the equation to solve for v₂:
v₂ = sqrt(v₁² + 2(P₁ - P₂)/ρ)
Plugging in the given values, we get:
v₂ = sqrt(100² + 2(0 - P₂)/(1.2))
Since we don't have the specific values for P₁ and P₂, we cannot calculate the exact speed of the air over the upper surface of the wing. However, we can determine that it must be greater than 100 m/s in order for the plane to stay in the air.
Learn more about Bernoulli's principleYou design an experiment to see how light effects plant growth. You put one plant in a room with no
windows and one lamp. You put another plant bought from the same store that was the same size, on a
windowsill in your kitchen. Two weeks later, the plant in the room with no windows has died. The plant on
the windowsill is growing great! What was the dependent variable?
The temperature in the room.
The life or death of the plant.
The light the plants received
The kind of plants that were bought
Answer:
dependent variable is the light the plant received
Explanation:
Answer:
kjhgfd
Explanation:
At an instant a traffic light turns green an automobile that has been waiting at an intersection of the road accelerates with 5m/s². At the same instant a truck travelling with a constant velocity of 20m/s overtake and passes the automobile. How far beyond the starting point will the automobile overtake the truck and how fast will it be travelling.
1) The car overtakes the truck at a distance of 160 m far from the intersection.
2) The velocity of the car is 40 m/s
Explanation:
1)
The car is travelling with a constant acceleration starting from rest, so its position at time t (measured taking the intersection as the origin) is given by
[tex]x_c(t) = \frac{1}{2}at^2[/tex]
where
[tex]a=5 m/s^2[/tex] is the acceleration
t is the time
On the other hand, the truck is travelling at a constant velocity, therefore its position at time t is given by
[tex]x_t(t) = vt[/tex]
where
v = 20 m/s is the velocity of the truck
t is the time
The car overtakes the truck when the two positions are the same, so when
[tex]x_c(t) = x_t(t)\\\frac{1}{2}at^2 = vt\\t=\frac{2v}{a}=\frac{2(20)}{5}=8 s[/tex]
So, after a time of 8 seconds. Therefore, the distance covered by the car during this time is
[tex]x_c(8) = \frac{1}{2}(5)(8)^2=160 m[/tex]
So, the car overtakes the truck 160 m far from the intersection.
2)
The motion of the car is a uniformly accelerated motion, so the velocity of the car at time t is given by the suvat equation
[tex]v=u+at[/tex]
where
v is the velocity at time t
u is the initial velocity
a is the acceleration
For the car in this problem, we have:
u = 0 (it starts from rest)
[tex]a=5 m/s^2[/tex]
And we know that the car overtakes the truck when
t = 8 s
Substituting into the equation,
[tex]v=0+(5)(8)=40 m/s[/tex]
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What has happened to scientists' estimate of the world's age over time?
a) They have become increasingly uncertain as to the age of the earth
b) their thoughts about the age of the earth has remained unchanged
c) they think the Earth is much younger
d) they think the world is much older
Scientists' estimate of the world's age has changed significantly over time, thanks to the discovery of radiometric dating techniques. They now estimate the Earth to be around 4.5 billion years old.
Explanation:Over time, scientists' estimate of the world's age has changed significantly. Initially, scientists believed that the Earth was only a few thousand years old, based on religious and historical texts. However, with the discoveries of radioactivity and the development of radiometric dating techniques, scientists now estimate the age of the Earth to be around 4.5 billion years. This estimate has become increasingly accurate and is supported by various lines of evidence, including the dating of rocks and the Moon's formation.
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2. A driver brings a car traveling at 22 m/s to a stop in 2.0 seconds. What is the car's acceleration?
Answer:
11 m/s²
Explanation:
Acc = v/t
Acc = 22 / 2.0
Acc = 11 m/s²
Answer: a= m/s divide by sec
22m/s divided by 2.0sec
11m/s
Explanation: dividing the meter per second [m/s] by the second [s
A billiard ball of mass 0.28 kg hits a second, identical ball at a speed of 7.2 m/s and comes to rest as the second ball flies off. The collision takes 250 μs. What is the average force on the first ball? The second ball?
