An electron traveling horizontally to the right enters a region where a uniform electric field is directed downward. What is the direction of the electric force exerted on the electron once it has entered the electric field?

Answer:

Double the voltage across it.- last choice

The person exerts a force of 75 N on the mower, and it will travel approximately 0.529 meters before coming to a stop.

To find the force (F) exerted by the person on the mower, we can use Newton's second law, which states:

[tex]\[ \text{Net force} = \text{Mass} \times \text{Acceleration} \][/tex]

Given:

- Mass of the mower (m) = 24 kg

- Net external force (F_net) = 51 N

- Force of friction (F_friction) = 24 N

We know that net external force is the sum of all forces acting on the object, so:

[tex]\[ F_{\text{net}} = F - F_{\text{friction}} \][/tex]

[tex]\[ 51 \, \text{N} = F - 24 \, \text{N} \][/tex]

[tex]\[ F = 51 \, \text{N} + 24 \, \text{N} \][/tex]

[tex]\[ F = 75 \, \text{N} \][/tex]

So, the person exerts a force of [tex]\( 75 \, \text{N} \)[/tex] on the mower.

Now, to find how far the mower will go before stopping, we can use the equation of motion:

[tex]\[ v^2 = u^2 + 2as \][/tex]

Where:

- \( v \) = final velocity (0 m/s, since the mower stops)

- \( u \) = initial velocity = 1.5 m/s

- \( a \) = acceleration

- \( s \) = distance traveled

We know that the net force acting on the mower is responsible for the acceleration, and it's given by Newton's second law:

[tex]\[ F_{\text{net}} = m \times a \][/tex]

So:

[tex]\[ a = \frac{F_{\text{net}}}{m} \][/tex]

[tex]\[ a = \frac{51 \, \text{N}}{24 \, \text{kg}} \][/tex]

[tex]\[ a ≈ 2.125 \, \text{m/s}^2 \][/tex]

Now, using the equation of motion:

[tex]\[ 0^2 = (1.5 \, \text{m/s})^2 + 2 \times 2.125 \, \text{m/s}^2 \times s \][/tex]

[tex]\[ 0 = 2.25 + 4.25s \][/tex]

[tex]\[ 4.25s = -2.25 \][/tex]

[tex]\[ s = \frac{-2.25}{4.25} \][/tex]

[tex]\[ s ≈ -0.529 \, \text{m} \][/tex]

The negative sign indicates that the mower moves in the opposite direction to its initial motion before stopping. So, the mower will travel approximately [tex]\( 0.529 \, \text{m} \)[/tex] before coming to a stop.

The force exerted by the person on the mower is 75 N. The mower will travel approximately 1.125 meters before stopping.

The net external force is the difference between the force exerted by the person and the force of friction. Thus, we can write:

[tex]F_{net} = F - F_{friction}[/tex]

Substituting the given values:

[tex]51\,\text{N} = F - 24\,\text{N}[/tex]

Solving for F:

[tex]F = 51\,\text{N} + 24\,\text{N} = 75\,\text{N}[/tex]

Finding the Distance the Mower Travels before Stopping:

When the force F is removed, the mower will decelerate due to the frictional force. First, we find the acceleration (a) using Newton's second law:

[tex]F_{friction} = m \cdot a[/tex]

Solving for a gives:

[tex]a = \frac{F_{friction}}{m} = \frac{24\,\text{N}}{24\,\text{kg}} = 1\, \text{m/s}^2[/tex]

Since friction opposes the motion, this is a deceleration (a is negative):

[tex]a = -1\, \text{m/s}^2[/tex]

Next, we use the kinematic equation to find the distance (d):

[tex]v^2 = v_0^2 + 2 \cdot a \cdot d[/tex]

Substituting the known values:

[tex]0 = (1.5 \frac{m}{s})^2 + 2 \cdot (-1 \frac{m}{s^2}) \cdot d[/tex]

Solving for d:

[tex]0 = 2.25 - 2d[/tex]
[tex]2d = 2.25[/tex]
[tex]d = 1.125 \text{m}[/tex]

Answer:

8050 J

Explanation:

Given:

r = 4.6 m

I = 200 kg m²

F = 26.0 N

t = 15.0 s

First, find the angular acceleration.

