Work is the amount of energy transferred
Explanation:
In physics, work is a measure of the energy transfer occurring in a process. Typically, we talk about work when energy is converted from one form into another.
For instance, work is done when a force is applied on an object. The work done on the object is given by:
[tex]W=Fd cos \theta[/tex]
where
where
F is the magnitude of the force
d is the displacement
[tex]\theta[/tex] is the angle between the direction of the force and of the displacement
We notice the following:
No work is done when the force is perpendicular to the displacement ([tex]cos 90^{\circ}=0[/tex])The work is maximum when the force is parallel to the displacementWhenever work is done, there is also an energy transfer taking place. For instance, in the previous example, when the force is applied to the object, the object will accelerate (assume there is no friction), and will gain kinetic energy: therefore, there is a transfer of energy to the object.
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If we increase the distance traveled when doing work, and keep all other factors the same, what will happen?
A) The amount of force will go down
B) The amount of force will increase.
C) The amount of work will decrease
D) The amount of work will increase
Answer: D
Explanation:
A -amount of work
F-force
s-distance
If you keep same Force(F) and increase distance(s), amount of work will increase, according to:
A=F*s
Answer:
D) The amount of work will increase
Explanation:
Took test
Please help me people
Answer:
1 ) density = 1200[kg/m^3]
2)
a) wide = 0.35[m]
long = 1,1 [m]
thick = 0.015 [m]
b)
Volume = 0.005775[m^3]
mass = 15.6 [kg]
Explanation:
1)
This problem can be solved by knowing the definition of density which is expressed as the relation between mass and volume, so that the following equation can give us the density value.
[tex]density=\frac{m}{V}\\ where:\\m= mass [kg]\\V= volume [m^3]\\replacing\\density=\frac{90}{0.075} \\density=1200 [kg/m^3][/tex]
2)
The following data is equal to:
a)
wide = 35 [cm] = 0.35[m]
long = 11 [dm] = 11 [decimeters] = 1,1 [m]
thick = 15 [mm] = 0.015 [m]
b)
We know that density is the relationship between mass and volume of a body.
So the volume will be:
volume = wide * long * thick = (0.35*1.1*0.0015) = 0.005775[m^3]
[tex]the mass will be:\\m=density*volume\\m=2700[kg/m^3]*0.005775[m^3]\\m=15.6[kg][/tex]
Answer :
(1) The density of asphalt is, [tex]1200kg/m^3[/tex]
(2) (a) Length, width and thickness of sheet in meter is, 0.35 m, 1.1 m and 0.015 m respectively.
(b) The volume and mass of slab is, 0.005775 m³ and 15.59 kg respectively.
Explanation :
Part 1 :
As we are given:
Mass of block = 90 kg
Volume of block = [tex]0.075m^3[/tex]
Formula used :
[tex]\text{Density of block}=\frac{\text{Mass of block}}{\text{Volume of block}}[/tex]
Now put all the given values in this formula, we get:
[tex]\text{Density of block}=\frac{90kg}{0.075m^3}=1200kg/m^3[/tex]
Thus, the density of asphalt is, [tex]1200kg/m^3[/tex]
Part 2(a) :
As we are given that:
Length of aluminium sheet = 35 cm
Width of aluminium sheet = 11 dm
Thickness of aluminium sheet = 15 mm
Now we have to convert these dimensions into meters.
Conversions used:
1 cm = 0.01 m
1 dm = 0.1 m
1 mm = 0.001 m
Length of aluminium sheet = 35 cm = 35 × 0.01 = 0.35 m
Width of aluminium sheet = 11 dm = 11 × 0.1 = 1.1 m
Thickness of aluminium sheet = 15 mm = 15 × 0.001 = 0.015 m
Part 2(b) :
First we have to calculate the volume of aluminium sheet.
Volume of aluminum sheet (cuboid) = Length × Width × Thickness
Volume of aluminum sheet (cuboid) = 0.35 m × 1.1 m × 0.015 m
Volume of aluminum sheet (cuboid) = 0.005775 m³
Now we have to calculate the mass of aluminium sheet.
