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3 years ago

A tennis ball is traveling at 50 m/s and has a kinetic energy of 75J. Calculate the mass of the tennis ball.

Physics

1 answer:

Vitek1552 [10]3 years ago

8 0

Couldn’t fit all of it , but here luv

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3 years ago

Write down the addres of earth in as much detail as possible.

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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:

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 :

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

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.

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,

$E_y=0$

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,

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

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:

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4 0

3 years ago

For light of wavelength 589 nm, calculate the critical angles for the following substances when surrounded by air.

NeX [460]

(1) The critical angles for the fused quartz is 44⁰,

(2) The critical angles for the polystyrene is 39⁰.

(3) The critical angles for the sodium chloride is 41⁰.

<h3>What is Critical angles?</h3>

In optics, the critical angle is the topmost angle at which a shaft of light traveling in one transparent medium can strike the boundary between that medium and another with a lower refractive indicator without being fully reflected within the first

Critical angles for the different medium.

θ $=sin^{-1}\frac{n_{2} }{n_{1} }$

where;

$n_{2}$ is the refractive index of air = 1

$n_{1}$ is the refractive index of the given substances.

Refractive index of quartz glass = 1.46

Refractive index of polystyrene = 1.59

Refractive index of sodium chloride = 1.54

Critical angles for quartz glass:

$\theta_{\text {crit }}=\sin ^{-1} \frac{1.00029}{1.46}\\$

$\theta_{\text {crit }}=44^{o}$

Critical angles for polystyrene

$$\theta_{\text {crit }}=\sin ^{-1} \frac{1.00029}{1.59}\\\\$

$\theta_{\text {crit }}=39^{\circ}$

Critical angles for Sodium Chloride

$$\theta_{\text {crit }}=\sin ^{-1} \frac{1.00029}{1.54}\\$

$\theta_{\text {crit }}=41^{o}$

Thus, the critical angles for the quartz glass is 44⁰, the critical angles for the polystyrene is 39 ⁰ and the critical angles for the sodium chloride is 41⁰.

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I understand that the question you are looking for is:

"For light of wavelength 589 nm, calculate the critical angles for the following substances when surrounded by air.

(a) Fused Quartz

(b) Polystyrene

8 0

2 years ago

A tennis ball is traveling at 50 m/s and has a kinetic energy of 75J￼. Calculate the mass of the tennis ball.

A tennis ball is traveling at 50 m/s and has a kinetic energy of 75J. Calculate the mass of the tennis ball.