True or false The expression that relates emf, potential, current, and resistance is Ohm's law.

What are four factors that can affect the amount of friction between two objects

umka2103 [35]

As the Force of friction is equal to $F _{f} =$ µ $F_{n}$ by using the Law of Action Reaction; as the force normal is the conterforce of gravity

$F _{f} =$ µ $F_{n}$
$F_{f}=$ µ $F_{n}$

Therefore, by looking at the equation we can infer that

The force of friction is directly proportional to the mass, gravitational constant, and the co-efficent of friction

Because the gravitational constant is dependant on gravitation, a planet's mass and radius also affect the force of friction

But DO NOTE:
That the co-effecient of friction is only applicable between two rubbing surfaces and unaffected by gravitational constants.

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7 0

4 years ago

Two cars have the same weight, but one of the cars has an engine that provides twice the power of the other. which car can make

Korvikt [17]

The car with the engine that has the most power

4 0

4 years ago

If I wave my arm does it cause a convection wave

lana [24]

Answer:

no mam

Explanation:

the answer to the question is

7 0

3 years ago

A roller-coaster car has a mass of 1080 kg when fully loaded with passengers. As the car passes over the top of a circular hill

Dmitriy789 [7]

Answer:

(a): The normal force on the car from the track when the car's speed is v= 7.6 m/s is FN= -6696 N.

(b): The normal force on the car from the track when the car's speed is v= 17 m/s is FN= 8912.7 N.

Explanation:

m= 1080 kg

r= 16m

v1= 7.6 m/s

v2= 17 m/s

g= 9.81 m/s²

v1= w1*r

w1= v1/r

w1= 0.475 rad/s

ac1= w1² * r

ac1= 3.61 m/s²

FN= m * (ac1 - g)

FN= -6696 N (a)

-----------------------------------------------------

v2= w2*r

w2= v2/r

w2= 1.06 rad/s

ac2= w2² * r

ac2= 18.06 m/s²

FN= m * (ac2 - g)

FN= 8912.7 N (b)

4 0

3 years ago

A 4 kg block is launched up a 30° ramp with an initial speed of 5 m/s. The coefficient of kinetic friction between the block and

zlopas [31]

Answer:

The speed of the block when it has returned to the bottom of the ramp is 6.56 m/s.

Explanation:

Given;

mass of block, m = 4 kg

coefficient of kinetic friction, μk = 0.25

angle of inclination, θ = 30°

initial speed of the block, u = 5 m/s

From Newton's second law of motion;

F = ma

a = F/m

Net horizontal force;

∑F = mgsinθ + μkmgcosθ

$a = \frac{F_{NET}}{m} = \frac{mgsin \theta + \mu_kmgcos \theta}{m} \\\\a = gsin \theta + \mu_kgcos \theta\\\\a = 9.8sin (30) + 0.25*9.8cos(30)\\\\a = 4.9 + 2.1217\\\\a = 7.022 \ m/s^2$

At the top of the ramp, energy is conserved;

Kinetic energy = potential energy

¹/₂mv² = mgh

¹/₂ v² = gh

¹/₂ x 5² = 9.8h

12.5 = 9.8h

h = 12.5/9.8

h = 1.28 m

Height of the ramp is 1.28 m

Now, calculate the speed of the block (in m/s) when it has returned to the bottom of the ramp;

v² = u² + 2ah

v² = 5² + 2 x 7.022 x 1.28

v² = 25 + 17.976

v² = 42.976

v = √42.976

v = 6.56 m/s

Therefore, the speed of the block when it has returned to the bottom of the ramp is 6.56 m/s.

4 0

4 years ago