Answer:
(a) -8064 N
(b) 8064 N
Explanation:
(a)
From Newton’s law of motion, Force, F=ma where m is mass and a is acceleration.
Since acceleration is the rate of change of velocity per unit time, then where v is velocity and the subscripts f and I denote final and initial
For the first ball, the mass is 0.28 Kg, final velocity is zero since it finally comes to rest, t is 0.00025 s and initial velocity is given as 7.2 s. Substituting these values we obtain
[tex]F=0.28\times \frac {0-7.2}{0.00025}=-8064 N[/tex]
(b)
For the second ball, the mass is also 0.28 Kg but its initial velocity is taken as zero, the final velocity of the second ball will be equal to the initial velocity of the second ball, that is 7.2 m/s and the time is also same, 0.00025 s. By substitution
[tex]F=0.28\times \frac {7.2-0}{0.00025}=8064 N[/tex]
Here, we prove that action and reaction are equal and opposite
The average force on the first ball is -8,064 N
The average force on the first ball is 8,064 N
The formula for calculating the average force according to Newton's second law is expressed as:
[tex]F=ma\\F=m(\frac{v-u}{t} )[/tex]
m is the mass of the ball
v is the final velocity
u is the initial velocity
t is the time taken
Given the following parameters
m = 0.28kg
v = 0m/s
u = 7.2m/s
t = 250×10⁻⁶secs
Substitute the given parameters into the formula:
[tex]F=0.28(\frac{0-7.2}{0.00025} )\\F=0.28(\frac{-7.2}{0.00025} )\\F=0.28(-28,800)\\F=-8,064N[/tex]
Hence the average force on the first ball is -8,064 N
For the average force on the second ball:
[tex]F=0.28(\frac{7.2-0}{0.00025} )\\F=0.28(\frac{7.2}{0.00025} )\\F=0.28(28,800)\\F=8,064N[/tex]
Hence the average force on the first ball is 8,064 N
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the energy of an onject as it is in motion is defined as?
The energy of an object as it is in motion is defined as Kinetic energy.
Explanation:
The energy that is attained by an object when it is moving is called as Kinetic energy. It is the amount of energy that is essential for inducing an acceleration in an object and making it to displace from its idle position to the destination. When an object attains the acceleration it can have this kinetic energy until there is a change in the speed of the object with which it moves.
The forms of energy changes and it can take any form like thermal, electrical, electromagnetic,etc. Potential and kinetic energy are the two things under which these forms are energy are grouped. There can be a transferring of Kinetic energy from one object to another. The kinetic energy can also take any form of energy.
10 centimeter represents 1 meter. If a class room is 5 centimeters by 7 centimeters. What are the actual dimensions of the classroom?
Answer:
The actual dimensions of the classroom are 50 cm x 70 cm
Explanation:
Scaling
When we need to represent real-world dimensions into small spaces, we use scaling. Distance scaling tells us what is the equivalence between the real units and the scaled units. In this case, we are told that 10 cm is equivalent to 1 meter. As 1 meter is 100 cm, it means that the scale is 100/10=10. Thus, each centimeter in the paper is equivalent to 10 cm in the real distance.
The classroom is 5 cm x 7 centimeters. Scaling back to the real values, the classroom has measures of 50 cm x 70 cm.
2. A 3 kg ball is thrown downward at 4 m/s from a height of 1.5 m. a. What is the kinetic energy of the ball as it leaves the thrower’s hand? b. What force is doing work on the ball as it falls? c. How much work is done on the ball as it falls? d. What is the final kinetic energy of the ball? e. What is the velocity of the ball as it strikes the ground?
Answer:
a) 24 J
b) Gravitational Force
c) 45 J
d) 0
e) 6.782m/s
Explanation:
a) m = 3kg
v = 4m/s
h = 1.5m
KE = ?
