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Juliette [100K]
4 years ago
11

Q. 5 A bullet of 10 g strikes a sandbag at a speed of 103 ms-1 and gets embedded after travelling 5 cm. Calculate (i) the resist

ive force exerted by the sand on the bullet (ii) the time taken by the bullet to come to rest.
Physics
1 answer:
Xelga [282]4 years ago
3 0

Answer:

(i) the resistive force exerted by the sand on the bullet is - 1 x 10⁵ N

(ii) the time taken by the bullet to come to rest is 1 x 10⁻⁴ s

Explanation:

Given;

mass of bullet, m = 10 g = 0.01 kg

speed of the bullet, v = 10³ m/s

distance traveled by the bullet, d = 5 cm = 0.05 m

(i) the resistive force exerted by the sand on the bullet

the acceleration of the bullet is given by;

v² = u² + 2as

0 = (10³)² + a(2 x 0.05)

-0.1a = (10⁶)

-a = (10⁶) / 0.1

a = -1 x 10⁷ m/s²

The resistive force is given by;

F = ma

F = (0.01)(-1 x 10⁷)

F = - 1 x 10⁵ N

(ii) the time taken by the bullet to come to rest.

v = u + at

0 = 10³ + (-1 x 10⁷ t)

1 x 10⁷ t = 10³

t = 10³  / 10⁷

t = 1 x 10⁻⁴ s

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Answer:

The acorn hasn't hit the ground because it only falsl half of the branch distance from the ground

Explanation:

given information:

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   = 4.8 m

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   =

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3 years ago
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Myopia

Explanation:

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3 years ago
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Zina [86]

Answer:

Solar and nuclear power generate more than 99 percent of our civilization's energy. Every other important source of energy is a combination of these two. The majority of them are solar in nature. We discharge previously collected solar energy when we burn wood.

and

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Explanation:

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3 years ago
A cannon, positioned on a hill, shoots a cannonball horizontally at 23 m/s. The cannonball hits the stone wall 1.96 m below the
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Answer: 14. 49 m

Explanation:

We can solve this problem with the following equations:

x=V_{o} cos \theta t (1)

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Where:

x is the horizontal distance between the cannon and the ball

V_{o}=23 m/s is the cannonball initial velocity

\theta=0\° since the cannonball was shoot horizontally

t is the time

y=0 is the final height of the cannonball

y_{o}=1.96 m is the initial height of the cannonball

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Isolating t from (2):

t=\sqrt{-\frac{2(y-y_{o})}{g}} (3)

t=\sqrt{-\frac{2(0 m-1.96 m)}{9.8 m/s^{2}}} (4)

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Artyom0805 [142]

Answer:

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Explanation:

The first law of thermodynamics relates the heat transfer into or out of a system to the change of internal and the work done on the system, through the following equations.

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ΔU  is the change in internal energy

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Given;

ΔU = 155 J

W = 28.5 J

Q = ?

155 = Q - 28.5

Q = 155 + 28.5

Q = 183.5 J

Therefore, the heat transferred into the system is 183.5 J.

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