Please someone help me lol

A force of 100N moves a body on a horizontal frictionless surface when......

zalisa [80]

Answer:

I. Friction force exerted on the body is less than 100N

Explanation:

For a body to be static, the moving force must be equal to the frictional force. Since the frictional force is a force of opposition. It tends to oppose the moving force acting on an object.

Hence if the moving force is greater than the force of friction, the Force of fiction will not be able to overcome the moving hence the body will tend to move.

Therefore, for a body to move, Fm > Ff or Ff < Ff

Fm is the moving force

Ff is the force of friction

Given

Fm = 100N

For the 100N body to move the frictional force must be less than 100N

4 0

3 years ago

Two manned satellites approaching one another at a relative speed of 0.550 m/s intend to dock. The first has a mass of 2.50 ✕ 10

NNADVOKAT [17]

Answer: Their final relative velocity is -0.412 m/s.

Explanation:

According to the law of conservation,

$m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v$

Putting the given values into the above formula as follows.

$m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v$

$2.50 \times 10^{3} kg \times 0 m/s + 7.50 \times 10^{3} kg \times -0.550 m/s = (2.50 \times 10^{3} kg + 7.50 \times 10^{3} kg)v$

$-4.12 \times 10^{3} kg m/s = (10^{4} kg) v$

v = $\frac{-4.12 \times 10^{3} kg m/s}{10^{4} kg}$

= -0.412 m/s

Thus, we can conclude that their final relative velocity is -0.412 m/s.

8 0

3 years ago

The stopping distances associated with slower speeds approximate the forward visibilities provided by low beam lights. However,

aleksandr82 [10.1K]

Answer:

a. True

Explanation:

Illumination distance is the distance, up to which the light of the vehicle can reach. Hence, it is a maximum distance from the, that driver can see.

Stopping distance is the minimum distance required by the car to stop after brakes are applied.

So, in order to avoid any accident the illumination distance must be greater than the stopping distance. So, the driver can stop the vehicle in time, when he sees something in front of it.

Since, the stopping distance in this case is two or three times longer than illumination distance. Therefore, low beam light does not provide enough visibility in high speed driving situations.

Hence, the correct option is:

a. True

5 0

3 years ago

You are an engineer helping to design a roller coaster that carries passengers down a steep track and around a vertical loop. Th

vova2212 [387]

Answer:

h >5/2r

Explanation:

This problem involves the application of the concepts of force and the work-energy theorem.

The roller coaster undergoes circular motion when going round the loop. For the rider to stay in contact with the cart at all times, the roller coaster must be moving with a minimum velocity v such that at the top the rider is in a uniform circular motion and does not fall out of the cart. The rider moves around the circle with an acceleration a = v²/r. Where r = radius of the circle.

Vertically two forces are acting on the rider, the weight and normal force of the cart on the rider. The normal force and weight are acting downwards at the top. For the rider not to fall out of the cart at the top, the normal force on the rider must be zero. This brings in a design requirement for the roller coaster to move at a minimum speed such that the cart exerts no force on the rider. This speed occurs when the normal force acting on the rider is zero (only the weight of the rider is acting on the rider)

So from newton's second law of motion,

W – N = mv²/r

N = normal force = 0

W = mg

mg = ma = mv²/r

mg = mv²/r

v²= rg

v = √(rg)

The roller coaster starts from height h. Its potential energy changes as it travels on its course. The potential energy decreases from a value mgh at the height h to mg×2r at the top of the loop. No other force is acting on the roller coaster except the force of gravity which is a conservative force so, energy is conserved. Because energy is conserved the total change in the potential energy of the rider must be at least equal to or greater than the kinetic energy of the rider at the top of the loop

So

ΔPE = ΔKE = 1/2mv²

The height at the roller coaster starts is usually higher than the top of the loop by design. So

ΔPE =mgh - mg×2r = mg(h – 2r)

2r is the vertical distance from the base of the loop to the top of the loop, basically the diameter of the loop.

In order for the roller coaster to move smoothly and not come to a halt at the top of the loop, the ΔPE must be greater than the ΔKE at the top.

So ΔPE > ΔKE at the top. The extra energy moves the rider the loop from the top.

ΔPE > ΔKE

mg(h–2r) > 1/2mv²

g(h–2r) > 1/2(√(rg))²

g(h–2r) > 1/2×rg

h–2r > 1/2×r

h > 2r + 1/2r

h > 5/2r

5 0

3 years ago

Read 2 more answers

Correct Number Formats Numbers may be entered in several formats - including scientific notation and numerical expressions.

Illusion [34]

Answer:

a. Valid

b. Invalid

c. Invalid

d. Invalid

e. Valid

f. Valid

g. Invalid

Explanation:

a.

Since, it follows all the four rules,

Therefore, this number is Valid

b.

It uses upper case E, instead of lower case e.

Therefore, it is Invalid

c.

It violates third rule for including unit.

Therefore, it is Invalid

d.

It violates the first rule. As, there is a space between 6 and e. While in the example given in question, there is no gap.

Therefore, it is Invalid

e.

Since, it follows all the four rules,

Therefore, this number is Valid

f.

Since, it follows all the four rules,

Therefore, this number is Valid

g.

It violates third rule for including unit.

Therefore, it is Invalid

6 0

3 years ago