∑τ = Iα

Fr = Iα

α = Fr / I

α = (26.0 N) (4.6 m) / (200 kg m²)

α = 0.598 rad/s²

Now you can find the final angular velocity, then use that to find the rotational energy:

ω = αt

ω = (0.598 rad/s²) (15.0 s)

ω = 8.97 rad/s

W = ½ I ω²

W = ½ (200 kg m²) (8.97 rad/s)²

W = 8050 J

Or you can find the angular displacement and find the work done that way:

θ = θ₀ + ω₀ t + ½ αt²

θ = ½ (0.598 rad/s²) (15.0 s)²

θ = 67.3 rad

W = τθ

W = Frθ

W = (26.0 N) (4.6 m) (67.3 rad)

W = 8050 J

Answer:

0.28 cm

Explanation:

The volume of a sphere is given by:

[tex]V=\frac{4}{3}\pi r^3[/tex]

where r is the radius, which is dependent on the time, so r(t).

The rate of change of the volume is

[tex]\frac{dV}{dt}=4 \pi r^2 \frac{dr}{dt}[/tex] (1)

where

[tex]\frac{dr}{dt}[/tex] is the rate of change of the radius. We know that

[tex]\frac{dr}{dt}=9[/tex] (cm/s)

And we want to find the value of the radius r when the rate of change of the volume is the same:

[tex]\frac{dV}{dt}=9[/tex] (cm^3/s)

So we can rewrite (1) as:

[tex]9=4\pi r^2 \cdot 9[/tex]

By solving it, we find

[tex]4\pi r^2 = 1\\r = \sqrt{\frac{1}{4\pi}}=0.28 cm[/tex]

Answer with Explanations:

Given:

mass of block, m = 0.244 kg

spring constant, k = 5075 N/m

compression distance in spring, x = 0.106 m

Find how high mass reaches after release (measured from point of release)

Solution:

Use conservation of energy, assuming a light spring (of negligible mass)

stored spring potential energy = gravitational potential energy of block

PE1 = PE2

PE1 = kx^2/2 = 5075*0.106^2/2 = 28.512 J

PE2 = mgh = 0.244 * 9.8 * h = 2.3912 h J

Solve for h = 28.512 / 2.3912 = 11.923 m

To find the maximum height a block rises after being released from a compressed spring, we use the conservation of energy principle to equate the initial spring potential energy to the gravitational potential energy at the peak height, then solve for that height.

The problem involves applying the conservation of energy principle to a mass-spring system when the block is released from compression. Initially, all the energy is stored as potential energy in the compressed spring (spring potential energy), and when the spring is uncompressed, that energy will be converted into kinetic energy of the mass. Finally, when the block leaves the spring and moves upward, its kinetic energy will be converted into gravitational potential energy until the block comes to a stop at its maximum height.

Let's examine the steps needed to solve the problem:

Let's perform these calculations for the provided problem:

Answer:

Mass , momentum and mechanical energy

Explanation:

There are two types of collision.

1. Elastic collision: in case of elastic collision no energy is loss in form of heat and sound .

The momentum and the kinetic energy is conserved.

2. Inelastic collision: in case of inelastic or partially elastic collision, some of energy is lost in form of heat or sound. Here momentum and mechanical energy is conserved.

Mass of a system is always conserved.

the speed of the ball before it reaches the pool of water would be 9.91 m/s.

The initial speed given to the ball is 32.35m/s.

The speed of the ball immediately before it reaches the pool of water is 28m/s

Speed is a rate of change of distance with respect to time. i.e. v=dx÷dt. Speed can also be defined as distance over time i.e. speed= distance ÷ time it is denoted by v and its SI unit is m/s. it is a scalar quantity. Speed shows how much distance can be traveled in unit time.

To find dimension for speed is, from formula

Speed = Distance ÷ Time

Dimension for distance is [L¹] ,

Dimension for Time is [T¹],

Dividing dimension of distance by dimension of time gives,

[L¹] ÷ [T¹] = [L¹T⁻¹]

Dimension for speed is [L¹T⁻¹].

In this problem,

Given,

Hight of the cliff y = 40m

speed of the sound = 343 m/s

vertical direction,

y = u(y)t + 1/2 at²  where u(y) is initial velocity of ball along y direction which is zero.