0.5 * 3 * 16 = 24J
b) Gravitational force
c) F = ma = 3 * 10 = 30N
Work done = Force * distance = 30 * 1.5 = 45J
d) Final Kinetic Energy of the ball is zero because the ball eventually stops moving
e) velocity of ball as it strikes the ground = v
[tex]v^{2} - u^{2} = 2as[/tex] where
v is the velocity as it strikes the ground
u is the initial velocity
a is acceleration
s is the distance
Now since the ball is thrown downwards, a is positive because the velocity of the ball is increasing as the gravitational force acts on it
u = 4m/s
a = 10
s = 1.5
=> [tex]v = \sqrt{2as + u^{2} }[/tex]
= [tex]\sqrt{(2*10*1.5) + 16}[/tex]
= [tex]\sqrt{46} = 6.782m/s[/tex]
What is a combination of two or more simple machines?
Answer:
B) A Compound machine
Explanation: I just took this quiz
9. Calculate the distance (in km) that Charlie runs if he maintains the average
speed from question 8 for 1 hour.
Correct Question:
Calculate the distance (in km) charlie runs if he maintains an average speed of 8 km/hr for 1 hour
Answer:
The total distance covered by Charlie is 8 km in 1 hour.
Explanation:
The average velocity as given in the question is,
v = 8 km/hr
Total time taken,
[tex]$t=1 hour[/tex]
As we know the formula to evaluate the total distance d when the average velocity and time is given;
[tex]v=\frac{d}{t}[/tex]
[tex]d=v \times t[/tex]
[tex]d=8 \times 1[/tex]
[tex]d=8 k m[/tex]
Hence, the total distance covered by Charlie in 1 hour will be 8 km.
Charlie runs a distance of 60 kilometers in 1 hour if he maintains an average speed of 60 km/h; this is calculated using the formula Distance = Speed × Time.
Explanation:To calculate the distance that Charlie runs if he maintains his average speed for 1 hour, we follow a simple relation, which is: Distance = Speed × Time.
In question 8, an average speed calculation was mentioned, but since we don't have the exact number here, let's assume (based on the given information elsewhere) that the average speed we calculated previously was 60 km/h. To find the distance Charlie runs at this average speed over the course of 1 hour, we simply use the formula:
Distance = Average Speed × Time
Substituting the values we have:
Distance = 60 km/h × 1 hour = 60 km
Hence, Charlie runs a distance of 60 kilometers in 1 hour if he maintains the aforementioned average speed.
A container is formed by revolving the region bounded by the graph of y=x^2, and the x-axis, 0 < x < 2, about the y-axis. How much work is required to fill the container with a liquid from a source 2 units below the x-axis by pumping through a hole in the bottom of the container? (Assume pg=1)
The given question requires calculating the work done to pump liquid into a specifically shaped container by integrating the weight of the liquid over the height it is lifted, taking gravity into account.
Explanation:The question concerns calculating the work required to pump liquid into a container. This container has been formed by revolving the region bounded by y=x^2 and the x-axis around the y-axis, between x=0 and x=2. When calculating the work done in a physics context, the scenario typically involves considerations of force, distance, and energy. For this particular case, you would use the concept of work done against gravity to fill the container with a liquid from a source below its base, which involves integrating the weight of the liquid elements being moved over the height they are lifted. The fact that pg=1 implies we're assuming a uniform density of the liquid and a gravitational field strength of 1, probably to simplify calculations. However, without additional information such as vector force fields or specific motion paths like in other provided excerpts, we can't calculate an exact numerical answer here.
What controls the traits an organism has?
The Traits an organism displays are ultimately determined by the genes it inherited from its parents, in other words by its genotype. Variant copies of a gene are called alleles, and an individuals genotype is the sum of all the alleles inherited from the parents.
On the coast of Georgia, warm air usually picks up water vapor over the Atlantic Ocean. What will happen when the warm, moist air moves inland over hot, dry land?
A
The warm air will become cool, causing water vapor to evaporate and be absorbed by the land.
B
The warm air will rise, causing water vapor to evaporate from the air and generate winds.
C
The warm air will rise, causing water vapor to condense and form clouds.
D
The warm air will sink, causing water vapor to condense and form snow.
