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3 years ago

11

An astronaut weighs 700 N on Earth what is the astronaut's Mass on Earth?

1 answer:

ch4aika [34]3 years ago

3 0

Answer:

70 kg

Explanation:

On earth, g = 9.8 m/s².

W = mg

700 N = m (9.8 m/s²)

m ≈ 71 kg

Rounded to 1 significant figure, the astronaut's mass is 70 kg.

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why it's impossible for baseball players to make sharp turns when moving form base to base? explain using Newton 1st law "object

AVprozaik [17]

Answer:

An object that is moving wants to stay moving in a straight line. It takes an outside force acting upon it to change its direction or cause an acceleration.

Explanation:

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3 years ago

An athlete kicks a soccer ball that starts at rest so that it leaves their foot with a speed of 10m/s from the top o f a rectang

kirza4 [7]

Answer:

$a=500m/s^2$

Explanation:

We need only to apply the definition of acceleration, which is:

$a=\frac{v_f-v_i}{t_f-t_i}$

In our case the final velocity is $v_f=10m/s$ , the initial velocity is $v_i=0m/s$ since it departs from rest, the final time is $t_f=0.02s$ and the initial time we are considering is $t_i=0s$

So for our values we have:

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3 years ago

An ac generator consists of a coil with 40 turns of wire, each with an area of 0.06 m2 . The coil rotates in a uniform magnetic

mezya [45]

Answer:

d. 332 V

Explanation:

Given;

number of turns in the wire, N = 40 turns

area of the coil, A = 0.06 m²

magnitude of the magnetic field, B = 0.4 T

frequency of the wave, f = 55 Hz

The maximum emf induced in the coil is given by;

E = NBAω

Where;

ω is angular velocity = 2πf

E = NBA(2πf)

E = 40 x 0.4 x 0.06 x (2 x π x 55)

E = 332 V

Therefore, the maximum induced emf in the coil is 332 V.

The correct option is "D"

d. 332 V

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3 years ago

A 60-kg rollerblader rolls 10 m down a 30? incline. When she reaches the level floor at the bottom, she applies the brakes. The

VARVARA [1.3K]

Answer:

s = 20 m

Explanation:

given,

mass of the roller blader = 60 Kg

length = 10 m

inclines at = 30°

coefficient of friction = 0.25

using conservation of energy

$\dfrac{1}{2}mu^2 = m g d sin \theta$

$u^2 = 2 g d sin30^0$

$u= \sqrt{2\times 9.8 \times 10 sin30^0}$

u = 9.89 m/s

Using second law of motion

ma =μ mg

a = μ g

a = 0.25 x 9.8

a = 2.45 m/s²

Using third equation of motion ,

v² - u² = 2 a s

0² - 9.89² = 2 x 2.45 x s

s = 20 m

the distance moved before stopping is 20 m

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3 years ago

Answers the questions <br> 1. At what distance did the object start to move ?

Citrus2011 [14]

Answer:

it started to move a 1 second

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3 years ago

Read 2 more answers