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balu736 [363]
4 years ago
8

I'm not sure about these 2 questions

Physics
1 answer:
Natali [406]4 years ago
7 0
2) Newton's third law<span>: For every action, there is an equal and opposite reaction.

3) Please provide the diagram </span>
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If a car has a momentum of 2.04 x 104 kgm/s and a velocity of 18 m/s, what is its mass?
muminat

Answer:

Mass = 1133.33 kg (Approx.)

Explanation:

Given:

Momentum = 2.04 x 10⁴ kg[m/s]

Velocity = 18 m/s

Find:

Mass

Computation:

Mass = Momentum / Velocity

Mass = [2.04 x 10⁴] / 18

Mass = 1133.33 kg (Approx.)

5 0
3 years ago
a student standing between two walls shouts once.he hears the first echo after 3 seconds and the next after 5 seconds. calculate
Karolina [17]

Explanation:

It took t_1 =1.5\:\text{s} for the sound to reach the 1st wall and at the same time time, the same sound took t_2 = 2.5\:\text{s} to reach the 2nd wall. Assuming that the sound travels at 343 m/s, then let x_1 be the distance of the person to the 1st wall and x_2 be the distance to the 2nd wall. So the distance between the walls X is

X = x_1 + x_2 = v_st_1 + v_st_2 = v_s(t_1 + t_2)

\:\:\:\:\:= (343\:\text{m/s})(4.0\:\text{s}) = 1372\:\text{m}

6 0
3 years ago
In the nucleus, the a)...... have b)...... charge.
Inga [223]
C:carry 
I hope I help
8 0
4 years ago
Find the speed required to throw a ball straight up and it return 6 seconds later. Neglect air resistance
Nadya [2.5K]

Answer:

the ball will go up 3s and down 3s

v=gt

where t=3s and g=9.8m/s^2

distance=v0(t)+(1/2)gt^2

where initial velocity (v0)=0

Explanation:

8 0
3 years ago
A ball is thrown horizontally from the top of a building 100m high. The ball strikes the ground at a point 120 m horizontally aw
pshichka [43]

Answer:

44.3 m/s

Explanation:

Given that a ball is thrown horizontally from the top of a building 100m high. The ball strikes the ground at a point 120 m horizontally away from and below the point of release.

What is the magnitude of its velocity just before it strikes the ground ?

The parameters given are:

Height H = 100m

Since the ball is thrown from a top of a building, initial velocity U = 0

Let g = 9.8m/s^2

Using third equation of motion

V^2 = U^2 + 2gH

Substitute all the parameters into the formula

V^2 = 2 × 9.8 × 100

V^2 = 200 × 9.8

V^2 = 1960

V = 44.27 m/s

Therefore, the magnitude of its velocity just before it strikes the ground is 44.3 m/s approximately

6 0
3 years ago
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