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Delicious77 [7]
2 years ago
9

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begin

s to move. Then the force (in newtons) exerted by the expanding gas on the bullet is 14,000 + 10,000x − 26,000x2, where x is in meters.
Physics
1 answer:
Mashutka [201]2 years ago
5 0

The given question is incomplete. The complete question is as follows.

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begins to move. Then the force (in newtons) exerted by the expanding gas on the bullet is 14,000 + 10,000x − 26,000x^{2}, where x is in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.

Explanation:

We will calculate the work done as follows.

     W = \int_{0}^{0.54} F dx

         = \int_{0}^{0.54} (14,000 + 10,000x - 26,000x^{2}) dx

         = [14000x + 5000x^{2} - 8666.7x^{3}]^{0.54}_{0}

         = 7560 + 1458 - 1364.69

         = 7653.31 J

or,      = 7.65 kJ       (as 1 kJ = 1000 J)

Thus, we can conclude that the work done by the gas on the bullet as the bullet travels the length of the barrel is 7.65 kJ.

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As the external magnetic field decreases, an induced current flows in the coil. The direction of the induced magnetic field would be pointing to the screen. The flux through the coil is said to decrease. In order to counter this change, the coil would generate or produce a magnetic field that is induced that would be pointing to the same direction as the external field that is flowing which is into the the screen. This is according to Lenz's law or the right hand rule. It states that an induced current in a circuit that is due to the change or motion in   magnetic field should be directed opposing to the change in the flux.
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"if a stream flow measures 12 meters in 60 seconds, what is the stream's average rate of flow?"
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<span>Discharge is the volume of water moving down a stream or river per unit of time, commonly expressed in cubic feet per second or gallons per day. In general, river discharge is computed by multiplying the area of water in a channel cross section by the average velocity of the water in that cross section: discharge = area * velocity. In this case, the answer is 0.2 m/s.</span>
7 0
2 years ago
What is the absolute value of the horizontal force that each athlete exerts against the ground?
alexandr402 [8]
Refer to the diagram shown below.

When an athlete is in motion, he/she exerts a vertical force (the person's weight, W) on the ground. The ground exerts an equal and opposite force, N, the normal reaction on the athlete, so that W = N.

At the same time, the ground exerts a horizontal force, F, o n the athlete so that he/she does not slip.
The magnitude of the horizontal force is
F = μN = μW
where μ = the dynamic coefficient of friction.

Answer:  
The horizontal force is μW,
where
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6 0
2 years ago
Find the kinetic of a 0.1 kilogram toy truck moving at a speed of 1.1 meters per second
omeli [17]
KE = kinetic energy
PE = potential energy
GPE = gravitational potential energy
energy is always measured in Joules (J)

KE = (0.5) times the mass times the velocity^2
square the velocity first

Mass = (KE x 2) / v^2
square the velocity first, then double the kinetic energy, then divide
mass is measured in kg

velocity = sqrt(KE x 2 / m)
velocity can be called speed, like in the the second problem
remember to find the square root after you double the KE and divide that by the mass.
for example: if after you doubled KE and divided it by the mass you got sqrt(20), the answer would be about 4.47

GPE = mass x gravitational pull (about 9.8 m/s^2 on earth) x height

height = (PE) / (g x m)
do g x m first

So for question 1:
KE = (0.5)0.1 x 1.1^2
always square the velocity first:
KE = (0.5)0.1 x 1.21
KE = 0.0605
so if you rounded it to the nearest hundreths you would get KE = 0.06 J
don't forget the unit of energy is in Joules
5 0
2 years ago
A 2530-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force so that
gavmur [86]

Answer:

A = 1.4 m/s²

B = -0.10493 m/s³

a = 1.29507 m/s²

T = 28095.8271 N

T = 1.13198 W

Explanation:

t = Time taken

g = Acceleration due to gravity = 9.81 m/s²

The equation

v(t)=At+Bt^2

Differentiating with respect to time

\frac{dv}{dt}=\frac{d(At+Bt^2)}{dt}\\\Rightarrow 1.4=A+2Bt

At t = 0

1.4=A

Hence, A = 1.4 m/s²

B=\frac{v-At}{t^2}\\\Rightarrow B=\frac{2.18-1.4\times 1.8}{1.8^2}\\\Rightarrow B=-0.10493\ m/s^3

B = -0.10493 m/s³

At t = 5 seconds

a=1.4+2\times -0.010493\times 5=1.29507\ m/s^2

a = 1.29507 m/s²

T=m(a+g)\\\Rightarrow T=2530(1.29507+9.81)\\\Rightarrow T=28095.8271\ N

T = 28095.8271 N

Weight of rocket

W=2530\times 9.81=24819.9\ N

\frac{T}{W}=\frac{28095.8271}{24819.9}\\\Rightarrow \frac{T}{W}=1.13198\\\Rightarrow T=1.13198W

T = 1.13198 W

3 0
2 years ago
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