Ask question

Ask question

All categories

2 years ago

9

Find the Horizontal (x) vector for these forces.

1 answer:

Anestetic [448]2 years ago

6 0

Answer:

See the explanation below.

Explanation:

The force is a vector therefore we can decompose the force into components x & y. as we need the horizontal component of the force, we must use the cosine function of the angle.

$F_{1x}=30.8*cos(20)\\F_{1x}=28.94[N]\\F_{2x}=34.3*cos(20)\\\\F_{2x}= 32.23[N]$

You might be interested in

What happens when air from the balloon has exited the balloon? <br><br><br> How will motion change?

tiny-mole [99]

Answer:

When you release the opening of the balloon, gas quickly escapes to equalize the pressure inside with the air pressure outside of the balloon. The escaping air exerts a force on the balloon itself. ... That opposing force—called thrust, in this case—propels the rocket forward.

6 0

3 years ago

What is the current if a charge of 25 coulombs passes in 5 seconds?

Jobisdone [24]

Answer: 5 amps

Explanation: I = Q/t

8 0

3 years ago

Like magnetism, static electricity can attract and repel.<br> True<br> False

earnstyle [38]

The answer to the question is True

5 0

3 years ago

Read 2 more answers

Toon Train is traveling at the speed of 10 m/s at the top of a hill. Five seconds later it reaches the bottom of the hill and is

Naddika [18.5K]

Answer:

the rate of acceleration of the train is 4 m/s²

Explanation:

Given;

initial velocity of the train, u = 10 m/s

change in time of motion, dt = 5 s

final velocity of the train, v = 30 m/s

The rate of acceleration of the train is calculated as;

$a = \frac{dv}{dt} = \frac{v-u}{dt} = \frac{30-10}{5} = \frac{20}{5} = 4 \ m/s^2$

Therefore, the rate of acceleration of the train is 4 m/s²

5 0

3 years ago

Before colliding, the momentum of block A is -100 kg*m/, and block B is -150 kg*m/s. After, block A has a momentum -200 kg*m/s.

rjkz [21]

Answer:

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

Explanation:

Given

Before collision:

Momentum of block A = $p_{A1}$ = $-100\ kg\ ms^{-1}$

Momentum of block B = $p_{B1}$ = $-150\ kg\ ms^{-1}$

After collision:

Momentum of block A = $p_{A2}$ = $-200\ kg\ ms^{-1}$

Applying law of conservation of momentum to find momentum of block B after collision $p_{B2}$ .

$p_{A1}+p_{B1}=p_{A2}+p_{B2}$

Plugging in the given values and simplifying.

$-100-150=-200+p_{B2}$

$-250=-200+p_{B2}$

Adding 200 to both sides.

$200-250=-200+p_{B2}+200$

$-50=p_{B2}$

∴ $p_{B2}=-50\ kg\ ms^{-1}$

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

6 0

3 years ago