[tex]\text{Density of aluminium}=\frac{\text{Mass of aluminium}}{\text{Volume of aluminium}}[/tex]
[tex]2700kg/m^3=\frac{\text{Mass of aluminium}}{0.005775m^3}[/tex]
[tex]\text{Mass of aluminium}=15.59kg[/tex]
Thus, the volume and mass of slab is, 0.005775 m³ and 15.59 kg respectively.
Δ→ Δ→
5N 6N
what is the net force?
Answer:
The net force acting on the body, F = 11 N
Explanation:
Given data,
The force acting on the body, f = 5 N
Another force acting on the body, f' = 6 N
The direction of both forces is in the same direction.
The net force is given by the resultant vector,
[tex]F= \sqrt{f^{2} + f'^{2} + 2 ff' Cos \theta}[/tex]
[tex]F= \sqrt{f^{2} + f'^{2} + 2 ff'[/tex] (∵ θ = 0)
[tex]F=\sqrt{(f + f')^{2} }[/tex]
F = f + f'
= 5 N + 6 N
= 11 N
Hence, the net force acting on the body, F = 11 N
Match the vocabulary terms to their definitions.
1. distance a measurement that has both magnitude and direction
2. velocity the net change in position of an object
3. displacement the rate and direction at which an object moves
4. vector the length along a path between two points
5. speed measure of the rate at which an object changes position
6. momentum the mass of an object multiplied by its veloci
Answer:
Distance- The length along a path between two points .
Velocity- The rate and direction at which an object moves .
Displacement- The net change in position of an object .
Vector- A measurement that has both magnitude and direction .
Speed- Measure of the rate at which an object changes position .
Momentum- The mass of an object multiplied by its velocity.
Explanation:
Vector: A physical quantity having both magnitude and direction is called Vector quantity. Example: Displacement, Velocity, Acceleration, Force etc.
On the other hand , quantity having only magnitude is called Scalar. Example: Distance, Speed, Time, Current etc.
Distance: The totle length of path traversed by a body in certain time is called distance. It is scalar quantity as it has only magnitude. SI unit of distance is meter.
Displacement: It is defined as change in position of a body in given period of time. Actually it is the distance between initial and final position along with given direction. So it is vector quantity. SI unit is meter.
Speed: The rate at which distance is traveled with respect to time is called speed.It is scalar quantity. SI unit is meter/second.
[tex]Speed=\frac{Distance}{Time}[/tex]
Velocity: It can be defined as rate of change in position or displacement with respect to time.It is vector quantity. SI unit is meter/second.
[tex]Velocity=\frac{Displacement}{Time}[/tex]
Momentum: The amount of motion contain in a body is the momentum. It is defined as product of mass and velocity. It is vector quantity. SI unit is kg.meter/second
Momentum=Mass×Velocity
Answer: The answer above is correct! I got 100% :))
Explanation:
Object A attracts object B with a gravitational force of 10 newtons from a given distance. If the distance between the two objects is doubled,
what is the new force of attraction between them?
A. 2.5 newtons
B. 5newtons
C. 20 newtons
D. 100 newtons
Answer:
20N option C
Explanation:
10× 2
What is the shortest possible time for a sprinter moving at 4 m/s to finish a 100 meter race?
The time needed is 25 s
Explanation:
Assuming that the sprinter is moving by uniform motion, then we can use the following equation that relates distance, time and speed:
[tex]v=\frac{d}{t}[/tex]
where
v is the speed of the sprinter
d is the distance covered
t is the time elapsed
In this problem, we know that
v = 4 m/s is the speed of the sprinter
d = 100 m is the distance that must be covered
Solving for t, we find the minimum time needed:
[tex]t=\frac{d}{v}=\frac{100}{4}=25 s[/tex]
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1. Complete the table with the appropriate equivalences
Scientific notation
Order of magnitude
Look the picture there’s a example
Answer:
The different values are in the located in the word document
Explanation:
In the attached document in word, we can find each of the conversion for each one of the quantities.
