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3 years ago

What will be the final velocity of a 5.0 g bullet starting from rest, if a net force of 45 N is applied over a time of 0.02s?

1 answer:

3 0

We know there’s a change in momentum due to a force applied over a time interval. Ft= m[v(final)-v(initial)]. Now simply plug in know values: (45)(0.02)=.005[v(final)-0]. Remember converting grams to kilograms. Solve for v final

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Check the correctness of P=√2mE/t where the symbols have usual meaning.

uysha [10]

Answer:

Check the correctness of the relation c=1 √μ0∈0 where the symbols have their usual meaning. check-circle.

5 0

3 years ago

A spinning wheel has a rotational inertia of 2 kg•m². It has an angular velocity of 6.0 rad/s. An average counterclockwise torqu

ira [324]

Answer:

$-20.0 kg m^2/s$

Explanation:

The angular momentum of an object in rotation is given by

$L=I \omega$

where

I is the moment of inertia

$\omega$ is the angular speed

In this problem, initially we have

$I=2 kg m^2$ is the moment of inertia of the wheel

$\omega_i = 6.0 rad/s$ is the initial angular speed

So the initial angular momentum is

$L_i = I\omega_i = (2)(6.0)=12 kg m^2/s$

Later, a counterclockwise torque of

$\tau=-5.0 Nm$ is applied

So the angular acceleration of the wheel is:

$\alpha = \frac{\tau}{I}=\frac{-5.0}{2}=-2.5 rad/s^2$ in the direction opposite to the initial rotation.

As a result, the final angular velocity of the wheel will be:

$\omega_f = \omega_i + \alpha t$

where

t = 4.0 is the time interval

Solving,

$\omega_f = +6.0 +(-2.5)(4.0)=-4 rad/s$

which means that now the wheel is rotating in the counterclockwise direction.

Therefore, the new angular momentum of the wheel is:

$L_f = I\omega_f =(2)(-4.0)=-8.0 kg m^2/s$

So, the change in angular momentum is:

$\Delta L=L_f - L_i = (-8.0-(12))=-20.0 kg m/s^2$

7 0

3 years ago

A 17-mm-wide diffraction grating has rulings of 530 lines per millimeter. White light is incident normally on the grating. What

Sedbober [7]

Answer:

377 nm

Explanation:

Number of lines per meter is, $N &=530 \times 1000 \\ &=530000 \text { lines } / \mathrm{m} \end{aligned}$

Grating element is, $d=\frac{1}{N}$

$=1.8868 \times 10^{-6} \mathrm{~m}[tex]
Order is, n=5
Condition for maximum intensity is, [tex]d \sin \theta=n \lambda$

$\lambda &=\frac{1.8868 \times 10^{-6}}{5(\sin 90)} \\
&=0.377 \times 10^{-6} \mathrm{~m} \\
&=377 \mathrm{~nm}$

7 0

3 years ago

1. You are designing a new solenoid and experimenting with material for each turn. The particular turn you are working with is a

svetoff [14.1K]

Answer:

Explanation:

1 ) Magnetic field due to a circular coil carrying current

= μ₀I / 2r

I is current , r is radius of the wire , μ₀ = 4π x 10⁻⁷

= 4π x 10⁻⁷ x 15 / (2 x 3.5 x 10⁻²)

= 26.9 x 10⁻⁵ T

2 )

Negative z direction .

The direction of magnetic field due to a circular coil having current is known

with the help of screw rule or right hand thumb rule.

3 )

If we decrease the radius the magnetic field will:__increase _____.

A. Increase.

Magnetic field due to a circular coil carrying current

B = μ₀I / 2 r

Here r is radius of the coil . If radius decreases magnetic field increases.

So magnetic field will increase.

4 0

3 years ago

A place that limits a wave’s motion.

Yuki888 [10]

Answer:

boundary

Explanation:

7 0

3 years ago