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3 years ago

Before a rifle is fired, the linear momentum of the bullet-rifle system is zero.

Physics

1 answer:

marusya05 [52]3 years ago

3 0

Hi there!

II. Linear momentum of the system is zero.

This is an example of a RECOIL collision. With the Law of Conservation of Momentum, momentum remains constant before and after the collision.

Thus, the total momentum would also be equivalent to zero after the collision.

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Jayne lifts the barbell 120 cm upwards. She has a mass of 60kg. How much work does she do?

Rasek [7]

Work = Force x displacement
Force = mass x gravitational acceleration
Force = 60 x 9.81 = 588.6N
Work = 588.6 x 0.12m =70.632 J

3 0

3 years ago

You are a member of an alpine rescue team and must get a box of supplies, with mass 3 kg , up an incline of constant slope angle

denis23 [38]

Answer:

v = 9.04 m / s

Explanation:

For this exercise we can use the relation that the work of the non-conservative force (friction) is equal to the variation of the mechanical energy of the system.

W = Em_f - Em₀ (1)

Starting point. Lower slope

Em₀ = K = ½ m v²

highest point. Where is the skier at a height h

Em_f = U = m g h

The work of rubbing

W = -fr x

the negative sign is because the friction force opposes the movement.

Let's set a reference system where the x axis is parallel to the slope and the y axis is perpendicular

let's use trigonometry to break down the weight

cos θ = W_y / W

sin θ = Wₓ / W

W_y = W cos θ

Wₓ = W sin θ

Y axis

N - Wₓ = 0

N = mg sin θ

X axis

fr = m a

the friction force has the expression

fr = μ N

fr = μ mg sin θ

we look for the job

W = - μ mg sin θ x

where x is the distance along the slope

we substitute in 1

-μ mg sin θ x = mg h - ½ m v²

let's use trigonometry to find the distance x

tan 30 = h / x

x = h / tan 30

we substitute

- $\mu \ mg \ sin \theta \ \frac{h}{tan 30} \ x$ = m gh - ½ m v²

we use

tan 30 = sin30 / cos30

v² = 2g h + 2 μ g h cos 30

v = $\sqrt{ 2gh \ (1+ cos 30}$

let's calculate

v = $\sqrt{ 2 \ 9.8 \ 4 \ (1 + 0.05 \ cos \ 30)}$

v = 9.04 m / s

4 0

3 years ago

What are the 2 types of electricity

tamaranim1 [39]

The two are AC and DC.

4 0

3 years ago

A liquid of density 1270 kg/m3 flows steadily through a pipe of varying diameter and height. At Location 1 along the pipe, the f

FinnZ [79.3K]

Answer:

${P_2}-P_1=49.99\ KPa$

Explanation:

$v_1=9.43\ m/s$

$d_1=11.7\ cm/s$

$d_2=17.5\ cm/s$

From continuity equation

$A_1v_1=A_2v_2$

$v_1d_1^2=v_2d_2^2$

$v_2=\dfrac{9.43\times 11.7^2}{17.5^2}$

$v_2=4.21\ m/s$

$\dfrac{P_1}{\rho g}+\dfrac{v_1^2}{2g}+y_1=\dfrac{P_2}{\rho g}+\dfrac{v_2^2}{2g}+y_2$

${P_1}+\rho\dfrac{v_1^2}{2}+\rho y_1g={P_2}+\rho\dfrac{v_2^2}{2}+\rho y_2g$

$\rho\dfrac{v_1^2}{2}+\rho y_1g -\rho\dfrac{v_2^2}{2}-\rho y_2g={P_2}-P_1$

$1270\times \dfrac{9.43^2}{2}+1270\times 0\times 10 -1270\times\dfrac{4.21^2}{2}-1270\times 0.175\times 10={P_2}-P_1$

${P_2}-P_1=49.99\ KPa$

4 0

3 years ago

A cord is wrapped around the rim of a solid uniform wheel 0.280m in radius and of mass 8.80kg. A steady horizontal pull of 32N t

Grace [21]

Answer: $25.97 rad/s^2$

Explanation:

Given

radius of wheel $r=0.28 m$

mass of wheel $m=8.80 kg$

Force $F=32 N$

Moment of Inertia of solid wheel $I=\frac{mr^2}{2}$

$I=\frac{8.8\times 0.28^2}{2}$

$I=0.344 kg-m^2$

Torque is given by

$\tau =F\times r=I\times \alpha$

$32\times 0.28=0.344\times \alpha$

$\alpha =25.97 rad/s^2$

Force on the axle is 32 N since there is no linear acceleration of the system

using Third law F=32 N

Torque of the axle applied to the wheel is zero because force of axle imparted at the center of axle

3 0

3 years ago