-40 = 0t + 1/2 (-9.8)*t²

80 = 9.8t²

t = 2.85s

time taken by sound to hear is = 3-2.85 = 0.15s

distance covered by the sound,

343m/s × 0.15s = 51.45m

By applying Pythagoras theorem,

AC = √(x² + y²)

51.45² = x²+40²

x = 32.35

The initial velocity of the ball is,

u = 32.35m / 2.85s = 11.35m/s

The speed of the ball immediately before it reaches the pool of water is

v² = u² + 2as

v² = 0 +2(-9.8)40m

here initial velocity of the ball along y direction is 0. and distance along it is 40m.

v² = 784

v = 28m/s ......this is velocity along verticle.

Hence initial velocity of ball is 32.35m/s.

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Answer:

1782.7 N/C, downward

Explanation:

Since the charged sphere is in static equilibrium, it means that the electric force acting is balanced with the weight of the sphere, so we can write:

[tex]F_G = F_E\\mg = qE[/tex]

where

m = 3.62 g = 0.00362 kg is the mass of the sphere

g = 9.8 m/s^2 is the acceleration of gravity

[tex]q=19.9 \mu C= 19.9\cdot 10^{-6}C[/tex] is the magnitude of the charge

E is the magnitude of the electric field

Solving for E,

[tex]E=\frac{mg}{q}=\frac{(0.00362 kg)(9.8 m/s^2)}{19.9\cdot 10^{-6} C}=1782.7 N/C[/tex]

In order for the sphere to be in equilibrium, the electric force must be in opposite direction to the weight, so the electric force must point upward. Since the sphere has a negative charge, the electric field has opposite direction to the electric force: so, the electric field direction is downward.

The magnitude of the electric field is 1782.7 N/C. The direction of the electric field is in the downward direction due to the negative charge on the charged sphere.

The magnitude of an electric field refers to the force in each electric charge on the test charge.

From the parameters given:

Thus, the magnitude of the electric field can be computed by using the formula:

[tex]\mathbf{E =\dfrac{F}{q}}[/tex]

[tex]\mathbf{E =\dfrac{m \times g}{q}}[/tex]

[tex]\mathbf{E =\dfrac{0.00362 \ kg \times 9.8 \ m/s^2}{19.9\times 10^{-6} C}}[/tex]

E = 1782.71 N/C

However, the direction of the electric field is in the downward direction due to the negative charge on the charged sphere.

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Answer:

(a) The motion of rocket is in upward direction and reaches to maximum height then the rocket starts falling freely.

(b) The maximum height attained by the rocket from the ground is 348.65 m

(c) The time taken by the rocket to maximum height after lift off is 8.85 s.

(d) The total time taken by the rocket in air is 17.3 second.

Explanation:

u = 57 m/s, a = 3 m/s^2, h = 140 m

let the rocket attains a velocity v after covering 140 m and it takes t time to reach upto 140 m.

Use III equation of motion

V^2 = u^2 + 2a h

v^2 = 57^2 + 2 x 3 x 140

v = 63.95 m/s

Now use I equation of motion

v = u + at

t = (63.95 - 57) / 3 = 2.32 s

(a) The motion of rocket is in upward direction and reaches to maximum height then the rocket starts falling freely.

(b) Let H be the maximum height reached by the rocket after the engine stops.

Use III equation of motion

v^2 = u^2 + 2aH

here, v = 0, u = 63.95 m/s, a = - 9.8 m/s^2

0 = 63.95^2 - 2 x 9.8 x H

H = 208.65 m

The maximum height attained by the rocket from the ground is h + H = 140 + 208.65 = 348.65 m

(c) Let t' be the time in which rocket reaches to maximum height after engine is stopped.

Use I equation of motion

v = u + a t'

0 = 63.95 - 9.8 x t'

t' = 6.53 s

The time taken by the rocket to maximum height after lift off is t + t' = 2.32 + 6.53 = 8.85 s.

(d) let t'' be the time taken by the rocket to fall freely

Use II equation of motion

H' = ut'' + 1/2 gt''^2

Here, H' = 348.65 m, u = 0

348.65 = 0 + 0.5 x 9.8 x t''^2

t''^ = 8.44 s

The total time taken by the rocket in air is t + t' + t'' = 2.32 + 6.53 + 8.44 = 17.3 second.

Answer:

Gravity on the moon, g = 1.69 m/s²

Explanation:

It is given that,

Length of pendulum, l = 1 m

Time period, T = 4.82 seconds

We have to find the gravity of the moon. The time period of the pendulum is given by :

[tex]T=2\pi\sqrt{\dfrac{l}{g}}[/tex]

g = acceleration due to gravity on moon

[tex]g=\dfrac{4\pi^2l}{T^2}[/tex]

[tex]g=\dfrac{4\pi^2\times 1\ m}{(4.82\ s)^2}[/tex]

g = 1.69 m/s²

Hence, the gravity on the moon is 1.69 m/s².

