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3 years ago

How many nano-meters thick is a nickel

Physics

1 answer:

marishachu [46]3 years ago

3 0

<span>2,000,000 nm, algebraic steps below: 2 mm*1 nm 1.0E-6 mm=2,000,000 nm</span>

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A 70 kg hunter, standing on frictionless ice, shoots a 42 g bullet horizontally at a speed of 590 m/s . Part A Part complete Wha

Kisachek [45]

Answer:

The recoil velocity is 0.354 m/s.

Explanation:

Given that,

Mass of hunter = 70 kg

Mass of bullet = 42 g = 0.042 kg

Speed of bullet = 590 m/s

We need to calculate the recoil speed of hunter

Using conservation of momentum

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

Where, $m_{1}$ = mass of hunter

$m_{2}$ = mass of bullet

u = initial velocity

v = recoil velocity

Put the value in the equation

$0+0=70\times v_{1}+0.042\times590$

$v_{1}=-\dfrac{0.042\times590}{70}$

$v=-0.354\ m/s$

Hence, The recoil velocity is 0.354 m/s.

8 0

3 years ago

There are stars located in the center bulge of the Milky Way and the spiral arms of the Milky Way. What is the difference betwee

nadya68 [22]

Answer:

The stars at the center bulge are bigger and brighter than the stars in the arms.

Explanation:

5 0

3 years ago

Two families meet at the park at 10:00am. Each family uses a different way of transportation to get there. The Gonzalez family l

pav-90 [236]

Firstly, we need to make the times and distance equal to compare.

1hr=60mins
Gonz drove 90km in 60mins,so they would drive 45km in 30 mins dividing it by 2.
Because the Rivs had gone 30km in 30mins, Gonz were faster.

For the adverage speed, we should first add the speed of each family in mph , so 90+60, which equals 150 and divide that by 2 because there are 2 speeds so the average speed is 75mph.

P.S English is my second language so tell me if you don't understand something and I will try and explain.

5 0

3 years ago

He has 54 cds and every box has 9 cds.how many boxes are in 54 cds?

Gemiola [76]

Solution:
54 / 9 = 6 boxes.

6 0

3 years ago

A tuning fork vibrating at 508 Hz falls from rest and accelerates at 9.80 m/s^2. How far below the point of release is the tunin

JulijaS [17]

Answer:

Explanation:

given,