How far away are the Stars?

The free-body diagram of a crate is shown. A free body diagram with 4 force vectors. The first vector is pointing downward, labe

satela [25.4K]

The net force acting on the crate is determined as 176 N to the left.

<h3>Net force acting on the crate</h3>

The net force acting on the crate is calculated as follows;

∑F = F1 + F2 + F3 + F4

F(net) = -440y + 176x + 440y - 352x

F(net) = -176 x

The resultant force is pointing in negative x direction.

Thus, the net force acting on the crate is determined as 176 N to the left.

Learn more about net force here: brainly.com/question/14361879

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8 0

2 years ago

An automobile starter motor has an equivalent resistance of 0.0500Ω and is supplied by a 12.0-V battery with a 0.0100-Ω internal

alexandr1967 [171]

Answer

given,

resistance = 0.05 Ω

internal resistance of battery = 0.01 Ω

electromotive force = 12 V

a) ohm's law

V = IR

and volage

$V = \epsilon - Ir$

now,

$IR = \epsilon - Ir$

$I(R+r) = \epsilon$

$I= \dfrac{\epsilon}{R+r}$

inserting the values

$I= \dfrac{12}{0.05+0.01}$

I = 200 A

b) Voltage

V = I R

V = 200 x 0.05

V = 10 V

c) Power

P = I V

P = 200 x 10 = 2000 W

d) total resistance = 0.05 + 0.09 = 0.14 Ω

$I= \dfrac{\epsilon}{R+r}$

$I= \dfrac{12}{0.14+0.01}$

I = 80 A

V = 80 x 0.05 = 4 V

P = 4 x 80 = 320 W

6 0

3 years ago

Read 2 more answers

a 70 kg skydiver opens her parachute. The force due to air resistance is now 1200 N. what is the acceleration of the skydiver

Natasha_Volkova [10]

Around 50 I think so

7 0

3 years ago

You are on a school bus driving north at 20 m/s. You walk toward the back of the bus with a velocity of 6 m/s. What is your velo

Ganezh [65]

I answered the question but it got deleted?? why?

3 0

3 years ago

A hollow cylinder that is rolling without slipping is given a velocity of 5.0 m/s and rolls up an incline to a vertical height o

inysia [295]

Answer:

The hollow cylinder rolled up the inclined plane by 1.91 m

Explanation:

From the principle of conservation of mechanical energy, total kinetic energy = total potential energy

$M.E_T = \frac{1}{2}mv^2 + \frac{1}{2} I \omega^2 + mgh$

The total energy at the bottom of the inclined plane = total energy at the top of the inclined plane.

$\frac{1}{2}mv_i^2 + \frac{1}{2} I \omega_i^2 + mg(0) = \frac{1}{2}mv_f^2 + \frac{1}{2} I \omega_f^2 + mgh$

moment of inertia, I, of a hollow cylinder = ¹/₂mr²

substitute for I in the equation above;

$\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}mr^2 \omega_i^2) = \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}mr^2 \omega_f^2) + mgh\\\\ but \ v = r \omega\\\\\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}m v_i^2 ) = \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}m v_f^2) + mgh\\\\\frac{1}{2}mv_i^2 +\frac{1}{4}mv_i^2 = \frac{1}{2}mv_f^2 +\frac{1}{4}mv_f^2 +mgh\\\\\frac{3}{4}mv_i^2 = \frac{3}{4}mv_f^2 +mgh\\\\mgh = \frac{3}{4}mv_i^2 - \frac{3}{4}mv_f^2\\\\gh = \frac{3}{4}v_i^2 - \frac{3}{4}v_f^2\\\\$

$h = \frac{3}{4g}(v_1^2 -v_f^2)$

given;

v₁ = 5.0 m/s

vf = 0

g = 9.8 m/s²

$h = \frac{3}{4g}(v_1^2 -v_f^2) =\frac{3}{4*9.8}(5^2 -0) = 1.91 \ m$

Therefore, the hollow cylinder rolled up the inclined plane by 1.91 m

5 0

3 years ago