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andreev551 [17]
3 years ago
9

How do you do this? Plz help answer

Physics
1 answer:
Contact [7]3 years ago
7 0

Answer: Really

Explanation:

Just look it up for this page and maybe you will find an anwser sheet.

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A rifle is aimed horizontally at a target 50.0 m away. The bullet hits the target 2.90 cm below the aim point. . . Whats the bul
Marina CMI [18]
The problem ask to calculate the bullet's flight time and the bullet's speed as it left the barrel. So base on the problem, the answer would be that the flight time is 0.076 seconds and the speed of the bullet is 657.9 m/s. I hope you are satisfied with my answer and feel free to ask for more if you have questions and further clarifications. 
5 0
3 years ago
I need help thanks :)))))))))
dsp73

°C = (5/9) · (°F-32)

The "wet" thermometer is the upper one ... you can see the wet cloth wrapped around the bulb at the end.  It's reading 70° F.

°C = (5/9) · (38) = 21.1° C

The "dry" thermometer is the lower one.  It's reading 80° F.

°C = (5/9) · (48) = 26.7° C

So it looks like choice-A is your answer.

6 0
3 years ago
In a carrom game, a striker weighs three times the mass of the other pieces, the carrom men and the queen, which each have a mas
Mila [183]

Answer:

- The final velocity of the queen is (3/2) of the initial velocity of the striker. That is, (3V/2)

- The final velocity of the striker is (1/2) of the initial velocity of the striker. That is, (V/2)

Hence, the relative velocity of the queen with respect to the striker after collision

= (3V/2) - (V/2)

= V m/s.

Explanation:

This is a conservation of Momentum problem.

Momentum before collision = Momentum after collision.

The mass of the striker = M

Initial Velocity of the striker = V (+x-axis)

Let the final velocity of the striker be u

Mass of the queen = (M/3)

Initial velocity of the queen = 0 (since the queen was initially at rest)

Final velocity of the queen be v

Collision is elastic, So, momentum and kinetic energy are conserved.

Momentum before collision = (M)(V) + 0 = (MV) kgm/s

Momentum after collision = (M)(u) + (M/3)(v) = Mu + (Mv/3)

Momentum before collision = Momentum after collision.

MV = Mu + (Mv/3)

V = u + (v/3)

u = V - (v/3) (eqn 1)

Kinetic energy balance

Kinetic energy before collision = (1/2)(M)(V²) = (MV²/2)

Kinetic energy after collision = (1/2)(M)(u²) + (1/2)(M/3)(v²) = (Mu²/2) + (Mv²/6)

Kinetic energy before collision = Kinetic energy after collision

(MV²/2) = (Mu²/2) + (Mv²/6)

V² = u² + (v²/3) (eqn 2)

Recall eqn 1, u = V - (v/3); eqn 2 becomes

V² = [V - (v/3)]² + (v²/3)

V² = V² - (2Vv/3) + (v²/9) + (v²/3)

(4v²/9) = (2Vv/3)

v² = (2Vv/3) × (9/4)

v² = (3Vv/2)

v = (3V/2)

Hence, the final velocity of the queen is (3/2) of the initial velocity of the striker and is in the same direction.

The final velocity of the striker after collision

= u = V - (v/3) = V - (V/2) = (V/2)

The relative velocity of the queen withrespect to the striker after collision

= (velocity of queen after collision) - (velocity of striker after collision)

= v - u

= (3V/2) - (V/2) = V m/s.

Hope this Helps!!!!

3 0
3 years ago
Read 2 more answers
The table below shows some common animals and their hearing range in Hertz (Hz). Animal Hearing Range in Rate of Vibrations per
kap26 [50]

Answer:

D

Explanation:

If you list and compare, the elephants hearing range is the closest to a humans.

Humans: 20-20,000

Elephant: 16-12,000

While the others are either too much more or less than a humans.

Bat: 2,000-110,000

Dolphin: 90-105,000

Chicken: 125-2,000

5 0
2 years ago
A 10-kg rock and 20-kg rock are dropped from the same height and experience no significant air resistance. if it takes the 20-kg
bulgar [2K]
<span>Mass of rock 1 is m1 = 10 kg Mass of rock 2 is m2 = 20 kg 10-kg rock takes T (the same time ) to reach the ground as similar to 20-kg rock that takes time T to reach the ground. If no air resistance is present, the rate of descent depends only on how far the object has fallen, no matter how heavy the object is. This means that two objects will reach the ground at the same time if they are dropped simultaneously from the same height. This statement follows from the law of conservation of energy and has been demonstrated experimentally by dropping a feather and a lead ball in an airless tube. When air resistance plays a role, the shape of the object becomes important. THUS 10-KG & 20-KG ROCK reaches the ground in T-time</span>
3 0
3 years ago
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