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AURORKA [14]
3 years ago
10

A 140-kg box is being lowered at constant velocity. A rope is attached to the top of the box, and Jimmy is slightly supporting t

he box from the bottom. If Jimmy is applying 50 N of force to the box, how much force is the rope applying?
Physics
1 answer:
Zinaida [17]3 years ago
4 0

Answer:

the anwser is 90

Explanation:

because 140-50=90

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10 rectangular object's mass from greatest to least. You can choose books, sandwiches, phones, pictures - as long as the shape i
Greeley [361]

Answer: 1 is phone 2 is sandwich, Last is picture.

Explanation: I hoped That Helped !!

7 0
3 years ago
A rock is projected upward from the surface of the moon, at time t = 0.0 s, w a velocity of 30 m/s. The acceleration due to grav
Vinvika [58]
<h2>Answer: 277.777 m</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told that the rock was<u> projected upward from the surface</u>, we will only use the equations related to the Y axis.

In this sense, the movement equations in the Y axis are:

y-y_{o}=V_{o}.t+\frac{1}{2}g.t^{2}    (1)

V=V_{o}-g.t    (2)

Where:

y  is the rock's final position

y_{o}=0  is the rock's initial position

V_{o}=30\frac{m}{s} is the rock's initial velocity

V is the final velocity

t is the time the parabolic movement lasts

g=1.62\frac{m}{s^{2}}  is the acceleration due to gravity at the surface of the moon

As we know y_{o}=0 , equation (2) is rewritten as:

y=V_{o}.t+\frac{1}{2}g.t^{2}    (3)

On the other hand, the maximum height  is accomplished when V=0:

V=V_{o}-g.t=0    (4)

V_{o}-g.t=0    

V_{o}=g.t    (5)

Finding t:

t=\frac{V_{o}}{g}    (6)

Substituting (6) in (3):

y=V_{o}(\frac{V_{o}}{g})+\frac{1}{2}g(\frac{V_{o}}{g})^{2}    (7)

y_{max}=\frac{{V_{o}}^{2}}{2g}    (8)  Now we can calculate the maximum height of the rock

y_{max}=\frac{{(30m/s)}^{2}}{(2)(1.62m/s^{2})}   (9)

Finally:

y_{max}=277.777m  

4 0
3 years ago
How do you say scuba dive in spanish
Brrunno [24]

submarinismo is scuba diving but Bueco is dive

3 0
3 years ago
Read 2 more answers
An old grindstone, used for sharpening tools, is a solid cylindrical wheel that can rotate about its central axle with negligibl
krok68 [10]

(a) The moment of inertia of the wheel  is 78.2 kgm².

(b) The mass (in kg) of the wheel is 1,436.2 kg.

(c) The angular speed (in rad/s) of the wheel at the end of this time period is 3.376 rad/s.

<h3>Moment of inertia of the wheel</h3>

Apply principle of conservation of angular momentum;

Fr = Iα

where;

  • F is applied force
  • r is radius of the cylinder
  • α is angular acceleration
  • I is moment of inertia

I = Fr/α

I = (200 x 0.33) / (0.844)

I = 78.2 kgm²

<h3>Mass of the wheel</h3>

I = ¹/₂MR²

where;

  • M is mass of the solid cylinder
  • R is radius of the solid cylinder
  • I is moment of inertia of the solid cylinder

2I = MR²

M = 2I/R²

M = (2 x 78.2) / (0.33²)

M = 1,436.2 kg

<h3>Angular speed of the wheel after 4 seconds</h3>

ω = αt

ω = 0.844 x 4

ω = 3.376 rad/s

Thus, the moment of inertia of the wheel  is 78.2 kgm².

The mass (in kg) of the wheel is 1,436.2 kg.

The angular speed (in rad/s) of the wheel at the end of this time period is 3.376 rad/s.

Learn more about moment of inertia here: brainly.com/question/14839816

#SPJ1

7 0
1 year ago
I will Give Brainliest To whoever actually answers A 500 kg satellite experiences a gravitational force of 3000 N, while moving
snow_lady [41]

Answer:

9.7\times 10^{-4}\ rad/s

Explanation:

Given:

m=500 kg\\F=3000 N

Radius of earth , R=6371 \times 10^3\ m\\Angular speed =\omega\\We\  know\  that\ \\F= m\times \omega^{2} \times R\\\omega^{2}=\frac{F}{m*R} \\\\=\frac{3000}{500*6371 \times 10^3\ m}

=\frac{6}{6371 \times 10^3\ m}

=9.7\times 10^{-4}\ rad/s

7 0
3 years ago
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