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Step2247 [10]
3 years ago
13

What is the momentum of a 50 at a speed of 5m/s

Physics
1 answer:
aleksandr82 [10.1K]3 years ago
5 0
50*5=250
Momentum will be 250kgm/s^2
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Given the nuclear equation:
Leokris [45]
This is a nuclear fission reaction, in which a larger nucleus is bombarded with a neutron to make it break down into two smaller nuclei and release energy.
6 0
3 years ago
What is the function of the bronchioles?<br> please help
beks73 [17]

Answer:

The bronchioles function is to deliver air to tiny sacs called alveoli where oxygen and carbon dioxide are exchanged.

Explanation:

Bronchioles are air passages inside the lungs that branch off like tree limbs from the bronchi—the two main air passages into which air flows from the trachea (windpipe) after being inhaled through the nose or mouth. The bronchioles deliver air to tiny sacs called alveoli where oxygen and carbon dioxide are exchanged.

4 0
3 years ago
Suppose you push a hockey puck of mass m across frictionless ice for a time \Delta t, starting from rest, giving thepuck speed v
bazaltina [42]

Answer:

1. t_2 = 2t_1

2. t_2 = t_1\sqrt{2}

Explanation:

1. According to Newton's law of motion, the puck motion is affected by the acceleration, which is generated by the push force F.

In Newton's 2nd law: F = ma

where m is the mass of the object and a is the resulted acceleration. So in the 2nd experiment, if we double the mass, a would be reduced by half.

a_1 = 2a_2

Since the puck start from rest, in the 1st experiment, to achieve speed of v it would take t time

t = v / a_1

Now that acceleration is halved:

t = \frac{v}{2a_2}

\frac{v}{a_2} = 2t

You would need to push for twice amount of time t_2 = 2t_1

2. The distance traveled by the puck is as the following equation:

d = at^2

So if the acceleration is halved while maintaining the same d:

\frac{d_1}{d_2} = \frac{a_1t_1^2/2}{a_2t_2^2/2}

As d_1 = d_2, then d_1/d_2 = 1. Also a_1 = 2a_2

1 = \frac{2a_2t_1^2}{a_2t_2^2}

t_2^2 = 2t_1^2

t_2 = t_1\sqrt{2}\approx 1.14t_1

So t increased by 1.14

7 0
3 years ago
A reservoir located in the mountain 250 m above sea level flows through a pipe to a hydroelectric plant in a town at sea level.
Pavlova-9 [17]

Answer:

     v₂ = 70 m / s

Explanation:

For this exercise let's use Bernoulli's equation

where subscript 1 is for the top of the mountain and subscript 2 is for Tuesday's level

 

          P₁ + ½ ρ v₁² + ρ g y₁ = P₂ +1/2 ρ v₂² + ρ g y₂

indicate that the pressure in the two points is the same, y₁ = 250 m, y₂ = 0 m, the water in the upper part, because it is a reservoir, is very large for which the velocity is very small, we will approximate it to 0 (v₁ = 0), we substitute

         ρ g y₁ = ½ ρ v₂²

         v₂ = \sqrt {2g \ y_1}

let's calculate

         v₂ = √( 2 9.8 250)

         v₂ = 70 m / s

6 0
3 years ago
What is the frequency of the electromagnetic wave if the period is 1.54x10-15s
pogonyaev

answer✿࿐

I was not able to write it here

so I did it somewhere else and attached the picture

i hope it helps

have a nice day

#Captainpower

3 0
3 years ago
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