Ask question

Ask question

All categories

3 years ago

14

The force exerted by expanding gases is what propels a shell out of a gun barrel. Suppose a force that has an average magnitude

F propels the shell so that it has a speed v at the instant it reaches the end of the barrel.
A) What magnitude force is required to double the speed of the shell in the same gun?

1 answer:

masya89 [10]3 years ago

8 0

Answer: 2F

Explanation:

You might be interested in

A car at the top of a ramp starts from rest and rolls to the bottom of the ramp, achieving a certain final speed. If you instead

zimovet [89]

Answer:

It must be 4 times high.

Explanation:

Assuming that the car can be treated as a point mass, and that the ramp is frictionless, the total mechanical energy must be conserved.
This means, that at any time, the following must be true:
ΔK (change in kinetic energy) = ΔU (change in gravitational potential energy)

⇒ $m*g*h = \frac{1}{2} * m*v^{2}$

Let's call v₁, to the final speed of the car, and h₁ to the height of the ramp.

So, at the bottom of the ramp, all the gravitational potential energy

must be equal to the kinetic energy of the car (Defining the bottom of

the ramp as our zero reference for the gravitational potential energy):

$m*g*h_{1} = \frac{1}{2} * m*v_{1} ^{2}$ (1)

Now, let's do v₂ = 2* v₁
Replacing in (1) we get:

$m*g*h_{2} = \frac{1}{2} * m*(2*v_{1}) ^{2}$ (2)

Dividing (2) by (1), and rearranging terms, we get:
h₂ = 4* h₁

8 0

2 years ago

A 5.0-kilogram sphere, starting from rest, falls freely 22 meters in 3.0 seconds near the surface of a planet. Compared to the a

Whitepunk [10]

Answer:

C) one-half as great

Explanation:

We can calculate the acceleration of gravity in that planet, using the following kinematic equation:

$\Delta x=v_0t+\frac{gt^2}{2}$

In this case, the sphere starts from rest, so $v_0=0$ . Replacing the given values and solving for g':

$g'=\frac{2\Delta x}{t^2}\\g'=\frac{2(22m)}{(3s)^2}\\g'=4.89\frac{m}{s^2}$

The acceleration due to gravity near Earth's surface is $g=9.8\frac{m}{s^2}$ . So, the acceleration due to gravity near the surface of the planet is approximately one-half of the acceleration due to gravity near Earth's surface.

5 0

2 years ago

A particle with mass 1.81×10−3 kg and a charge of 1.22×10−8 C has, at a given instant, a velocity v⃗ =(3.00×104m/s)j^. What are

slava [35]

Answer:

The magnitude and direction of the acceleration of the particle is $a= 0.3296\ \hat{k}\ m/s^2$

Explanation:

Given that,

Mass $m = 1.81\times10^{-3}\ kg$

Velocity $v = (3.00\times10^{4}\ m/s)j$

Charge $q = 1.22\times10^{-8}\ C$

Magnetic field $B= (1.63\hat{i}+0.980\hat{j})\ T$

We need to calculate the acceleration of the particle

Formula of the acceleration is defined as

$F = ma=q(v\times B)$

$a =\dfrac{q(v\times B)}{m}$

We need to calculate the value of $v\times B$

$v\times B=(3.00\times10^{4}\ m/s)j\times(1.63\hat{i}+0.980\hat{j})$

$v\times B=4.89\times10^{4}$

Now, put the all values into the acceleration 's formula

$a =\dfrac{1.22\times10^{-8}\times(-4.89\times10^{4}\hat{k})}{1.81\times10^{-3}}$

$a= -0.3296\ \hat{k}\ m/s^2$

Negative sign shows the opposite direction.

Hence, The magnitude and direction of the acceleration of the particle is $a= 0.3296\ \hat{k}\ m/s^2$

7 0

2 years ago

Read 2 more answers

Lasers can be constructed that produce an extremely high intensity electromagnetic wave for a brief time—called pulsed lasers. T

alex41 [277]

Answer:

30643 J

Explanation:

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

t = Time taken = 1 ns

c = Speed of light = $3\times 10^8\ m/s$

$E_0$ = Maximum electric field strength = $1.52\times 10^{11}\ V/m$

A = Area = $1\ mm^2$

Magnitude of magnetic field is given by

$B_0=\dfrac{E_0}{c}\\\Rightarrow B_0=\dfrac{1.52\times 10^{11}}{3\times 10^8}\\\Rightarrow B_0=506.67\ T$

Intensity is given by

$I=\dfrac{cB_0^2}{2\mu_0}\\\Rightarrow I=\dfrac{3\times 10^8\times 506.67^2}{2\times 4\pi \times 10^{-7}}\\\Rightarrow I=3.0643\times 10^{19}\ W/m^2$

Power, intensity and time have the relation

$E=IAt\\\Rightarrow E=3.0643\times 10^{19}\times 1\times 10^{-6}\times 1\times 10^{-9}\\\Rightarrow E=30643\ J$

The energy it delivers is 30643 J

4 0

2 years ago

List three reasons that rain predictions made by meteorologists are sometimes incorrect. Explain your response.

Agata [3.3K]

1. <span>the low pressure is moving slower than expected.
This make the meteorologist receive premature data which make them fail to interpret the data correctly and make the wronf prediction.
2. Sudden change in wind direction, which transfer the natural occurence into other region than where it initially predicted
3. We still haven't developed the methodology to 100% predict natural occurence</span>

5 0

3 years ago