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liberstina [14]
4 years ago
6

A car is traveling on a level road at a constant rate of speed. What additional force is necessary to bring the car into equilib

rium? zero. greater than the normal force times the coefficient of static friction. the normal force times the coefficient of kinetic friction. equal to the normal force times the coefficient of static friction.​
Physics
1 answer:
iragen [17]4 years ago
5 0

Answer: Zero.

Explanation:

By the first Newton's law, we know that:

every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force.

Now, we know that the car is moving with constant speed, then there is no net force acting on the car, which means that the car is already in equilibrium.

Then if we add one force to the situation, the car will not be anymore in equilibrium.

The correct option is zero.

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What is the organ(s) that detect taste
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Your taste buds that are located on your tongue
6 0
4 years ago
A proton experiences a force of 4 Newton when it enters perpendicular to the direction of the magnetic field with a speed 100 m/
____ [38]

Answer:

F = 0 N

Explanation:

Force on a moving charge in constant magnetic field is given by the formula

F = q(\vec v\times \vec B)

so here it depends on the speed of charge, magnetic field and the angle between velocity of charge and the magnetic field

here when charge is moving with speed 100 m/s in a given magnetic field then the force on the charge is given as

F = 4 N

now when charge is moving parallel to the magnetic field with different speed then in that case

\vec v \time \vec B = 0

so here we have

F = 0

7 0
3 years ago
Hi,please i need help with this one in physics.hurry and correct i need it.<br>Topic:Equilibrium.​
Rudik [331]

The half-meter rule (easy math) is 0.5 meters or 50 centimeters since a meter is 1 meters long, which is equivalent to 100 centimeters. Therefore, we shall apply the 50 cm rule.

A 50 cm rule's center of mass is now 25 cm away.

Additionally, according to the data, the object is pivoted at 15 cm, while the 40 g object is hung at 2 cm from the rule's beginning. Using a straightforward formula, we can compare the two situations: the distance from the pivot to the center of the mass times the mass of the 40 g object divided by 2 cm must equal the distance from the pivot to the center of the mass times mass of the 10 x g object

The result of the straightforward computation must be 52g.

Most simplified version:

the center of mass of the rule is at the 25 cm mark

⇒ 40 g * (15 cm - 2 cm)

⇒ = M * (25 cm - 15 cm)

#SPJ2

5 0
2 years ago
Question 4 of 10 (1 point) Jump to Question: Choose the word that best completes this sentence. A personal fall arrest system is
lisabon 2012 [21]
B
is the most logical answer to me
6 0
3 years ago
Read 2 more answers
A student stretches a spring, attaches a 1.20 kg mass to it, and releases the mass from rest on a frictionless surface. The resu
Sphinxa [80]

Answer:

the speed of the mass when it is halfway to the equilibrium position is 1.26 m/s

Explanation:

Given that;

mass of the object m = 1.20 kg

period of oscillation = 0.750 s

Amplitude ( A/x) = 15.0 cm = 0.15 m

now;

a) Determine the oscillation frequency;

oscillation frequency f = 1/T

we substitute

f = 1 / 0.750 s

f = 1.33 Hz

Therefore, the oscillation frequency is 1.33 Hz

b) Determine the spring constant;

we solve for spring constant from the following expression;

T = 2π√(m/k)

k = 4π²m / T²

so we substitute

k = (4π² × 1.20) / (0.750)²

k = 47.3741 / 0.5625

k =  84.22 N/m

Therefore, the spring constant is 84.22 N/m

c) determine the speed of the mass when it is halfway to the equilibrium position

So, at equilibrium, the energy is equal to K.E

such that;

1/2mv² = 1/2kx²

mv² = kx²

v² = kx² / m

v = √( kx²/m)

we substitute

v = √( 84.22×(0.15 m)²/ 1.2 )

v = √( 1.89495 / 1.2 )

v = √ 1.579125

v = 1.26 m/s  

Therefore, the speed of the mass when it is halfway to the equilibrium position is 1.26 m/s

3 0
3 years ago
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