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3 years ago

13

An object moves with a constant speed of 20 m/s on a circular track of radius 100 m. What is the tangential acceleration of the

object?

1 answer:

MissTica3 years ago

4 0

Answer:4s^-1

Explanation:Tangential Acceleration=v^2/r=(20)²/100=400/100

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Displacement vectors of 10 m west and 14 m west make a resultant vector that is ?

goldenfox [79]

Resultant vector= 24 m west.

Explanation:

resultant vector= r1+ r2 when the two vectors are in same direction

r1=10m west

r2= 14 m west

so resultant displacement= 10 + 14

resultant displacement= 24 m west

4 0

3 years ago

a horizontal force of 100N is required to push a crate across a factory floor at a constant speed. What is the net force acting

Bas_tet [7]

If the crate is moving along the floor in the same direction with a constant speed, it is in dynamic equilibrium. Equilibrium means there is no net force acting on the crate.

Since there is no net force on the crate, there must be a friction force on the crate equal in magnitude and opposite in direction to the applied horizontal force. Therefore the force of friction acting on the crate is 100N.

8 0

3 years ago

Rachel has been reading her physics book. She takes her weighing scales into an elevator and stands on them. If her normal weigh

GrogVix [38]

Answer:

345 N

Explanation:

Given:

Normal weight of Rachel (mg) = 690 N

Case 1: Upward motion of elevator

Given:

Acceleration of elevator (a) = 0.25 g

The scale reading is given by the normal force acting on Rachel. Let N₁ be the normal force.

So, the net force acting on Rachel is given as:

$F_{net}=N_1-mg=N_1-690$

Now, from Newton's second law:

$F_{net}=ma\\\\N_1-690=m\times 0.25g\\\\N_1-690=0.25\times (mg)\\\\N_1-690=0.25\times 690\\\\N_1=690+172.5=862.5\ N------(1)$

Case 2: Downward motion of elevator

Given:

Acceleration of elevator (a) = 0.25 g

The scale reading is given by the normal force acting on Rachel. Let N₂ be the normal force.

So, the net force acting on Rachel is given as:

$F_{net}=mg-N_2=690-N_2$

Now, from Newton's second law:

$F_{net}=ma\\\\690-N_2=m\times 0.25g\\\\690-N_2=0.25\times (mg)\\\\690-N_2=0.25\times 690\\\\N_2=690-172.5=517.5\ N------(2)$

Now, the difference in the scale reading is obtained by subtracting equation (2) from equation (1). This gives,

$Difference=N_1-N_2=862.5-517.5=345\ N$

Therefore, the difference between the up and down scale readings is 345 N.

4 0

3 years ago

Following your push the ball rolls down the lane at 4.2m/s. What is the net force on the ball as it rolls down the lane at the c

son4ous [18]

Following your push the ball rolls down the lane at 4.2m/s. What is the net force on the ball as it rolls down the lane at the constant speed?

3 0

3 years ago

Alborosie

What is this on, is this on a test?

4 0

3 years ago

Read 2 more answers