Answer:
C. The warm air will rise, causing water vapor to condense and form clouds.
Explanation:
Consider Compton Scattering with visible light.A photon with wavelength 500nm scatters backward(theta=180degree) from a free electron initially at rest.What is the fractional shift in wavelength delta\lamda/\lamda for the photon?
Answer: [tex]4.86(10)^{-12}m[/tex]
Explanation:
The Compton Shift [tex]\Delta \lambda[/tex] in wavelength when photons are scattered is given by the following equation:
[tex]\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos\theta)[/tex] (1)
Where:
[tex]\lambda'=500 nm=500(10)^{-9} m[/tex] is the wavelength of the scattered photon
[tex]\lambda_{o}[/tex] is the wavelength of the incident photon
[tex]\lambda_{c}=2.43(10)^{-12} m[/tex] is a constant whose value is given by [tex]\frac{h}{m_{e}.c}[/tex], being [tex]h=4.136(10)^{-15}eV.s[/tex] the Planck constant, [tex]m_{e}[/tex] the mass of the electron and [tex]c=3(10)^{8}m/s[/tex] the speed of light in vacuum.
[tex]\theta=180\°[/tex] the angle between incident phhoton and the scatered photon.
[tex]\Delta \lambda=2.43(10)^{-12} m (1-cos(180\°))[/tex] (2)
[tex]\Delta \lambda=4.86(10)^{-12}m[/tex] (3) This is the shift in wavelength
A desktop computer and monitor together draw about 0.6 A of current. They
plug into a wall outlet that is 120 V. What is the resistance, in ohms, of the
computer-monitor combination?
Answer:
200 ohms
Explanation:
Answer:
200
Explanation:
An object of known mass M with speed v0 travels toward a wall. The object collides with it and bounces away from the wall in the opposite direction in which the object was initially traveling. The wall exerts an average force F0 on the object during the collision. A student must use the equation Δp⃗ =F⃗ Δt to determine the change in momentum of the object from immediately before the collision to immediately after the collision. Which side of the equation could the student use to determine the change of the object's momentum?
Δp⃗ , because the mass of the object and the initial speed of the object are known.
A
F⃗ Δt, because the average force exerted on the object during the collision is known.
B
Either side of the equation may be used because the mass of the object, the initial speed of the object, and the average force exerted on the object during the collision are known.
C
Neither side of the equation may be used because there are too many unknown quantities before, during, and after the collision.
Neither side of the equation may be used because there are too many unknown quantities before, during, and after the collision
Explanation:
The impulse theorem states that the change in momentum of an object is equal to the impulse, which is the product between the average force applied and the duration of the collision:
[tex]\Delta p = F \Delta t[/tex]
where
[tex]\Delta p[/tex] is the change in momentum
F is the average force
[tex]\Delta t[/tex] is the duration of the collision
In this problem, neither side of the equation can be used to measure the change in momentum. In fact:
- The change in momentum (left side) is given by
[tex]\Delta p = m(v-u)[/tex]
where
m is the mass of the object
u is the initial velocity
v is the final velocity
Here the final velocity is not known, so it's not possible to use this side of the equation
- The impulse (right side) is given by
[tex]F\Delta t[/tex]
here the average force is known, however the duration of the collision is not known, so it's not possible to use this side of the equation.
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Neither side of the equation may be used because there are too many unknown quantities before, during, and after the collision. Therefore, option (C) is correct.
According to the impulse-momentum theorem, "The change in momentum of an object is equal to the impulse produced by the object. Where the impulse is expressed as the product of average force on the object and the duration of collision (reaction time)".
The expression is given as,
..............................................(1)
Here, [tex]\delta p[/tex] is the change in momentum, [tex]F_{av.}[/tex] is the average force and t is the reaction time.
In equation (1), [tex]\delta p[/tex] is the change in momentum which is given as,
[tex]\delta p = m(v-u)[/tex]
Here, m is the mass, v and u are the final and initial velocities of object respectively.
Since, object's mass (m) and velocities are not given. Therefore, left hand side of equation (1) cannot be used to determine the change of object's momentum.Also, right hand side of equation (1) requires the duration of collision (t), which is missing in the problem.Thus, we can conclude that there are various unknown variables present in the problem, for which neither side of the equation may be used to determine change in momentum of object. Hence, option (C) is correct.
