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3 years ago

If a bullet loses 1/nth of its velocity while passing through a plank,then no of planks required to stop the bullet is?

1 answer:

8 0

Based on the answer provided, it seems the writer wanted you to assume that the energy loss per plank is constant. This is not the same as the bullet losing 1/nth1/nth of its velocity per plank (however, the fact that the question does not mention this assumption arguably makes the question ambiguous).

With this assumption, the energy loss becomes

ΔE = 1/2 mv2 − 1/2 m (v−v/n) 2

and the number of planks NN becomes

N = 1/2mv2 /ΔE = n2/ 2n−1

Otherwise, if you assume that the bullet loses 1/nth1/nth of its velocity per plank, then the answer is N=∞

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A block whose weight is 45 N rests on a horizontal table. A horizontal force of 36 N is applied to the block. The coefficient of

riadik2000 [5.3K]

Answer:

1.5 m/s²

Explanation:

For the block to move, it must first overcome the static friction.

Fs = N μs

Fs = (45 N) (0.42)

Fs = 18.9 N

This is less than the 36 N applied, so the block will move. Since the block is moving, kinetic friction takes over. To find the block's acceleration, use Newton's second law:

∑F = ma

F − N μk = ma

36 N − (45 N) (0.65) = (45 N / 9.8 m/s²) a

6.75 N = 4.59 kg a

a = 1.47 m/s²

Rounded to two significant figures, the block's acceleration is 1.5 m/s².

Usually the coefficient of static friction is greater than the coefficient of kinetic friction. You might want to double check the problem statement, just to be sure.

6 0

3 years ago

The dipole moment of HI is 0.42D. What is the dipole moment of HI in C⋅m?

sleet_krkn [62]

So in your question where ask to find the dipole moment of HI in C.M. So in my calculation by converting the given data from Debyes to Coulomb Meteres is that 1 Dyebes is equals to 3.33X10^(-30) C.M and the answer would be 1.40X10(-30)C.M. I hope you are satisfied with my answer and feel free to ask for more

5 0

3 years ago

What has more kinetic energy 15 kg ball rolling north at 15 m/s or a 15 kg ball rolling backwards at 7m/s

Setler79 [48]

Answer:

15 kg ball

Explanation:

6 0

3 years ago

A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.8 1010 m (inside the orbit

creativ13 [48]

Answer:

Explanation:

From the given information:

Distance $d_i = 4.8 \times 10^{10} \ m$

Speed of the comet $V_i = 9.1 \times 10^{4} \ m/s$

At distance $d_2 = 6 \times 10^{12} \ m$

where;

mass of the sun = $1.98 \times 10^{30}$

$G = 6.67 \times 10^{-11}$

To find the speed $V_f$ :

Using the formula:

$E_f = E_i + W \\ \\ where; \ \ W = 0 \ \ \text{since work done by surrounding is zero (0)}$

$E_f = E_i + 0 \\ \\ K_f + U_f = K_i + U_i \\ \\ = \dfrac{1}{2}mV_f^2 + \dfrac{-GMm}{d^2} = \dfrac{1}{2}mV_i^2+ \dfrac{-GMm}{d_i} \\ \\ V_f = \sqrt{V_i^2 + 2 GM \Big [ \dfrac{1}{d_2}- \dfrac{1}{d_i}\Big ]}$

$V_f = \sqrt{(9.1 \times 10^{4})^2 + 2 (6.67\times 10^{-11}) *(1.98 * 10^{30} ) \Big [ \dfrac{1}{6*10^{12}}- \dfrac{1}{4.8*10^{10}}\Big ]}$

$\mathbf{V_f =53.125 \times 10^4 \ m/s}$

3 0

2 years ago

Name human body organs where magnetism is significant

ser-zykov [4K]

It has been hypothesized, and some studies have supported the conjecture,
that certain species of insects and birds are able to sense the direction of external
magnetic fields.

I don't think there is any such notion where human beings are concerned.

5 0

3 years ago