All categories

3 years ago

The velocity of a 48.0 g shell leaving a 2.95 kg rifle is 391. m/s. What is the recoil velocity of the rifle?

Physics

1 answer:

Monica [59]3 years ago

8 0

Hi there!

$\large\boxed{ -6.36m/s}$

Use the equation:

$v_{2} = -\frac{m_{1}}{m_{2}} v_{1}$

Where m2 and v2 deal with the larger object, and m1 and v1 with the smaller object. Plug in the given values:

v2 = ?

m1 = 0.048 kg (converted)

m2 = 2.95

v1 = 391

$v_{2} = -\frac{0.048}{2.95} *391$

$v_{2} = -6.36m/s$

You might be interested in

Which of the following is not a characteristic of the source of rivers and streams?

MrRa [10]

<span>The water is clearer, has higher oxygen levels, and freshwater fish such as trout and heterotrophs can be found. In the middle of the river, the width increases, so do the species of aquatic green plants and algae. Toward the mouth of the river, the water becomes murky from all the sediments that it has picked up upstream.<span>
</span></span>

7 0

3 years ago

A potato is shot out of cylinder at an angle of 17 degrees above the horizontal with an initial speed of 20 m/s. What is its max

madreJ [45]

Answer:

Maximum height, h = 1.74 meters

Explanation:

It is given that,

A potato is shot out of the cylinder. It is a case of projectile motion. The potato makes an angle of 17 degrees above the horizontal.

Initial speed with which the potato is shot out, u = 20 m/s

We have to find the maximum height of the potato. The maximum height of a projectile (h) is given by the following formula as :

$h=\dfrac{u^2sin^2\theta}{2g}$

Where

$\theta$ = angle between the projectile and the surface

g = acceleration due to gravity

$h=\dfrac{(20\ m/s)^2sin^2(17)}{2\times 9.8\ m/s^2}$

h = 1.74 m

or h = 1.74 meters

Hence, this is the required solution.

8 0

3 years ago

A light ray incident from medium 1 to medium 2, where n1>n2. When the incident angle exceed the critical angle ac, the refrac

vovikov84 [41]

Explanation:

(a)

Critical angle is the angle at the angle of refraction is 90°. After the critical angle, no refraction takes place.

Using Snell's law as:

$n_1\times {sin\theta_i}={n_2}\times{sin\theta_r}$

Where,

${\theta_i}$ is the angle of incidence

${\theta_r}$ is the angle of refraction = 90°

${n_2}$ is the refractive index of the refraction medium

${n_1}$ is the refractive index of the incidence medium

Thus,

$n_1\times {sin\ \theta_{critical}}={n_2}\times{sin\ 90^0}$

The formula for the calculation of critical angle is:

${sin\theta_{critical}}=\frac {n_2}{n_1}$

Where,

${\theta_{critical}}$ is the critical angle

(b)

No it cannot occur. It only occur when the light ray bends away from the normal which means that when it travels from denser to rarer medium.

7 0

3 years ago

Not sure if it went through last time. Please help asap!

Olin [163]

The equation for force is F=ma. Because we have the value of mass (0.42 kg) and the acceleration (14.8 m/s^2), simply plug them into the equation for force to get
$0.42 \times 14.8 = 6.22$
The answer is 6.22 N because newtons are the unit used to measure force.

8 0

3 years ago

Read 2 more answers

What’s the unit for velocity?

torisob [31]

Answer:

meter per second

Explanation:

It could be any other unit such as yard or feet, put it will be whatever measure per second or whatever time.

Examples

feet per second

miles per hour

4 0

3 years ago