Question

1. What is the mass of the cart with a constant net force of 200N is exerted to accelerate from rest to a velocity of 40m/s in 10 s.


2. What is the accelerate of a ball with a mass of 0.40 kg is hit with a force of 16N?


3. What is the external net force exerted on a 3.5 kg papaya, which is being pushed across a table and has an accelerate of 2.2 m/s² to the left?


4. What is the mass of a crate with a net force of 300N and accelerate it by 0.750 m/s²?

Pa answer po need ko na kasi ngayon bukas na kasi pasahan namin

Answer 1

Answer :

1. Mass is 50 kg.

2. 400m/s

3. 7.7N

4. 400kg

Step-by-step explanation :

1. Force applied on the cart = 200N.

Initial velocity = 0m/s.

Final velocity = 40m/s.

Time taken = 10s.

Mass of the cart = ?

Acceleration of the cart = v-u/t

So, a = 40 - 0/10

==> a = 40/10

==> a = 4 m/s²

Now, Force = mass × acceleration

==> F = ma

==> m = F/a

==> m = 200/40

Hence, the mass of the cart is 50kg.

2. F = m × a

16N = 0.04 × a

16/0.04 = a

40m/sec² = a

400m/s

The accelerate of a ball with a mass of 0.04kg is 400m/s.

3. Given, mass = 3.5kg, acceleration = 2.2m/s²

force = F

We know that,

F = m × a

F = 3.5kg × 2.2m/s²

F = 7.7N

The force exerted on 3.5kg papaya is 7.7N.

4. Force = m × a

300N = m × 0.750m/s²

m = 300/0.750

m = 400

The mass is 400kg.