A parallel circuit contains four identical lamps. The current through the energy source is 4 A. The total resistance of the circuit is 10 Ω. What is the voltage across the energy source and across each lamp?
a)2.5 V; 40 V
b)40 V; 40 V
c)2.5 V; 2.5 V
d)40 V; 2.5 V
Answer:
b. 40V , 40V
Explanation:
Connections are as per the figure.
As total current through source is 4A , current through each lamp is 1A.
As total resistance of the circuit is 10Ω ,resistance of each bulb is 40Ω because in case of a parallel circuit in which identical objects are connected , [tex]R_{eff} = \frac{R}{n}[/tex] where R is the resistance of each bulb and n is the number of bulbs.
As per Ohm's law , voltage of the source =IR = 4×10 =40V.
We can see from the figure that if the voltage across the source is 40V , the voltage across each bulb is also 40V.
in which type of wave are vibrations at right angles to the direction in which the wave is travelling
Answer:
longitudinal waves have those properties
An object islaunched at velocity of 20 m/s in a direction making an angle of 25 degree upward with the horizontal. what is the maximum height reached?
Answer:
[tex]\displaystyle y_m=3.65m[/tex]
Explanation:
Motion in The Plane
When an object is launched in free air with some angle respect to the horizontal, it describes a known parabolic path, comes to a maximum height and finally drops back to the ground level at a certain distance from the launching place.
The movement is split into two components: the horizontal component with constant speed and the vertical component with variable speed, modified by the acceleration of gravity. If we are given the values of [tex]v_o[/tex] and [tex]\theta\\[/tex] as the initial speed and angle, then we have
[tex]\displaystyle v_x=v_o\ cos\theta[/tex]
[tex]\displaystyle v_y=v_o\ sin\theta-gt[/tex]
[tex]\displaystyle x=v_o\ cos\theta\ t[/tex]
[tex]\displaystyle y=v_o\ sin\theta\ t -\frac{gt^2}{2}[/tex]
If we want to know the maximum height reached by the object, we find the value of t when [tex]v_y[/tex] becomes zero, because the object stops going up and starts going down
[tex]\displaystyle v_y=o\Rightarrow v_o\ sin\theta =gt[/tex]
Solving for t
[tex]\displaystyle t=\frac{v_o\ sin\theta }{g}[/tex]
Then we replace that value into y, to find the maximum height
[tex]\displaystyle y_m=v_o\ sin\theta \ \frac{v_o\ sin\theta }{g}-\frac{g}{2}\left (\frac{v_o\ sin\theta }{g}\right )^2[/tex]
Operating and simplifying
[tex]\displaystyle y_m=\frac{v_o^2\ sin^2\theta }{2g}[/tex]
We have
[tex]\displaystyle v_o=20\ m/s,\ \theta=25^o[/tex]
The maximum height is
[tex]\displaystyle y_m=\frac{(20)^2(sin25^o)^2}{2(9.8)}=\frac{71.44}{19.6}[/tex]
[tex]\displaystyle y_m=3.65m[/tex]
What is the acceleration of this object ?
Answer:
-2.5 m/s^2
Explanation:
From a velocity-time graph, the acceleration can be determined by finding the gradient of the straight line graph.
Acceleration=(Change in velocity)/(time)
Where change in velocity = final velocity-initial velocity.
Hence, from the graph,
we can see (10-40)/12 = -2.5
Answer:
-2.5
Explanation:
just did this one
what is the work done by a force equivalent to the weight of 100kg that moves an object in 7 minutes?