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The u.s army’s parachuting team, the Golden Knights, are on a routine
Jumping mission over a deserted beach. On a jump, a 65-kg Knight lands on the beach with a speed of 4.0 m/s, making a 0.20-m deep indentation in the sand with what average force did the parachuter hit the sand?
The average force is -2600 N
Explanation:
First of all, we need to calculate the acceleration of the man during the collision, which is given by the suvat equation:
[tex]v^2-u^2=2as[/tex]
where:
v = 0 is his final velocity (he comes to a stop)
u = 4.0 m/s is the initial velocity
a is the acceleration
s = 0.20 m is the distance covered
Solving for a,
[tex]a=\frac{v^2-u^2}{2s}=\frac{0-4.0^2}{2(0.20)}=-40 m/s^2[/tex]
The negative sign indicates that it is a deceleration.
Now we can find the average force on the man by using Newton's second law of motion:
[tex]F=ma[/tex]
where
m = 65 kg is the mass
[tex]a=-40 m/s^2[/tex]
And substituting,
[tex]F=(65)(-40)=-2600 N[/tex]
where the negative sign indicates the force is in the direction opposite to the motion.
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To calculate the average force with which the parachuter hits the sand, we can use the equation F = ma, where F is the force, m is the mass, and a is the acceleration. The average force is approximately 67 N (in magnitude), with a negative sign indicating the direction of the force exerted on the parachuter.
Explanation:To calculate the average force with which the parachuter hits the sand, we can use the equation F = ma, where F is the force, m is the mass, and a is the acceleration. Since the parachuter lands with a certain speed, we can assume that the acceleration is equal to the change in velocity divided by the time taken to come to a stop. The change in velocity can be calculated by subtracting the final velocity from the initial velocity, and the time taken to come to a stop can be found using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
Given that the mass of the parachuter is 65 kg, the initial velocity is 4.0 m/s, and the final velocity is 0 m/s, we can calculate the acceleration using a = (v - u)/t. Assuming the time taken is the same as the time taken to come to a stop, we can rearrange the equation to solve for t: t = (v - u)/a. Substituting the given values into the equation, we can calculate the time taken. Finally, we can substitute the mass and acceleration into the equation F = ma to calculate the average force.
So, the average force with which the parachuter hits the sand can be calculated as F = (65 kg) * (-(0 m/s) - (4.0 m/s))/(3.84 s), which gives a result of -67 N (or approximately 67 N in magnitude).
Human diseases such as tetanus syphilis and cholera are caused by small cells that lack nucleus the cells can enter the human body and cause disease. Which common name is associated with these cells.
A antibody
B toxin
C bacteria
D virus
11) A 0.2 kg apple on an apple tree has a potential energy of 10 J. It falls to
the ground, converting all of its PE to kinetic energy. What is the velocity
of the apple just before it hits the ground?
Now choose from one of the following options Why?
Om/s
2 m/s
10 m/s
50 m/s
The velocity of the apple just before it hits the ground is 10 m/s.
The given parameters;
mass of the apple, m = 0.2 kgpotential energy of the apple, P.E = 10 JApply the principle of conservation of mechanical energy to determine the velocity of the apple before it hits the ground.
K.E = P.E
¹/₂mv² = 10
mv² = 2(10)
mv² = 20
0.2v² = 20
[tex]v^2 = \frac{20}{0.2} \\\\v^2 = 100\\\\v = \sqrt{100} \\\\v = 10 \ m/s[/tex]
Thus, the velocity of the apple just before it hits the ground is 10 m/s.
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What is the best approximate value for the elastic potential energy (EPE) of the spring elongated by 7.0 meters?
