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kykrilka [37]
1 year ago
11

a bullet fired straight through a board 0.10 mm thick strikes the board with a speed of 480 m/sm/s, has constant acceleration th

rough the board, and emerges with a speed of 345 m/sm/s.
Physics
1 answer:
vekshin11 year ago
5 0

the average acceleration of the bullet through the board   -55657×10⁵ m/s²

acceleration-  Rate of change of velocity with respect to time.

S = 0.10 mm = 10⁻⁵ m  ( distance)          [∵ 1mm = 10⁻³]

u =  480 m/s (initial velocity)

v = 345 m/s (final velocity)

As we know the 3rd equation of motion.  

v² = u² + 2aS

a = ? ( acceleration)

using these values in equation we get

(345)² = (480)² + 2×a× 10⁻⁵

a = (345)² -  (480)² / 2× 10⁻⁵

a =  -55657×10⁵ m/s²

The negative sign shows that the direction of acceleration is opposite of velocity thus bullet is slowing down.

the average acceleration of the bullet through the board  -55657×10⁵ m/s²

The given question is incomplete. The complete question is.

a bullet fired straight through a board 0.10 mm thick strikes the board with a speed of 480 m/sm/s, has constant acceleration through the board, and emerges with a speed of 345 m/sm/s.

What is the average acceleration of the bullet through the board/?

to know more about  the equation of motion :

brainly.com/question/13269040

#SPJ4

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2 years ago
Help please! Will give brainly! and thanks.
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7 0
3 years ago
A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t
Yuliya22 [10]

Answer:

a) W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

b) W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

c) V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

d)  d_1 =0.183m or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

Part b

For this case first we can convert the spring constant to N/m like this:

2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}

And the work donde by the spring on this case is given by:

W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i

And if we solve for the initial velocity we got:

V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

-1/2mV^2_i = W_g + W_{spring}

And replacing we got:

-1/2mV^2_i =mg d_1 -1/2 k d^2_1

And we can put the terms like this:

\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0

If we multiply all the equation by 2 we got:

k d^2_1 -2 mg d_1 -m V^2_i =0

Now we can replace the values and we got:

200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0

200 d^2_1 -2.058 d_1 -6.3525=0

And solving the quadratic equation we got that the solution for d_1 =0.183m or 18.3 cm because the negative solution not make sense.

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Answer:

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6 0
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7 0
3 years ago
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