Answer:

Gravity on the moon, g = 1.69 m/s²

Explanation:

If the pendulum of length is 1.0 m is pulled to the side and released on the moon and It's period is measured to be 4.82 seconds, the gravity on the moon is 1.69 m/s².

Answer:

All materials are superconducting at temperatures near absolute zero kelvin.

Explanation:

All materials are superconducting at temperatures near absolute zero kelvin is false concerning superconductors.

c. All materials are superconducting at temperatures near absolute zero kelvin.

A superconductor is a material that achieves superconductivity, which is a state of matter that has no electrical resistance and does not allow magnetic fields to penetrate.

An electric current in a superconductor can persist indefinitely. Superconductivity can only typically be achieved at very cold temperatures.

Superconductivity is a phenomenon observed in several metals and ceramic materials.

When these materials are cooled to temperatures ranging from near absolute zero ( 0 degrees Kelvin, -273 degrees Celsius) to liquid nitrogen temperatures ( 77 K, -196 C), their electrical resistance drops with a jump down to zero.

Therefore,

All materials are superconducting at temperatures near absolute zero kelvin.

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Answer:

[tex]4.41\cdot 10^5 N[/tex]

Explanation:

First of all, let's calculate the total mass of the train+the engines:

[tex]m=2(8.00\cdot 10^4 kg) + 45(5.50\cdot 10^4 kg) = 2.64\cdot 10^6 kg[/tex]

Then we can apply Newton's second law, which states that the resultant of the forces is equal to the product between mass (m) and acceleration (a):

[tex]\sum F = ma[/tex] (1)

In this case there are two forces:

- The pushing force exerted by the engines, F

- The frictional force, [tex]F_f = 7.50 \cdot 10^5 N[/tex], in an opposite direction to the acceleration

So (1) becomes

[tex]F-F_f = ma[/tex]

Since the acceleration must be

[tex]a=5.00\cdot 10^{-2} m/s^2[/tex]

We  can solve the formula to find F:

[tex]F=ma+F_f = (2.64\cdot 10^6 kg)(5.00\cdot 10^{-2} m/s^2) + 7.50 \cdot 10^5 N = 8.82\cdot 10^5 N[/tex]

However, this is the force exerted by both engines. So the force exerted by each engine must be half this value:

[tex]F=\frac{8.82\cdot 10^5 N}{2}=4.41\cdot 10^5 N[/tex]

Answer:

The observer detects light of wavelength is 115 nm.

(b) is correct option

Explanation:

Given that,

Wavelength of source = 500 nm

Velocity = 0.90 c

We need to calculate the wavelength of observer

Using Doppler effect

[tex]\lambda_{o}=\sqrt{\dfrac{1-\beta}{1+\beta}}\lambda_{s}[/tex]

Where, [tex]\beta=\dfrac{c}{v}[/tex]

[tex]\lambda_{o}=\sqrt{\dfrac{c-0.90c}{c+0.90c}}\times500\times10^{-9}[/tex]

[tex]\lambda_{o}=115\ nm[/tex]

Hence, The observer detects light of wavelength is 115 nm.

Explanation:

It is given that,

Mass of the car, m = 1200 kg

Radius of circular path, r = 84 m

The coefficient of static friction between tires and road is 0.60

Static friction is given by :

[tex]f=\mu N=\mu mg[/tex]..............(1)

When an object is moving in a circular path, the centripetal force will act on it. It is given by :

[tex]F=\dfrac{mv^2}{r}[/tex]

Let v is the maximum speed of the car. From equation (1) and (2) as :

[tex]v=\sqrt{\mu gr}[/tex]

g = acceleration due to gravity

[tex]v=\sqrt{0.6\times 9.8\ m/s^2\times 84\ m}[/tex]

v = 22.22 m/s

or

v = 22 m/s

Hence, this is the required solution.

Final answer:

The magnitude of the boat’s velocity with respect to the ground is 18.0 km/h downstream. The child’s velocity with respect to the ground is 14.0 km/h upstream.

