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4 years ago

A rock thrown to the right with an initial horizontal velocity of 4 m/s lands after 5

Physics

1 answer:

pickupchik [31]4 years ago

6 0

Answer:

20m

Explanation:

Using the equation s=d/t and re arranging it to solve for distance, d=St, where d=distance, s=speed, ant t=time, we can plug in known variables.

d=4*5

d=20m

there is a distance of 20m traveled

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If a 50-kg Person is running at a rate of 2m/s, the person's momentum is _____ kg• m/s.

sergiy2304 [10]

Answer:

100

Explanation:

Use equation p=mv

p=(50kg)(2m/s)= 100

8 0

3 years ago

Question: A 12 volt car battery is connected to a 3 ohm brake light. What is the current carrying energy to the lights?

Natali [406]

Answer:

4 A

Explanation:

V = IR, where V=voltage, I=current, R=resistance. This is Ohm's Law. (remember that for units V = volts, Ω = ohms, A = amperes.)

V = IR

12 V = I * 3 Ω

12/3 = I

8 0

4 years ago

Two astronauts, each having a mass of 74 kg, are connected by a 8.53 m rope of negligible mass. They are isolated in space, orbi

AveGali [126]

Answer:

a) 2575 kgm²/s

b) 1.23 kJ

c) 0.478 rad/s

Explanation:

Given

Mass of astronauts, m = 74 kg

Length of rope, l = 8.53 m

Speed of orbit, v = 4.08 m/s

L = m1v1.x1(i) + m2v2.x2(i) = 2mv(d/2)

Thus, L = 2.m.v.(d/2)

L = 2 * 74 * 4.08 * (8.53/2)

L = 2 * 74 * 4.08 * 4.265

L = 2575.38 kgm²/s

Rotational Energy of the system

K(i) = 1/2m1v1(i)² + 1/2m2v2(i)²

K(i) = 2(1/2) * 74 * 4.08²

K(i) = 74 * 16.6464

K(i) = 1231.83 J = 1.23 kJ

Angular momentum is conserved, thus, angular velocity, w = v/r

w = 4.08 / 8.53

w = 0.478 rad/s

7 0

3 years ago

. True or False: A bystander intervention is the term used to define when an individual actively moves in and changes the outcom

notsponge [240]

Answer:

true

Explanation:

8 0

2 years ago

A rotating space station is said to create "artificial gravity" - a loosely-defined term used for an acceleration that would be

FrozenT [24]

Answer:

$\omega=0.31\frac{rad}{s}$

Explanation:

The artificial gravity generated by the rotating space station is the same centripetal acceleration due to the rotational motion of the station, which is given by:

$a_c=\frac{v^2}{r}(1)$

Here, r is the radius and v is the tangential speed, which is given by:

$v=\omega r(2)$

Here $\omega$ is the angular velocity, we replace (2) in (1):

$a_c=\frac{(\omega r)^2}{r}\\\\a_c=\omega^2r$

Recall that $r=\frac{d}{2}=\frac{200m}{2}=100m$ .

Solving for $\omega$ :

$\omega=\sqrt{\frac{a_c}{r}}\\\omega=\sqrt{\frac{9.8\frac{m}{s^2}}{100m}}\\\omega=0.31\frac{rad}{s}$

3 0

4 years ago