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nikdorinn [45]
4 years ago
13

The muzzle velocity of a 50.0g shell leaving a 3.00 kg rifle is 400m/s what is the recoil velocity of the rifle

Physics
1 answer:
serg [7]4 years ago
6 0

Here if we consider bullet and gun as a system then we can say that momentum of the system will remain conserved

so we will have

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

now we know that

m_1 = 50 g = 0.050 kg

m_2 = 3 kg

v_{1i} = v_{2i} = 0

v_{1f} = 400 m/s

now we will have

0.050(0) + 3(0) = 0.050(400) + 3(v)

20 + 3v = 0

v = - \frac{20}{3} = - 6.67 m/s

so gun will recoil with speed 6.67 m/s

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An airplane flies at 40 m/s at an altitude of 50 meters. The pilot drops a heavy package which falls to the ground. Where, appro
igor_vitrenko [27]

Answer:

128 m

Explanation:

From the question given above, the following data were obtained:

Horizontal velocity (u) = 40 m/s

Height (h) = 50 m

Acceleration due to gravity (g) = 9.8 m/s²

Horizontal distance (s) =?

Next, we shall determine the time taken for the package to get to the ground.

This can be obtained as follow:

Height (h) = 50 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

50 = ½ × 9.8 × t²

50 = 4.9 × t²

Divide both side by 4.9

t² = 50 / 4.9

t² = 10.2

Take the square root of both side

t = √10.2

t = 3.2 s

Finally, we shall determine where the package lands by calculating the horizontal distance travelled by the package after being dropped from the plane. This can be obtained as follow:

Horizontal velocity (u) = 40 m/s

Time (t) = 3.2 s

Horizontal distance (s) =?

s = ut

s = 40 × 3.2

s = 128 m

Therefore, the package will land at 128 m relative to the plane

6 0
3 years ago
A rectangular gasoline tank can hold 50.0 kg of gasoline when full. What is the depth of the tank if it is 0.500-m wide by 0.900
sweet-ann [11.9K]

Answer:

0.16 m

Explanation:

A rectangular gasoline tank can hold 50.0 kg of gasoline when full, and the density of gasoline is 6.8 × 10² kg/m³. We can find the volume occupied by the gasoline (volume of the tank).

50.0 kg × (1 m³/6.8 × 10² kg) = 0.074 m³

The volume of the rectangular tank is:

volume = width × length × depth

depth = volume / width × length

depth = 0.074 m³ / 0.500 m × 0.900 m

depth = 0.16 m

3 0
3 years ago
At one instant, the center of mass of a system of two particles is located on the x-axis at 2.0 cm and has a velocity of (5.0 m/
Nata [24]

Answer:

Explanation:

Given that,

At one instant,

Center of mass is at 2m

Xcm = 2m

And velocity =5•i m/s

One of the particle is at the origin

M1=? X1 =0

The other has a mass M2=0.1kg

And it is at rest at position X2= 8m

a. Center of mass is given as

Xcm = (M1•X1 + M2•X2) / (M1+M2)

2 = (M1×0 + 0.1×8) /(M1 + 0.1)

2 = (0+ 0.8) /(M1 + 0.1)

Cross multiply

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b. Total momentum, this is an inelastic collision and it momentum after collision is given as

P= (M1+M2)V

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Using conversation of momentum

Momentum before collision is equal to momentum after collision

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We are told that M2 is initially at rest, then, V2=0

So, P(before) = 0.3V1

We already got P(after) = 2 •i kgm/s in part b of the question

Then,

P(before) = P(after)

0.3V1 = 2 •i

V1 = 2/0.3 •i

V1 = 6 ⅔ •i m/s

V1 = 6.667 •i m/s

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egoroff_w [7]

the answer is radioactive processes within the Earth.

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