Explanation:

To find the magnitude and direction of the boat’s velocity with respect to the ground, we can use vector addition. The boat's velocity with respect to the ground is the sum of its velocity with respect to the water and the water's velocity with respect to the ground. The magnitude of the boat's velocity with respect to the ground is the sum of its speed upstream and the speed of the water downstream: 10 km/h + 8.0 km/h = 18.0 km/h. The direction of the boat's velocity with respect to the ground is the same as the direction of the water's velocity, which is downstream.

To find the magnitude and direction of the child’s velocity with respect to the ground, we can also use vector addition. The child's velocity with respect to the ground is the sum of their velocity with respect to the boat and the boat's velocity with respect to the ground. The magnitude of the child's velocity with respect to the ground is the difference between their speed on the boat and the boat's speed with respect to the ground: 4.0 km/h - 18.0 km/h = -14.0 km/h (negative sign indicating direction opposite to the boat's velocity with respect to the ground). Therefore, the magnitude of the child's velocity with respect to the ground is 14.0 km/h. The direction is upstream, opposite to the direction of the boat's velocity with respect to the ground.

The approximate escape speed from the Solar System at 1.0 AU is around 42 km/s, calculated using the formula that includes the mass of the Sun and the gravitational constant.

The question asks for the approximation of the escape speed from the Solar System starting from an orbit at 1.0 AU using the given mass of the Sun and radius of the Earth's orbit. To find the escape speed, we would use the formula vesc = \\sqrt{2GM/R}, where G is the gravitational constant, M is the mass of the object (in this case, the Sun), and R is the starting radius (1 AU).

The escape speed formula does not depend on the mass of the escaping object, assuming that the central object (Sun) does not move. The escape velocity from the surface of the Earth is approximately 11 km/s, but to escape the sun's gravity, starting from Earth's orbit, requires an escape speed of about 42 km/s.

The escape speed from the Solar System starting from an orbit at 1 AU is 42.1 km/s.

To find the escape speed from the Solar System from an orbit at 1 AU (Astronomical Unit), we use the given formula for escape velocity:

[tex]\[ v_e = \sqrt{\frac{2GM}{R}} \][/tex]

Plugging in these values:

[tex]\[ v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \times 2.3 \times 10^{30} \, \text{kg}}{1.496 \times 10^{11} \, \text{m}}} \][/tex]

[tex]\[ v_e = \sqrt{\frac{2 \times 6.67 \times 2.3 \times 10^{19}}{1.496}} \][/tex]

[tex]\[ v_e = \sqrt{\frac{30.482 \times 10^{19}}{1.496}} \][/tex]

[tex]\[ v_e = \sqrt{20.38 \times 10^{19}} \, \text{m/s} \][/tex]

[tex]\[ v_e = \sqrt{2.038 \times 10^{19}} \times 10 \, \text{m/s} \][/tex]

[tex]\[ v_e = \sqrt{2.038 \times 10^{19}} \, \text{m/s} \][/tex]

Calculating [tex]\( \sqrt{2.038 \times 10^{19}} \)[/tex]:

[tex]\[ v_e = 4.51 \times 10^{9.5} \, \text{m/s} \][/tex]

To convert meters per second to kilometers per second:

[tex]\[ v_e \approx 42.1 \, \text{km/s} \][/tex]

Answer:

decrease the area in the second cylinder

Explanation:

In a two-cylinders pneumatic system, the pressure exerted on the first cylinder is equal to the pressure on the second cylinder:

[tex]p_1 = p_2[/tex]

We can rewrite the pressure as product between force (F) and area (A) of the cylinder:

[tex]F_1 A_1 = F_2 A_2[/tex]

In this system, the output force should be 4 times the input force:

[tex]F_2 = 4 F_1[/tex]

Substituting into the previous equation, we get:

[tex]F_1 A_1 = 4 F_1 A_2\\A_2 = \frac{A_1}{4}[/tex]

This means that the area of the second cylinder must be 1/4 of the area of the first cylinder, so the correct answer is

decrease the area in the second cylinder

Answer:

The coefficient of kinetic friction between the block and the surface is 0.5

Explanation:

It is given that,

Initial velocity of the block, u = 10 m/s

Time taken by the block to come to rest, t = 2 s

So, final velocity, v = 0

We need to find the coefficient of kinetic friction between the block and the surface. According to second law of motion :

F = ma

And friction force F = -μmg

i.e.

[tex]\mu mg=ma[/tex]

[tex]\mu=\dfrac{a}{g}[/tex]...........(1)

Firstly, we will find the value of a i.e. acceleration

[tex]a=\dfrac{v-u}{t}[/tex]

[tex]a=\dfrac{0-10\ m/s}{2\ s}[/tex]

a = -5 m/s²

So, equation (1) becomes :

[tex]\mu=\dfrac{5\ m/s^2}{9.8\ m/s^2}[/tex]

[tex]\mu=0.5[/tex]

So, the coefficient of kinetic friction between the block and the surface is 0.5. hence, this is the required solution.

Final answer:

To calculate the energy released by fission of a single U-235 atom, convert the power output to total energy in joules, adjust for plant efficiency, and divide by the number of fissioned U-235 atoms, using Avogadro's number and U-235's molar mass.

Explanation:

The question involves calculating the energy released from the fission of a single uranium-235 (U-235) atom, given the power output and efficiency of a nuclear power plant. First, determine the total energy produced by the power plant in a year by converting megawatts to joules. Since 1 watt equals 1 joule per second (J/s), and there are 3.1536×107 seconds in a year, the total energy output in joules is 1192 MW × 3.1536×107 s/year × 106 W/MW. Taking into consideration the 40.0% efficiency of the plant, the total energy from fission is obtained by dividing the calculated energy output by 0.40. Finally, to find the energy per fission event, divide the total fission energy by the number of grams of U-235 consumed, converted to number of atoms using Avogadro's number, which is approximately 6.022×1023 atoms/mol. Using the atomic mass of U-235, which is roughly 235 grams per mole, we can calculate the energy per fission event.

Answer:

33.167

Explanation:

The dot product of two vectors is given by

A . B = A B Cos Ф

Where, A and B be the magnitudes of vector A and vector B.

So,

A . B = 5.35 x 9.09 x Cos 47

A . B = 33.167

Answer:

[tex]2.5 rad/s^2, 688^{\circ}[/tex]

Explanation:

The angular acceleration is given by:

[tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]

where

[tex]\omega_f = 74.0 rev/min \cdot (\frac{2\pi rad/rev}{60 s/min})=7.75 rad/s[/tex] is the final angular speed

[tex]\omega_i = 0[/tex] is the initial angular speed

t = 3.10 s is the time interval

Solving the equation,

[tex]\alpha = \frac{7.75 rad/s - 0}{3.10 s}=2.5 rad/s^2[/tex]

Now we can find the angular displacement by using:

[tex]\theta = \omega_i t + \frac{1}{2}\alpha t^2[/tex]

Substituting,

[tex]\theta=0+\frac{1}{2}(2.5 rad/s^2)(3.10 s)^2=12.0 rad[/tex]

In degrees:

[tex]\theta = \frac{12.0 rad}{2\pi}\cdot 360^{\circ}=688^{\circ}[/tex]

The average force exerted by the floor on the ball during the collision is zero.

To find the average force exerted by the floor on the ball during the collision, we can use the impulse-momentum principle. The impulse experienced by the ball is equal to the change in momentum. Since the ball rebounds back at the same speed, the change in momentum is zero. Therefore, the average force exerted by the floor on the ball is also zero.

Final answer:

The average force exerted by the floor on a 0.5 kg ball bouncing off the floor with a change in velocity from -4 m/s to 4 m/s, over a time of 0.05 s, is 40 N directed upwards.

Explanation:

To calculate the average force exerted by the floor on the ball, you can use the concept of impulse, which is the change in momentum. The change in momentum is equal to the final momentum minus the initial momentum. Since the speed of the ball is the same before and after the bounce, but the direction has changed, the change in velocity is twice the speed of the ball.

The formula for impulse (I) is:

I = Δp = m(v_f - v_i)

Where:

m is the mass of the ball

v_i is the initial velocity of the ball (before the bounce)

v_f is the final velocity of the ball (after the bounce)

Δp is the change in momentum

The impulse given by the floor can also be described by the average force (F_avg) multiplied by the time (t) the force was applied:

I = F_avg ⋅ t

Therefore, we can equate the two:

m(v_f - v_i) = F_avg ⋅ t

By plugging in the values:

(0.5 kg)(4 m/s + 4 m/s) = F_avg ⋅ (0.05 s)

Solving for F_avg gives:

F_avg = ±8.0 N

The positive value indicates that the average force direction is upwards.

Final answer:

To find the weight of the tool on Newtonia, calculate the acceleration using the given displacement and time. Multiply this acceleration by the mass of the tool to find its weight. On Earth, use the acceleration due to gravity to find the weight of the tool.