7000 Joules
Explanation:We are given;
The mass of an object = 100 kgBut, Force = mass × g (taking g as 10 N/kg)
Then, Force = 1000 NAssuming we are given the distance moved by the object as 7 m
We can calculate the work done by the force
We need to know that;
Work done = Force × distance
Therefore;
Work done = 1000 N × 7 m
= 7000 Joules
Therefore, the work done by the force is 7000 Joules
The neck and bottom of a bottle are 0.5cm and 0.4cm respectively. If the cork in the neck is pressed with a force of 1kgwt and the bottle is full of oil. What is the force in te bottom due to the load
Answer:
0.64 kgwt
Explanation:
You should check that the diameters of the neck and bottom are correct. That is a very small bottle, and the neck shouldn't be bigger than the bottom.
The pressure on the cork = the pressure on the bottom
P₁ = P₂
F₁ / A₁ = F₂ / A₂
F₁ / (π r₁²) = F₂ / (π r₂²)
F₁ / r₁² = F₂ / r₂²
(1 kgwt) / (0.5 cm)² = F / (0.4 cm)²
F = 0.64 kgwt
Round as needed.
You apply a horizontal force of 200N against a 50kg couch. The force of friction between the couch and the floor is 180N. If the couch is pushed 2 metres, the kinetic energy gained by the couch is: a) 0 b) 40 J c) 100 J d) 360 J e) 400 J
Answer:
The kinetic energy gained is 40[J]
Explanation:
This is a problem that applies the work and energy conservation. We must analyze the initial and final positions as well as the forces that act on the couch during displacement.
The working and energy conservation theorem says:
[tex]E_{k1} + W_{1-2}= E_{k2}[/tex]
The above equation tells us that kinetic energy plus the work done to move an object from one position to another will be equal to kinetic energy in the last position.
The kinetic energy in the initial position is zero since the couch is at rest, Ek1=0
The work will be given by the force that drives the couch multiplied by the distance, and the friction force that acts in the opposite direction to the displacement.
[tex]0+(200-180)*2= Ek2\\Ek2=20*2= 40 [J][/tex]
7. Mildred was driving 12 m/s and slowed
to 8 m/s in three seconds. Mildred's
acceleration can be calculated as:
0 0
ID 111111
0 0
1223
33
Answer:
The acceleration of Mildred is, 1.33 m/s²
Explanation:
Given data,
The initial velocity of Mildred, u = 12 m/s
The final velocity of Mildred, v = 8 m/s
The time period of travel, t = 3 s
The acceleration of Mildred,
a = (v-u)/t
a = (8 - 12) /3
= -4/3
= -1.33 m/s²
The negative sign indicates that the Mildred is decelerating.
Hence, the acceleration of Mildred is, 1.33 m/s²
Two objects gravitationally attract with a force of 10N. If the distance between the two objects center is doubled, then the new force is
Answer:
20N
Explanation:
10×2
Answer:
Answer:
20N
Explanation:
10×2
You’ve made the hypothesis that the steeper the slope the faster a ball Will be rolling when it reaches the bottom
What should be your independent variable in an experiment testing this
hypothesis
A. How heavy the Ball is
B. How steep the slope is
C. How high the ball starts on the slope
D. How fast the ball is rolling at the bottom
What is the total displacement of the object ?
Answer:
562.5 m
Explanation:
Displacement is area under the graph of velocity.
d = 10*25 + 25^2/2 = 562.5 m
The displacement of the object is 550 meters.
How to solveThe velocity graph shows that the object is moving in a positive direction for the first 25 seconds.
The velocity is constant at 10 m/s for the first 10 seconds, and then it increases to 20 m/s for the next 15 seconds.
The area under the velocity graph is equal to the displacement of the object. The area of a triangle is equal to (1/2) * base * height. In this case, the base of the triangle is 25 seconds and the height is 20 m/s.
Therefore, the area of the triangle is (1/2) * 25 * 20 = 250 m^2.
The object is also moving in a positive direction for the next 10 seconds, but the velocity is constant at 20 m/s. Therefore, the area under the velocity graph for these 10 seconds is a rectangle with a base of 10 seconds and a height of 20 m/s.
