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JulsSmile [24]
2 years ago
11

Anna pushes a box with a force of 8.00 newtons. she generates a power of 3.00 watts. how much time does it take for anna to move

the box the distance of 9.72 meters? a. 12.4 seconds b. 20.1 seconds c. 25.9 seconds d. 30.8 seconds e. 35.8 seconds
Physics
1 answer:
mote1985 [20]2 years ago
7 0

The time  taken for anna to move the box the distance of 9.72 meters is 25.9 seconds. The correct option is c.

<h3>What is force?</h3>

The force is the action of push or pull to move or stop an object.

Anna pushes a box with a force of 8.00 newtons. she generates a power of 3.00 watts.

Force is related to power by the relation

Force = Power / Velocity

Put the values, we get

8 = 3/V

V = 0.375 m/s

Velocity is defined as the ratio of distance and the time taken.

The box is moved by the distance of 9.72 meters. So, time taken is

V = d/t

0.375 = 9.72/t

t =25.9 seconds.

Thus,  time  taken for anna to move the box the distance of 9.72 meters is 25.9 seconds.

Learn more about force.

brainly.com/question/21763

#SPJ1

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When serving a tennis ball, a player hits the ball when its velocity is zero (at the highest point of a vertical toss). The racq
Lilit [14]

Answer:60 gm

Explanation:

Given

initial velocity of ball u=0

Force exerted by racquet F=540 N

time period of force t=5\ ms

final velocity of ball v=45\ m/s

Racquet imparts an impulse to the ball which is given by

J=F\Delta t=\Delta P

J=540\times \Delta t=m(45-0)

m=\frac{540\times 5\times 10^{-3}}{45}

m=60\ gm

8 0
3 years ago
An infinite line of charge with linear density λ1 = 8.2 μC/m is positioned along the axis of a thick insulating shell of inner r
bixtya [17]

1) Linear charge density of the shell:  -2.6\mu C/m

2)  x-component of the electric field at r = 8.7 cm: 1.16\cdot 10^6 N/C outward

3)  y-component of the electric field at r =8.7 cm: 0

4)  x-component of the electric field at r = 1.15 cm: 1.28\cdot 10^7 N/C outward

5) y-component of the electric field at r = 1.15 cm: 0

Explanation:

1)

The linear charge density of the cylindrical insulating shell can be found  by using

\lambda_2 = \rho A

where

\rho = -567\mu C/m^3 is charge volumetric density

A is the area of the cylindrical shell, which can be written as

A=\pi(b^2-a^2)

where

b=4.7 cm=0.047 m is the outer radius

a=2.7 cm=0.027 m is the inner radius

Therefore, we have :

\lambda_2=\rho \pi (b^2-a^2)=(-567)\pi(0.047^2-0.027^2)=-2.6\mu C/m

 

2)

Here we want to find the x-component of the electric field at a point at a distance of 8.7 cm from the central axis.

The electric field outside the shell is the superposition of the fields produced by the line of charge and the field produced by the shell:

E=E_1+E_2

where:

E_1=\frac{\lambda_1}{2\pi r \epsilon_0}

where

\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m is the linear charge density of the wire

r = 8.7 cm = 0.087 m is the distance from the axis

And this field points radially outward, since the charge is positive .

And

E_2=\frac{\lambda_2}{2\pi r \epsilon_0}

where

\lambda_2=-2.6\mu C/m = -2.6\cdot 10^{-6} C/m

And this field points radially inward, because the charge is negative.

Therefore, the net field is

E=\frac{\lambda_1}{2\pi \epsilon_0 r}+\frac{\lambda_2}{2\pi \epsilon_0r}=\frac{1}{2\pi \epsilon_0 r}(\lambda_1 - \lambda_2)=\frac{1}{2\pi (8.85\cdot 10^{-12})(0.087)}(8.2\cdot 10^{-6}-2.6\cdot 10^{-6})=1.16\cdot 10^6 N/C

in the outward direction.

3)

To find the net electric field along the y-direction, we have to sum the y-component of the electric field of the wire and of the shell.

However, we notice that since the wire is infinite, for the element of electric field dE_y produced by a certain amount of charge dq along the wire there exist always another piece of charge dq on the opposite side of the wire that produce an element of electric field -dE_y, equal and opposite to dE_y.

Therefore, this means that the net field produced by the wire along the y-direction is zero at any point.

We can apply the same argument to the cylindrical shell (which is also infinite), and therefore we find that also the field generated by the cylindrical shell has no component along the y-direction. Therefore,

E_y=0

4)

Here we want to find the x-component of the electric field at a point at

r = 1.15 cm

from the central axis.

We notice that in this case, the cylindrical shell does not contribute to the electric field at r = 1.15 cm, because the inner radius of the shell is at 2.7 cm from the axis.

Therefore, the electric field at r = 1.15 cm is only given by the electric field produced by the infinite wire:

E=\frac{\lambda_1}{2\pi \epsilon_0 r}

where:

\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m is the linear charge density of the wire

r = 1.15 cm = 0.0115 m is the distance from the axis

This field points radially outward, since the charge is positive . Therefore,

E=\frac{8.2\cdot 10^{-6}}{2\pi (8.85\cdot 10^{-12})(0.0115)}=1.28\cdot 10^7 N/C

5)

For this last part we can use the same argument used in part 4): since the wire is infinite, for the element of electric field dE_y produced by a certain amount of charge dq along the wire there exist always another piece of charge dq on the opposite side of the wire that produce an element of electric field -dE_y, equal and opposite to dE_y.

Therefore, the y-component of the electric field is zero.

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

4 0
3 years ago
The center of gravity of an ax is on the centerline of the handle, close to the head. Assume you saw across the handle through t
andrew-mc [135]

Answer:

I believe it is they will weigh the same

Explanation:

Center of gravity is the axis on which the mass rotates evenly if I remember correctly from AP Physics

3 0
3 years ago
Which statement is correct?
Sedbober [7]

Answer:

The answer is "Choice C ".

Explanation:

The relationship between the E and V can be defined as follows:

\to E= -\Delta V

Let,

\to E= \frac{\delta V}{\delta x}

When E=0

\to   \frac{\delta V}{\delta x}=0

v is a constant value

Therefore, In the electric potential in a region is a constant value then the electric-field must be into zero that is everywhere in the given region, that's why in this question the "choice c" is correct.

7 0
3 years ago
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