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Ne4ueva [31]
4 years ago
10

a bullet of mass 25 g is fired from a pistol at a velocity of 200 m/s. if mass of the pistol is 2.5 kg calculate its recoil velo

city
Physics
1 answer:
My name is Ann [436]4 years ago
6 0
Do this type of thing but change it a little bit. This is a conservation of momentum problem. The initial momentum of the system (gun + bullet) is zero, as neither is moving.
There fore, after the gun is fired, the total momentum must still be zero.
m
g
×
v
g
+
m
b
×
v
b
=
0
Inserting what we know:
(
0.025
)
(
210
)
+
(
0.91
)
v
b
=
o
v
b
=
−
(
0.025
)
(
210
)
0.91
=
−
5.8
m
s
The negative sign tells us that the gun travels in a direction opposite to that of the bullet.
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Steel ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.45 m.
Amiraneli [1.4K]

Answer:

Vb= 5.42 m/s

Vc= 5.33 m/s

a=-1125\ m/s^2

Δ L= 0.43 mm

Explanation:

a)  A to B

Initial velocity Va = 0 m/s

As we know that

V^2=U^2+2aS

V_b^2=2\times 9.81\times 1.5

Vb= 5.42 m/s

b)

C to D

Vd= 0 m/s

0^2=V_c^2-2\times 9.81\times 1.45

Vc= 5.33 m/s

c)

Here

Vb= 5.42 m/s

Vc= 5.33 m/s

We know that

Acceleration is the rate of change of velocity

a=\dfrac{V_c-V_b}{t}

a=\dfrac{5.33-5.42}{8\times 10^{-5}}

a=-1125\ m/s^2

d)

Now length compressed given as

V_c^2=V_b^2+2a\Delta L

5.33^2=5.42^2-2\times 1125\times  \Delta L

  Δ L= 0.43 mm

3 0
4 years ago
19
djyliett [7]

Answer:

B 23m/s

Explanation:

From the above information we get,

Initial velocity= 3m/s

Acceleration by gravity = 10 m/s2 (approx)

Time taken =2 seconds

v=u+at (First Equation Of Motion)

v= 3 + 10 × 2

v= 23 m/s

5 0
3 years ago
The established value for the speed of light in a vacuum is 299,792,458 m/s. What is the order-of-magnitude of this number?
iogann1982 [59]
The order of magnitude is 10⁸ .
6 0
3 years ago
Explain two scenarios where a large truck can have the same momentum as a small car.
KengaRu [80]

The momentum, p, of any object having mass m and the velocity v is

p=mv\cdots(i)

Let M_L and M_S be the masses of the large truck and the car respectively, and V_L and V_S be the velocities of the large truck and the car respectively.

So, by using equation (i),

the momentum of the large truck = M_LV_L

and the momentum of the small car = M_SV_S.

If the large truck has the same momentum as a small car, then the condition is

M_LV_L=M_SV_S\cdots(ii)

The equation (ii) can be rearranged as

\frac {M_L}{M_S}=\frac {V_S}{V_L} \; or \; \frac{M_L}{V_S}=\frac{M_S}{V_L}

So, the first scenario:

\frac {M_L}{M_S}=\frac {V_S}{V_L}

\Rhghtarrow M_L:M_S=V_S:V_L

So, to have the same momentum, the ratio of mass of truck to the mass of the car must be equal to the ratio of velocity of the car to the velocity of the truck.

The other scenario:

\frac{M_L}{V_S}=\frac{M_S}{V_L}

\Rhghtarrow M_L:V_S= M_S:V_L

So, to have the same momentum, the ratio of mass of truck to the velocity of the car must be equal to the ratio of mass of the car to the velocity of the truck.

5 0
3 years ago
A 10 kg bowling ball sits at the top of a 10 m hill and then slides down its icy hillside.
Nataly_w [17]
We need more informtion
5 0
4 years ago
Read 2 more answers
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