The area of the rectangle is 10 * 20 = 200 m^2.
The total area under the velocity graph is 250 m^2 + 200 m^2 = 550 m^2. Therefore, the displacement of the object is 550 meters.
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A 275-kilogram object starts from rest and accelerates on level ground up to a speed of
41 meters per second over 49.3 meters. It continues at that speed for another 15
meters. At the end of the 15 meters, the object is launched forward at a 37° angle.
a. How much time does it take for the object to reach its final velocity?
b. How much force did the object apply in order to reach that final speed?
c. What are the maximum range and height of the object when it is launched?
d. At what angle would the object have to be launched to reach a height of 50
meters above the ground at the same initial velocity?
e. Describe a realistic scenario in which this might happen.
Answer:
Explanation:
a )
mass m = 275 kg , final velocity v = 41 m/s
distance, s = 49.3 m
acceleration a = ?
v² = u² + 2as
41² = 2 a x 49.3
a = 17.05 m/s²
time t = v - u / a
t = 41 / 17.05
= 2.4 s
b )
force = mass x acceleration
= 275 x 17.05
= 4688.75 N
c ) maximum range = v² / g
= 41²/9.8
= 171.5 m
height ( maximum )
v² sin²45 / 2g ( at angle of projectile of 45 , range and height is maximum.)
= 41²/ 2 x 9.8
= 85.76 m
d )
Let the angle be θ
height h = v² sin²θ / 2g
50 = 41² sin²θ /2g
sin²θ = .58
sinθ = .76
50°.
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When light is reflected, the "incoming" ray is also called the _____ ray.
A. incident
B. reflected
C. bouncing
D. electromagnetic
Answer: A (Incident ray).
Explanation:
What is the name and symbol of the element in the second row and fourteenth column of the periodic table? Hint: Review your periodic table. cobalt (Co) silicon (Si) carbon (C) aluminum (Al)
I need help with my physics work quickly please!!! I only need help with the circled questions!
Answer:
E11. 0.6 N
E12. 1.61×10⁻⁴ N
E13. 1.12 N
E14. 45 lb
Explanation:
The force due to gravity between two objects is:
F = Gm₁m₂ / r²
where G is the gravitational constant,
m₁ and m₂ are the masses of the objects,
and r is the distance between them.
E11. At a distance r, the force is 9.6 N.
9.6 = Gm₁m₂ / r²
At quadruple the distance (4r), the force is:
F = Gm₁m₂ / (4r)²
F = Gm₁m₂ / (16r²)
F = 9.6 / 16
F = 0.6
E12. Given m₁ = m₂ = 700 kg, and r = 0.45 m:
F = Gm₁m₂ / r²
F = (6.67×10⁻¹¹ m³/kg/s²) (700 kg) (700 kg) / (0.45 m)²
F = 1.61×10⁻⁴ N
E13. At a distance r, the force is 0.28 N.
0.28 = Gm₁m₂ / r²
At half the distance (r/2), the force is:
F = Gm₁m₂ / (r/2)²
F = 4 Gm₁m₂ / r²
F = 4 (0.28)
F = 1.12
E14. The acceleration due to gravity on the Moon is 1/6 that of Earth's. So an astronaut weighs 1/6 as much on the Moon.
W = 270 lb / 6
W = 45 lb
A punter kicks a football from the ground with a velocity of 19 m/s at an angle 32 degrees from the horizontal the football travels a horizontal distance of 33m before it hits the ground what is the maximum height the football reaches
Answer: 5.17 m
Explanation:
The football reaches its maximum height when the vertical component of the velocity is zero:
[tex]V_{y}=0 m/s[/tex]
This is exactly at the point where the football stops and then begins to fall thanks to the acceleration due gravity.
