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oee [108]
3 years ago
7

Kids are pelting a window with snowballs. On average, two snowballs of roughly 300-g mass hit the window each second, moving hor

izontally at some 10 m/s. The snowballs drop vertically to the ground after hitting the window.
Required:
Estimate the average force exerted on the window.
Physics
1 answer:
hram777 [196]3 years ago
5 0

Answer:

The average force exerted on the window due to two snowballs is 6 N

Explanation:

Given:

Mass of snowballs m = 300 \times 10^{-3} Kg

Velocity of snowball v = 10 \frac{m}{s}

For finding the average force,

Force is equal to the change in momentum,

   F = \frac{dP}{dt}

Here, final velocity is zero so we write,

 F = \frac{mv}{1}

Where dt = 1 sec

 F = 300 \times 10^{-30} \times 10

F = 3 N

Above value of force is due to one ball, but here given in question there are two ball,

F = 3 \times 2

F = 6 N

Therefore, the average force exerted on the window due to two snowballs is 6 N

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Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 1.00 s, it rotates 21.0 rad. Du
ELEN [110]

With constant angular acceleration \alpha, the disk achieves an angular velocity \omega at time t according to

\omega=\alpha t

and angular displacement \theta according to

\theta=\dfrac12\alpha t^2

a. So after 1.00 s, having rotated 21.0 rad, it must have undergone an acceleration of

21.0\,\mathrm{rad}=\dfrac12\alpha(1.00\,\mathrm s)^2\implies\alpha=42.0\dfrac{\rm rad}{\mathrm s^2}

b. Under constant acceleration, the average angular velocity is equivalent to

\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2

where \omega_f and \omega_i are the final and initial angular velocities, respectively. Then

\omega_{\rm avg}=\dfrac{\left(42.0\frac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)}2=42.0\dfrac{\rm rad}{\rm s}

c. After 1.00 s, the disk has instantaneous angular velocity

\omega=\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)=42.0\dfrac{\rm rad}{\rm s}

d. During the next 1.00 s, the disk will start moving with the angular velocity \omega_0 equal to the one found in part (c). Ignoring the 21.0 rad it had rotated in the first 1.00 s interval, the disk will rotate by angle \theta according to

\theta=\omega_0t+\dfrac12\alpha t^2

which would be equal to

\theta=\left(42.0\dfrac{\rm rad}{\rm s}\right)(1.00\,\mathrm s)+\dfrac12\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)^2=63.0\,\mathrm{rad}

5 0
3 years ago
Uniformly charged ring with 180 nC/m and radius R= 58 cm. Find the magnitude of the electric field in KN/C at a point P on the a
raketka [301]

Answer:

3.135 kN/C

Explanation:

The electric field on the axis of a charged ring with radius R and distance z from the axis is E = qz/{4πε₀[√(z² + R²)]³}

Given that R = 58 cm = 0.58 m, z = 116 cm = 1.16m, q = total charge on ring = λl where λ = charge density on ring = 180 nC/m = 180 × 10⁻⁹ C/m and l = length of ring = 2πR. So q = λl = λ2πR = 180 × 10⁻⁹ C/m × 2π(0.58 m) = 208.8π × 10⁻⁹ C and ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m

So, E = qz/{4πε₀[√(z² + R²)]³}

E = 208.8π × 10⁻⁹ C × 1.16 m/{4π8.854 × 10⁻¹² F/m[√((1.16 m)² + (0.58 m)²)]³}

E = 242.208 × 10⁻⁹ Cm/{35.416 × 10⁻¹² F/m[√(1.3456 m² + 0.3364 m²)]³}

E = 242.208 × 10⁻⁹ Cm/35.416 × 10⁻¹² F/m[√(1.682 m²)]³}

E = 6.839 × 10³ Cm²/[1.297 m]³F

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E = 3.135 × 10³ V/m

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2 years ago
A deep space probe travels in a straight line at a constant speed of over 16,000 m/s. Assuming there is no friction in space, if
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Answer:

I believe that the answer is d.

Explanation:

Because there is nothing to make the aircraft accelerate or decelerate, it is going to stay in constant motion with no acceleration.

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aleksley [76]

Answer:

1) The charge on the outer shell is +4·Q

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Explanation:

1) The given parameters of the spherical shell are;

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Let 'x' represent the charge on the outer shell, and let 'y', represent the charge on the inner shell, we have;

The net charge, 3·Q = -q + x

∴ x = 3·Q + Q = 4·Q

The charge on the outer shell, x = 4·Q

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