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3 years ago

What is the force required to produce an acceleration of 46.4

1 answer:

7 0

That depends on the mass of the object, and the unit of the '46.4' .

If the '46.4' is ' meters per second² ' , then the force required is

(mass of the object in kilograms) x (46.4) newtons .

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Identify the type of weathering seen in each picture

alukav5142 [94]

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5 0

3 years ago

Read 2 more answers

A Smart Car, which has a mass of 1000 kg, is going 20 m/s. When it hits the barrier, it stops with a time of 0.5 seconds. What i

antiseptic1488 [7]

Answer:

The change in momentum = -20000 kg m/s.

Explanation:

Mass m = 1000 kg

speed v₁ = 20 m/s

speed v₂ = 0 m/s

We know that,

The change in momentum

ΔP = m (Δv)

ΔP = m (v₂ - v₁)

= 1000 (0 - 20)

= 1000 (-20)

= -20000 kg m/s

Thus, the change in momentum = -20000 kg m/s.

Note: negative sign indicates that the velocity is reducing when it hits the barrier.

4 0

2 years ago

The Moon and Earth rotate about their common center of mass, which is located about RcM 4700 km from the center of Earth. (This

erica [24]

To solve this problem it is necessary to apply the concepts related to gravity as an expression of a celestial body, as well as the use of concepts such as centripetal acceleration, angular velocity and period.

PART A) The expression to find the acceleration of the earth due to the gravity of another celestial body as the Moon is given by the equation

$g = \frac{GM}{(d-R_{CM})^2}$

Where,

G = Gravitational Universal Constant

d = Distance

M = Mass

$R_{CM} =$ Radius earth center of mass

PART B) Using the same expression previously defined we can find the acceleration of the moon on the earth like this,

$g = \frac{GM}{(d-R_{CM})^2}$

$g = \frac{(6.67*10^{-11})(7.35*10^{22})}{(3.84*10^8-4700*10^3)^2}$

$g = 3.4*10^{-5}m/s^2$

PART C) Centripetal acceleration can be found throughout the period and angular velocity, that is

$\omega = \frac{2\pi}{T}$

At the same time we have that centripetal acceleration is given as

$a_c = \omega^2 r$

Replacing

$a_c = (\frac{2\pi}{T})^2 r$

$a_c = (\frac{2\pi}{26.3d(\frac{86400s}{1days})})^2 (4700*10^3m)$

$a_c = 3.34*10^{-5}m/s^2$

3 0

3 years ago

How to not get heart attacks ? And what is some plans/list for a healthy diet ?

Mamont248 [21]

Answer:

hey there

( i )we have to Choose good nutrition. A healthy diet is one of the best weapons you have to fight cardiovascular disease. and should be physically active and stop smoking if you do these are some

life style changes though which can prevent heart attacks

( ii )An healthy diet for heart

lots of fruits and vegetables.

nuts, beans, and legumes.

whole grains.

plant-based oils, such as olive oil.

low-fat dairy products.

Explanation:

Hope it helps you

have a nice day :)

6 0

2 years ago