So, we can use the following equation:
[tex](V_{y})^{2}=(V_{oy})^{2}-2gy_{max}[/tex]
Where:
[tex]V_{y}=0[/tex] is the final velocity
[tex]V_{oy}=V_{o} sin(32\°)[/tex] is the vertical component of the initial velocity
[tex]V_{o}=19 m/s[/tex] is the initial velocity
[tex]g=9.8 m/s^{2}[/tex] is the acceleration due gravity
[tex]y_{max}[/tex] is the football's maximum height
Isolating [tex]y_{max}[/tex]:
[tex]y_{max}=\frac{(V_{o} sin(32\°))^{2}}{2g}[/tex]
[tex]y_{max}=\frac{(19 m/s sin(32\°))^{2}}{2(9.8 m/s^{2})}[/tex]
Finally:
[tex]y_{max}=5.17 m[/tex] This is the football's maximum height
Add these measurements, using significant digit rules:
1.0090 cm + 0.02 cm = cm
Answer: " 1.03 " .
____________________________________
Explanation:
_____________________________________
When we add "measured" or "obtained values" ; the rules for "signficant digits" (also known as "significant figures" ; or "sig figs") are different from the rules we use when we multiply "measured or obtained values".
We are given 2 (two) "measured or obtained values" ; and we are asked to add these values together using the "significant digit rules.
In this case, we add up the "sum total value" of all the values—Not taking into account the number of significant figures. Then, we take that "sum value" ; and we round that value to the number of "decimal places" that corresponds to the the number of decimals places in the "measured/obtained valued" that occurs in the value using to calculate the sum —the one that also happens to have the least number of significant digits among all the values used to calculate the "sum total".
As such:
_____________________________________
1.0090 cm
+ 0.02 cm
________________
1.0290 cm.
→ Now, there were two (2) values used to calculate the "sum total" ;
1) 1.0090 cm ; which has 5 (five) significant digits.
2) 0.02 cm ; which as 1 (one) significant digits.
_____________________________________
So, " 1" is less than "5" ; so: " 0.02" is the value within the calculations used to find the sum total — that has the "least" number of significant digits.
→ " 0.02 " ; happens to have 2 (two) decimal spaces.
_____________________________________
So, we take the "sum total" :
→ " 1.0290 cm " ;
→ And round this number to: " 2 (two) decimal places:
→ 1.029... ;
→ If the number after the second decimal place is from "1 to 4" ; we round down. If the number after the second decimal place is
from "5 and 9" ; to round up. Since the third decimal place is: "9" ;
[Note: " 1.029.." ; is between "5 to 9" } ; we round up:
→ The answer is: " 1.03 " .
_____________________________________
Hope this is helpful to you!
Wishing you the best!
_____________________________________
A green dot in the middle of 4 concentric circles. The first away from the green dot is labeled n = 1. The second circle away from the green dot is labeled n = 2. The third circle away from the green dot is labeled n = 3 and the last is labeled n = 4. There is an arrow from n = 4 to n = 1 labeled A. There is an arrow from n = 3 to n = 2 labeled B. There is an arrow from n = 3 to n = 1 labeled C. There is an arrow from n = 2 to n = 1 labeled D.
Which transition by an electron will release the greatest amount of energy?
A
B
C
D
Answer:a??
Explanation:
Transition (A) from n=4 to n=1 by an electron will release the greatest amount of energy.
What is energy levels of energy?The fixed distances from an atom's nucleus where electrons may be found are referred to as energy levels (also known as electron shells). In an atom, the small, positively charged nucleus is surrounded by smaller, negatively charged electrons.
Energy levels resemble the rungs of a ladder in several ways. You cannot stand between the steps, only on one or the other. In the same way, electrons. They are able to occupy energy levels, but not the area between them.
In the given question there have four transitions of electron:
from n = 4 to n = 1 labeled A. from n = 3 to n = 2 labeled B. from n = 3 to n = 1 labeled Cfrom n = 2 to n = 1 labeled D.As for transition A; Δn = 4 -1 = 3 is the highest among the four transitions of electron, transition from n=4 to n=1 by an electron will release the greatest amount of energy.
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If for every 2 atoms of Chlorine-35 there are 5 atoms of Chlorine-38. What is the average mass?