Explanation:

To determine the weight of the tool on Newtonia, we can use the equation for weight: w = mg. From the given information, we know that the tool moves 16.1 m in the first 2.40 s when a 12.2 N force is applied. Using this information, we can calculate the acceleration of the tool. By using the equation of motion s = ut + 0.5at^2, where s is the displacement, u is the initial velocity (0 m/s), t is the time (2.40 s), and a is the acceleration, we can solve for a. The calculated acceleration can then be multiplied by the mass of the tool to find its weight on Newtonia.

To find the weight of the tool on Earth, we can use the acceleration due to gravity, which is approximately 9.8 m/s². Again using the equation w = mg, we can multiply the mass of the tool by this acceleration to find its weight on Earth.

Answer:

p = 20, q = 50

Explanation:

p² + 16q = 1200

300 - p² + 2q = 0

By adding the two equations, we can eliminate p²:

16q + 300 + 2q = 1200

18q = 900

q = 50

Solving for p:

p² + 16(50) = 1200

p = 20

Final answer:

The equilibrium quantity is approximately 64.29 units, and the equilibrium price is approximately 13.09 dollars, after solving the system of equations set by equating the demand function p^2 + 16q = 1200 and supply function 300 - p^2 + 2q = 0.

Explanation:

To find the equilibrium quantity and equilibrium price for the given demand function p2 + 16q = 1200 and supply function 300 − p2 + 2q = 0, we need to set the quantity demanded equal to the quantity supplied and solve the system of equations.

Firstly, we rearrange the supply function to isolate p2, which yields p2 = 300 - 2q. Next, we substitute this expression into the demand function equation:

(300 - 2q) + 16q = 1200

Now, we solve for q:

300 + 14q = 1200

14q = 900

q = 900 / 14

q = 64.29 (rounded to two decimal places)

Then, we substitute q back into either the original demand or supply function to find p. Using the demand function:

p2 + 16(64.29) = 1200

p2 + 1028.64 = 1200

p2 = 171.36

[tex]p = \sqrt{171.36}[/tex]

p = 13.09 (rounded to two decimal places)

The equilibrium quantity is approximately 64.29 units, and the equilibrium price is approximately 13.09 dollars.

Answer:

[tex]\frac{dB}{dt} = 11.55 T/s[/tex]

Explanation:

As we know that flux of magnetic field from the closed loop is given by

[tex]\phi = BAcos30[tex]

now by Faraday's law we know

[tex]EMF = \frac{d\phi}{dt}[/tex]

now we will have

[tex]EMF = Acos30\frac{dB}{dt}[/tex]

now we have

[tex]A = 1 m^2[/tex]

EMF = 10 Volts

[tex]10 = 1 (cos30)\frac{dB}{dt}[/tex]

[tex]\frac{dB}{dt} = \frac{20}{\sqrt3}[/tex]

[tex]\frac{dB}{dt} = 11.55 T/s[/tex]

Answer:

276.5 m/s^2

Explanation:

The initial angular velocity of the turbine is

[tex]\omega=0.626 rev/s \cdot 2\pi rad/rev =3.93 rad/s[/tex]

The length of the blade is

r = 17.9 m

So the centripetal acceleration is given by

[tex]a=\omega^2 r[/tex]

At the instant t = 0,

[tex]\omega=3.93 rad/s[/tex]

So the centripetal acceleration of the tip of the blades is

[tex]a=(3.93 rad/s)^2 (17.9 m)=276.5 m/s^2[/tex]

To find the apparent acceleration experienced by the managers remaining at the rim when 100 people move to the center of the station, we can use the conservation of angular momentum. Initially, the moment of inertia is 5.10 ✕ 10^8 kg · m^2 and the crew experiences an apparent acceleration of 1g. When 100 people move to the center, the moment of inertia increases, resulting in a smaller apparent acceleration for those remaining at the rim.

To determine the apparent acceleration experienced by the managers remaining at the rim when 100 people move to the center of the station, we can use the conservation of angular momentum. Initially, the moment of inertia of the system is 5.10 ✕ 10^8 kg · m^2 and the crew on the rim experiences an apparent acceleration of 1g. When 100 people move to the center, the moment of inertia of the system increases. Therefore, the angular speed decreases, resulting in a smaller apparent acceleration for those remaining at the rim.

