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3 years ago

American space pilots were called astronauts. What were soviet space pilots called?.

Physics

1 answer:

Mnenie [13.5K]3 years ago

4 0

Answer:

Cosmonauts

Explanation:

As the americans are called astronauts which means star sailors and the russians (soviet space pilots) are called cosmonauts which means sailors of the universe

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A light bulb is connected to a 2V supply and experiences a current of 6.4A. What is the power rating of the bulb?

Olenka [21]

Answer:

12.8 Watts

Explanation:

P = VI

P = (2 V) (6.4 A)

P = 12.8 W

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4 years ago

a 46 kilogram student climbs 11 meter up a rope at a constant speed if the student power output is 230 watts how long in seconds

Bezzdna [24]

230×46=10580÷11=961 second

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3 years ago

What type of stress is shown in the following diagram?

leva [86]

Here stress is parallel to the surface of the body. So it's a Shear stress.

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4 years ago

Read 2 more answers

Calculate the translational speed of a cylinder when it reaches the foot of an incline 7.05 mm high. Assume it starts from rest

mestny [16]

Height is 7.05 m and not 7.05 mm

Answer:

9.603 m/s

Explanation:

We are dealing with rotation, so velocity of centre of mass is given by;

v_cm = Rω

Since we are working with a solid cylinder, moment of inertia of the cylinder is; I = ½mR²

Since it is rolled from the top to the bottom, at the top it will have potential energy(mgh) while at the bottom it will have kinetic energy (rotational plus translational kinetic energy).

Using conservation of energy, we have:

P.E = K.E_t + K.E_r

Formula for rotational and kinetic energy here are;

K.E_t = ½mv²

K.E_r = ½Iω²

mgh = ½mv² + ½Iω²

Since we want to find translational speed(v), let's get rid of ω.

Earlier, we saw that v_cm = Rω

Thus; ω = v/R

Also, we know that I = ½mR².

Thus;

mgh = ½mv² + ½(½mR²)(v/R)²

This gives;

mgh = ½mv² + ¼mv²

Divide through by m to get;

gh = v²(½ + ¼)

gh = ¾v²

Making v the subject gives;

v = √(4gh/3)

v = √((4 × 9.81 × 7.05)/3)

v = 9.603 m/s

6 0

3 years ago

A person hits a tennis ball with a mass of 0.058 kg against a wall.

horrorfan [7]

Explanation:

Mass of the ball, m = 0.058 kg

Initial speed of the ball, u = 11 m/s

Final speed of the ball, v = -11 m/s (negative as it rebounds)

Time, t = 2.1 s

(a) Let F is the average force exerted on the wall. It is given by :

$F=\dfrac{m(v-u)}{t}$

$F=\dfrac{0.058\times (11-(-11))}{2.1}$

F = 0.607 N

(b) Area of wall, $A=3\ m^2$

Let P is the average pressure on that area. It is given by :

$P=\dfrac{F}{A}$

$P=\dfrac{0.607\ N}{3\ m^2}$

P = 0.202 Pa

Hence, this is the required solution.

8 0

3 years ago