Answer:
37.1
Explanation:
mass of 2 atoms of chlorine-35 = 2× 35 = 70
mass of 5 atoms of chlorine-38 = 5 × 38 = 190
average mass = total mass / total number of atoms
= (70 +190) / (2 +5)
= 260 / 7
= 37.143
= 37.1
Final answer:
The average mass of chlorine-35 and chlorine-38 can be calculated using their respective atomic masses and percent abundances. The average mass of chlorine is 35.46 amu.
Explanation:
The average mass of chlorine can be calculated using the information provided. Chlorine-35 has an atomic mass of 34.969 amu and a percent abundance of 75.77%. Chlorine-38 has an atomic mass of 36.966 amu and a percent abundance of 24.23%. To calculate the average mass, we multiply the fraction of chlorine-35 (0.7577) by its mass, and add it to the fraction of chlorine-38 (0.2423) multiplied by its mass.
Therefore, the average mass of chlorine is: (0.7577 * 34.969) + (0.2423 * 36.966) = 35.46 amu.
Pls help ASAP this questions are science questions ( biology chemistry physics)
46 I think a
47 I think d
49 I think a
50 I think a
51 idk
54 I think d
55 idk
56 idk
57 idk
58 I think d
59 idk
61 idk
62 I think a
Answer:
It would be A.
Explanation:
The scale goes from 0 to 14, With 0 being acidic and 14 being basic.
So if 7 is neutral, then anything less than 7 is moving more towards being more acidic. Anything higher than 7 is moving towards being more basic.
Answer:
100, 50, 40, 30, 20, 10 butter
Explanation:
What is the horizontal acceleration of a ball taht is launched horizontally with a velocity of 5.6 m/s?
Answer:
Horizontal acceleration=0
Explanation:
Horizontal Launching
When an object is launched horizontally at a certain speed [tex]v_o[/tex], only one force will be acting throughout the whole duration of the motion: the force of gravity. If no effect of the wind is to be considered, then the horizontal speed will not change over time. It means the horizontal acceleration is zero. Please note the vertical acceleration is not zero, it is in fact, the acceleration of gravity (0.8 m/sec^2)
A projectile is launched into the air with an initial speed of 40 m/s and a launch angle of 20° above the horizontal. The projectile lands on the ground five seconds later. Neglecting air resistance, calculate the projectile’s range and draw a projectile path.
Explanation:
V=40m/s
Vy=V.sina=40.sin20=40 . 0.342=13.68m/s
Vx=V.cosa=40.cos20=40 . 0.766=30.64m/s
Projectile travels during 5 seconds and the ramge becomes:
x=V.t=30.64 . 5=153.2m
Suppose you have a pendulum clock which keeps correct time on Earth(acceleration due to gravity = 1.6 m/s2). For ever hour interval (on Earth) the Moon clock:
A) (9.8/1.6)h
B) 1 h
C) the square root of 9.8/1.6 h
D) (1.6/9.8)h
E) the square root of 1.6/9.8 h
A pendulum clock runs at a rate proportional to the inverse square root of the acceleration due to gravity. Since gravity is weaker on the Moon, a clock calibrated for Earth's gravity will run slower on the Moon, taking the square root of 9.8/1.6 hours to complete one revolution, which is answer option C the square root of 9.8/1.6 hours.
The rate at which a pendulum clock runs is inversely proportional to the square root of the acceleration due to gravity. On Earth, the acceleration due to gravity is approximately 9.8 m/s², while on the Moon, it is about 1.63 m/s². According to the formula for the period of a pendulum (T = 2π √(L/g) where L is the length of the pendulum and g is the acceleration due to gravity), if a clock keeps correct time on Earth, on the Moon it would run at a rate that is the square root of the ratio of the accelerations (the square root of 9.8/1.63). This would mean, for every hour on Earth, the clock on the Moon would show a time of C) the square root of 9.8/1.6 hours.