1.5 J
11 J
37 J
74 J
Answer:
C. 37 J
Explanation:
This problem was given with a graph showing that when the spring was elongated to 7 m, the force was approximately 11 N. Seeing as the EPE is calculated by finding the area under the graph and that the area of a triangle is found by using A = 1/2bh, you can see that the givens are b = 7 m & h = 11 N. This means that the the EPE = 1/2 * 7 m * 11 N, which is 38.5N*m. Joules are also represented by N*m so that simplifies to J and seeing as they asked for an approximation, C is the best choice! :)
A car is moving with a velocity of 54km/h. The kinetic energy of a boy of mass 40kg sitting in the car is
Answer: KE = 4500 J
Explanation: Solution:
First convert 54 km/h to m/s
54 km/h x 1000m/ 1km x 1h /3600s
= 15 m/s
Use the equation for Kinetic Energy and substitute the values:
KE = 1/2 mv²
= 1/2 40 kg x ( 15 m/s)²
= 1/2 9000
= 4500 J
Compare how magnetic forces act through non-magnetic materials and
magnetic materials:
Compare how magnetic forces act through non-magnetic materials and magnetic materials:
Explanation:
Magnet
• Magnet :- is an object which attracts pieces of iron, steel etc towards itself.
Some facts about magnets:-
• When magnet is freely suspended it always align towards north-south direction
• Like poles always repel & opp. poles attract each other.
• Magnet always exist as dipole
• Two poles can never be separated : if we try to cut it then still both the poles will exist even ina small piece of magnet .it automatically develops the lost polarity
Magnet always develop certain area around it where its effect can be felt ie. magnetic field.
MAGNETIC Field
is studied by drawing imaginary lines called magnetic lines of forces.
Characteristics.
• They always originate from North pole & terminate at South pole. This shows that if north pole was free is move it would have mvre towards south pole.
•Place where they are closer indicate strong M. field i.e. at poles.
•Mag. Field lines gives the direction of magnetic force.
•Two magnetic lines will never intersect each other as they give direction of force & force can’t have 2 direction at a time.
M Field lines are closed continuous curves.This is what that happens in magnetic materials .
Non magnetic materials
Magnetic forces 'act through' non-magnetic materialsThese magnetic materials can be used as a shield around a magnet. The domain theory of magnetism tries to explain why metals get magnetised. The magnetic elements have little molecular magnets inside them.Magnets attract only the magnetic materialsTypes of magnetic materials
Soft magnetic materials (e.g. iron) have domains that easily move into line when the metal is placed in a magnetic field but as soon as the field is removed the domains take on a random pattern again. It returns to being unmagnetized straight away.
Hard magnetic materials (e.g. steel) have domains that do not easily move into line when the metal is placed in a magnetic field, a strong field is needed for some time, but then, when the field is removed the domains retain the magnetic pattern. The metal stays magnetic for a long time.
Suppose Earth's mass increased but Earth's diame-
ter did not change. How would the gravitational strength near
Earth's surface change?
Answer: It would increase.
Explanation:
The equation for determining the force of the gravitational pull between any two objects is:
[tex]F = G \frac{m1m2}{r^2}[/tex]
Where G is the universal gravitational constant, m1 is the mass of one body, m2 is the mass of the other body, and r^2 is the distance between the two objects' centers squared.
Assuming the Earth's mass but not its diameter increased, in the equation above m1 (the term usually indicative of the object of larger mass) would increase, while the r^2 would not.
Thus, it goes without saying that, with some simple reasoning about fractions, an increasing numerator over a constant denominator would result in a larger number to multiply by G, thus also meaning a larger gravitational strength between Earth and whatever other object is of interest.
Final answer:
If Earth's mass increased with its diameter unchanged, gravitational strength at its surface would also increase proportionally. The weight of objects on Earth would thus increase in direct relation to the mass increase.
Explanation:
If Earth's mass increased but its diameter remained unchanged, the gravitational strength near Earth's surface would increase proportionally to the mass increase. This is because gravitational force is directly proportional to the mass of the objects. For example, if Earth had 10 times its present mass but the same volume, a person's weight would increase by a factor of 10.
Conversely, with one-third of Earth's present mass, the gravitational force would reduce by a factor of 1/3, and a person would weigh only one-third as much as they currently do. Since the question implies that the gravitational force is what changes with the mass of Earth, we can infer that a greater mass leads to a stronger gravitational pull and thus an increase in weight for objects at the surface, assuming the radius stays constant.