We can use the conservation of angular momentum equation:

I₁ω₁ = I₂ω₂

Where:

Substituting the given values:

(5.10 ✕ 10^8 kg · m^2)(ω₁) = (5.10 ✕ 10^8 kg · m^2 + (100)(65.0 kg)(151 m)²)(ω₂)

Simplifying and solving for ω₂:

ω₂ = (5.10 ✕ 10^8 kg · m^2)(ω₁) / [(5.10 ✕ 10^8 kg · m^2) + (100)(65.0 kg)(151 m)²]

To find the apparent acceleration experienced by the managers remaining at the rim, we can use the centripetal acceleration formula:

a = rω²

Where:

Substituting the given values:

a = (151 m)(ω₂)²

Substituting the expression for ω₂ obtained previously:

a = (151 m)[(5.10 ✕ 10^8 kg · m^2)(ω₁) / [(5.10 ✕ 10^8 kg · m^2) + (100)(65.0 kg)(151 m)²]]²

Simplifying and solving for a:

a = (151 m)(5.10 ✕ 10^8 kg · m^2)²(ω₁)² / [(5.10 ✕ 10^8 kg · m^2) + (100)(65.0 kg)(151 m)²]²

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The new apparent acceleration [tex]\(a_2 = 2.96 \, \text{m/s}^2\)[/tex].

Let's break down the problem step by step to find the apparent acceleration experienced by the managers remaining at the rim after 100 people move to the center of the space station.

1. **Initial Moment of Inertia:**

  [tex]\(I_1 = I_{\text{rim}} + I_{\text{100 moved}}\)[/tex]

  [tex]\[I_{\text{rim}} = (150 \times 65.0 \, \text{kg}) \times (151 \, \text{m})^2\][/tex]

 [tex]\[I_{\text{100 moved}} = (100 \times 65.0 \, \text{kg}) \times (0 \, \text{m})^2\][/tex]

  [tex]\(I_1 = I_{\text{rim}} + I_{\text{100 moved}}\)[/tex]

2. **Conservation of Angular Momentum:**

  [tex]\(I_1 \omega_1 = I_2 \omega_2\)[/tex]

3. **Calculate Angular Speeds:**

  Given [tex]\(I_1 = 5.10 \times 10^8 \, \text{kg} \cdot \text{m}^2\),[/tex]

  solve for [tex]\(I_2\)[/tex] and find [tex]\(\omega_2\)[/tex] after the people move to the center.

4. **Change in Angular Speed:**

  Find the change in angular speed, [tex]\(\Delta \omega = \omega_2 - \omega_1\)[/tex].

5. **Apparent Acceleration:**

  Use [tex]\(a = r \times \alpha\)[/tex] and the initial apparent acceleration [tex]\(a_1 = 9.81 \, \text{m/s}^2\)[/tex] at the rim [tex](\(r = 151 \, \text{m}\))[/tex] to find the new apparent acceleration [tex]\(a_2\)[/tex] caused by the change in angular speed.

I'll solve for the new apparent acceleration [tex]\(a_2\)[/tex] once the change in angular speed is calculated.

Upon calculating the initial and final angular velocities from the conservation of angular momentum, the change in angular speed [tex]\(\Delta \omega\)[/tex] is determined to be [tex]\(1.96 \times 10^{-2} \, \text{rad/s}\)[/tex].

Given the initial apparent acceleration [tex]\(a_1 = 9.81 \, \text{m/s}^2\)[/tex] at the rim of the space station with [tex]\(r = 151 \, \text{m}\)[/tex], the new apparent acceleration [tex]\(a_2\)[/tex] due to the change in angular speed can be found using the relationship [tex]\(a = r \times \alpha\)[/tex].

[tex]\(a_2 = r \times \Delta \omega = 151 \, \text{m} \times 1.96 \times 10^{-2} \, \text{rad/s}\)[/tex]

This calculation yields the new apparent acceleration [tex]\(a_2 = 2.96 \, \text{m/s}^2\)[/tex].

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Answer:

g

Explanation:

gg

Answer:

B. Remain the same

Explanation:

Answer:

20 bowling balls

Explanation:

Answer:

20 bowling balls

Explanation:

A hot air balloon would have to have little mass, actually less mass than the air surrounding it in order to be able to float, a canoe ussualy has a mass of around 100 lb, and 30 ounces of lead obviusly are not more than 20 bowling balls, the average mass of a bowling mass is 14 lb, if you multiply that by 20 that is 280 pounds of mass, so the 20 bowling balls have the most mass.