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3 years ago

Which of the following are non-contact forces: friction, electrostatic force, magnetic force, gravity?

Physics

2 answers:

oee [108]3 years ago

4 0

Magnetic force and Gravity are non-contact forces among friction, electrostatic force, magnetic force and gravity.

CaHeK987 [17]3 years ago

4 0

Gravity and Magnetic force are non-contact forces

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Describe the difference between potencial and kinetic energy

avanturin [10]

Potential energy is energy stored in an object. kinetic energy is energy of motion

8 0

3 years ago

Two cars collide at an intersection. One car has a mass of 1600 kg and is

pickupchik [31]

The combined momentum is 4000 kg m/s south

Explanation:

The total combined momentum of the two cars is given by the vector addition of the momenta of the two cars.

For this problem, we choose north as positive direction and south as negative direction.

The momentum of the first car travelling north is given by:

$p_1 = m_1 v_1$

where

$m_1 = 1600 kg$ is the mass of the car

$v_1 = +8 m/s$ is its velocity

Substituting,

$p_1 = (1600)(8)=+12800 kg m/s$

The momentum of the second car travelling south is given by:

$p_2 = m_2 v_2$

where

$m_2 = 1400 kg$ is the mass of the car

$v_2= -12 m/s$ is its velocity (negative because the car travels south)

Substituting,

$p_2 = (1400)(-12)=-16800 kg m/s$

And therefore, the combined momentum is

$p=p_1 + p_2 = +12800 + (-16800)=-4000 kg m/s$

where the negative sign means the direction of the total momentum is south.

Learn more about momentum:

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4 0

3 years ago

Which of the following is an example of rolling friction?

il63 [147K]

Answer:

O bike tires on the road as you ride

Explanation:

is the rolling friction

7 0

3 years ago

Read 2 more answers

A mechanic needs to replace the motor for a merry-go-round. The merry-go-round should accelerate from rest to 1.5 rad/s in 6.0s

pashok25 [27]

Answer:

109656.25 Nm

Explanation:

$\omega_f$ = Final angular velocity = 1.5 rad/s

$\omega_i$ = Initial angular velocity = 0

$\alpha$ = Angular acceleration

t = Time taken = 6 s

m = Mass of disk = 29000 kg

r = Radius = 5.5 m

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{1.5-0}{6}\\\Rightarrow \alpha=0.25\ rad/s^2$

Torque is given by

$\tau=I\alpha\\\Rightarrow \tau=\dfrac{1}{2}mr^2\alpha\\\Rightarrow \tau=\dfrac{1}{2}29000\times 5.5^2\times 0.25\\\Rightarrow \tau=109656.25\ Nm$

The torque specifications must be 109656.25 Nm

5 0

3 years ago

Block B has mass 6.00 kg and sits at rest on a horizontal, frictionless surface. Block A has mass 2.50 kg and sits at rest on to

zzz [600]

Answer:

Explanation:

Block A sits on block B and force is applied on block A . Block A will experience two forces 1) force P and 2 ) friction force in opposite direction of motion . Block B will experience one force that is force of friction in the direction of motion .

Let force on block A be P . friction force on it will be equal to kinetic friction, that is μ mg , where μ is coefficient of friction and m is mass of block A

friction force = .4 x 2.5 x 9.8

= 9.8 N

net force on block A = P - 9.8

acceleration = ( P - 9.8 ) / 2.5

force on block B = 9.8

acceleration = force / mass

= 9.8 / 6

for common acceleration

( P - 9.8 ) / 2.5 = 9.8 / 6

( P - 9.8 ) / 2.5 = 1.63333

P = 13.88 N .

4 0

3 years ago