An 88 N forces pushes an object 15 m. how much work did the force do on the object

Acceleration due to gravity on the moon is 1.6m/s^2 or about 16% of the value of gg on Earth. If an astronaut on the moon threw

xxTIMURxx [149]

To solve this problem it is necessary to apply the concepts related to the conservation of Energy. Mathematically the conservation of kinetic energy must be paid in the increase of potential energy or vice versa. This expressed in algebraic terms is equivalent to

Kinetic Energy = Potential Energy

$\frac{1}{2}mv^2 = mgh$

Where

m = Mass

v = Velocity

g = Gravity

h = Height

As the mass is the same then we have to

$\frac{1}{2} v^2 = gh$

Rearrange to find v,

$v = \sqrt{2gh}$

Our values are given as

$g = 1.6m/s^2$

$h = 7.8m$

Therefore replacing we have

$v = \sqrt{2(1.6)(7.8)}$

$v = 4.99m/s$

Hence the velocity at the moon would be 4.99m/s

The only direct affectation is that concerning the Resistance or drag force generated by a fluid - such as air in the ground - that can diminish / sharpen the direct effects of gravity. Disregarding the resistance of the air, as we can see in the equation previously given, there should be no affectation because the speed depends on the gravity and height.

5 0

3 years ago

A hiker leaves her camp and walks 3.5 km in a direction of 55° south of west to the lake. after a short rest at the lake, she hi

3241004551 [841]

Let's choose east as positive x-direction and south as positive y-direction. We can resolve the two displacement along these two axes:

- Displacement 1 (3.5 km, $55^{\circ}$ south of west

$d_{1x}=-(3.5 km)( cos 55^{\circ})=-2.01 km$

$d_{1y}=(3.5 km)( sin 55^{\circ})=2.87 km$

- Displacement 2 (2.7 km, $16^{\circ}$ east of south

$d_{2x}=(2.7 km)( sin 16^{\circ})=0.74 km$

$d_{1y}=(2.7 km)( cos 16^{\circ})=2.60 km$

So, the total components on the two directions are

$d_x = -2.01 km+0.74 km=-1.27 km$

$d_y=2.87 km+2.60 km=5.47 km$

And the magnitude of the hiker's resultant displacement is

$d=\sqrt{(1.27 km)^2+(5.47 km)^2}=5.6 km$

8 0

3 years ago

Read 2 more answers

What does decelerate mean

fenix001 [56]

To slow down or reduce speed.

5 0

3 years ago

A 3,220 lb car enters an S-curve at A with a speed of 60 mi/hr with brakes applied to reduce the speed to 45 mi/hr at a uniform

Grace [21]

The magnitude of the total friction force exerted by the road on the tires at B will be 919.46 lb.

<h3>What is the friction force?</h3>

It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. Its unit is Newton (N).

Mathematically, it is defined as the product of the coefficient of friction and normal reaction.

The given data in the problem is;

The weight is,W= 3,220 lb

The speed is,u= 60 mi/hr

The reducing speed is,v= 45 mi/hr

The distance traveled is,d= 300 ft

The radius of curvature of the path of the car at B is,R= 600 ft.

1 mile = 5280 ft

From the Newtons' equation of motion;

$\rm v^2 = u^2 +2ad \\\\ \rm (45 \times \frac{5280}{3600} )^2 = (60 \times \frac{5280}{3600} )^2 +2a\times 300 \\\\$

The tangential accelerations are;

$\rm a_t = \frac{66^2 -88^2}{600} \\\\ \rm a_t = -5.65 ft/sec^2 \\\\$

The force is found as;

$\rm \sum F = ma \\\\ \frac{3220}{32.2} \times -5.65 \\\\ F_T= 565 \ lb$

The normal force is;

$\rm F_n = \frac{3220}{32.2} \times \frac{66^2}{600} \\\\ F_N =726 \ lb$

The net or the total friction force exerted by the road on the tires at B. is found as;

$\rm F = \sqrt{(565)^2+(726)^2} \\\ F = 919.946 \ lb$

Hence, the magnitude of the total friction force exerted by the road on the tires at B will be 919.46 lb.

To learn more about the friction force, refer to the link;

brainly.com/question/1714663

#SPJ1

6 0

1 year ago

PLEASE HELP<br><br> TOPIC - CONVECTION

STALIN [3.7K]

Answer:

SO ANSWER AGRE AS FOLLOWS:

Explanation:

A. Wind is the movement of air, caused by the uneven heating of the Earth by the sun and the Earth's own rotation. Winds range from light breezes to natural hazards such as hurricanes and tornadoes.

B.At night, the roles reverse. The air over the ocean is now warmer than the air over the land. The land loses heat quickly after the sun goes down and the air above it cools too. ... This causes the low surface pressure to shift to over the ocean during the night and the high surface pressure to move over the land.

<h2><em><u>HOPE THIS IS CORRECT </u></em></h2>

5 0

2 years ago