PHYSICS
COLLEGE

Answer:

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

Where, = mass of hunter

= mass of bullet

u = initial velocity

v = recoil velocity

Put the value in the equation

Hence, The recoil velocity is 0.354 m/s.

HIGH SCHOOL

The metric system is based on units of A: twenty five

B fives

C: sixteen

D: tens

The metric system is based on units of tens.

MIDDLE SCHOOL

A 60 KG skier’s racing down a course at 20 M/S what is the skiers momentum

Answer:1200kgm/s

Explanation:

Mass=60kg velocity=20m/s

Momentum=mass x velocity

Momentum=60 x 20

Momentum=1200kgm/s

MIDDLE SCHOOL

Energy can be thought of as stored

It can be thought as electricity that is stored

MIDDLE SCHOOL

In the figure, a 3.7 kg block slides along a track from one level to a higher level after passing through an intermediate valley. The track is frictionless until the block reaches the higher level. There a frictional force stops the block in a distance d. The block's initial speed is v0 = 7.4 m/s, the height difference is h = 1.2 m, and μk = 0.627. Find d.

Answer:

Use conservation of energy: Loss in kinetic

HIGH SCHOOL

A flat piece of glass covers the top of a vertical cylinder that is completely filled with water. If a ray of light traveling in the glass is incident on the interface with the water at an angle of 0a= 36.0 degrees , the ray refracted into the water makes an angle of 49.6 degrees with the normal to the interface.What is the smallest value of the incident angle 0a for which none of the ray refracts into the water? Express your answer with the appropriate units.

Answer:

Explanation:

As we know that the refractive index of the water is

also we know that refractive index of the glass is

now we know by Snell's law

so we have

now if no light will refract into the water then in that case

now we have

HIGH SCHOOL

A 4.0 kg object will have a weight of approximately 14.8 N on Mars. What is the gravitational field strength on Mars? 40 N/kg

3.7 N/kg

1.48 N/kg

0.27 N/kg

Answer:

The gravitational field strength on Mars is 3.7 N/kg.

Explanation:

Given that,

Mass of the object, m = 4 kg

Weight of the object on Mars, F = W = 14.8 N

We need to find the gravitational field strength on Mars. It is given by gravitational force per unit mass of an object. Mathematically, it is given by

G = 3.7 N/kg

So, the gravitational field strength on Mars is 3.7 N/kg. Hence, this is the required solution.

Field strength =force/unit mass

= 14.8N/ 4kg = 3.7N/kg

= 14.8N/ 4kg = 3.7N/kg

MIDDLE SCHOOL

Select the correct answer. Which healthy snack can provide protein after physical activity? A. an orange slice B. pretzels C. yogurt D. a sports drink

Answer:

From the given choices, the healthy snack that can provide protein after physical activity is the third option, yogurt. Yogurts are derived from fermentation of raw materials by the healthy bacteria, usually of lactobacillus origin. Other items in the choices are not able to provide protein.

Explanation:

Answer:

The answer is C. Yogurt

Explanation:

Have a nice day!

MIDDLE SCHOOL

Why does 100% relative humidity in the winter feel nothing like 100% relative humidity in the summer?

The 100% relative humidity in the winter feel nothing like 100% in summer because it depends on "the saturation of the temperature".

Explanation:

Temperature really makes a big difference. Even once warm, a cold winter air produces much less humidity than summer heat. One cubic unit of air needs 0.001 ounces of water to saturate it, to elevate its ratio to one hundred per cent.

Nevertheless, it takes 0.022 ounces of water to saturate the one cubic unit of air once the temperature is eighty, which is twenty-two times that amount of water. Air with a humidity of one hundred percent at eighty degrees holds twenty-two times as much water as air at zero with humidity at one hundred percent.

COLLEGE

Suppose that a 0.375 m radius 500 turn coil produces an average emf of 12000 V when rotated one-fourth of a revolution in 4.27 ms, starting from its plane being perpendicular to the magnetic field. what is the peak emf generated by this coil, in volts?

Answer:

epsilon_{peak}= 18833 V

Explanation:

Average emf

⇒

solving this we get

B=0.2320 T

No. of revolutions are 1/4 rev

Δθ=

angular velocity ω =

=

=367.68 rad/sec

The peak emf

putting values we get

solving we get

epsilon_{peak}= 18833 V

Answer:

Explanation:

Given:

no. of turns in the coil,

area of the coil,

average emf induced,

angle turned by the coil,

time taken to sweep the given angle,

peak emf,

We have the relation between peak emf and average emf as:

.............................(1)

where:

we already know the value:

From eq. (1) we have: