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harkovskaia [24]
3 years ago
6

Which of the following examples does not illustrate Newton's first law of motion?

Physics
1 answer:
Stella [2.4K]3 years ago
6 0

Answer: B

Explanation: The rocket is accelerating the the spacecraft. Accelerating  relates to Newtons Second Law of Motion.

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How is the moment magnitude scale used to describe earthquakes?
natka813 [3]
The Richter Scale<span> is not commonly </span>used<span> anymore, except for small </span>earthquakes<span>recorded locally, for which ML and Mblg are the only </span>magnitudes<span> that can be measured. For all other </span>earthquakes<span>, the </span>moment magnitude scale<span> is a more accurate measure of the </span>earthquake<span> size.</span>
3 0
3 years ago
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A raging bull of mass 700kg runs at 10 m/s. How much kinetic energy does it have?
Illusion [34]

Answer: 35000 J or 35kJ

Explanation:

Use equation for kinetic energy : Ek=mV²/2

m=700 kg

V=10m/s

-------------------

Ek= 700kg*100m²7s²/2

Ek=35000 J=35kJ

6 0
4 years ago
Do I divided or multiply
Papessa [141]

Weight is mass x gravity, so you'd multiply the mass of the astronaut by the gravitational pull.

6 0
4 years ago
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If a Ferrari, with an initial velocity of 10m/s, accelerates at at rate of 50m/s/s for 3 seconds, what will its final velocity b
seraphim [82]

Answer:

160 m/s

Explanation:

The Ferrari is moving by uniformly accelerated motion, with constant acceleration of a = 50 m/s^2, and initial velocity u = 10 m/s. The velocity at time t of the car is given by

v(t)= u +at

where

u = 10 m/s

a = 50 m/s^2

If we substitute t = 3 s into the equation, we can find the velocity of the car after 3 seconds:

v(3 s)=10 m/s + (50 m/s^2)(3 s)=160 m/s

5 0
3 years ago
Where is the near point of an normal eye when accidentally wear a contact lens with a power of +2.0 diopters?
Lerok [7]

Answer:

The near point of an eye with power of +2 dopters, u' = - 50 cm

Given:

Power of a contact lens, P = +2.0 diopters

Solution:

To calculate the near point, we need to find the focal length of the lens which is given by:

Power, P = \frac{1}{f}

where

f = focal length

Thus

f = \frac{1}{P}

f = \frac{1}{2} = + 0.5 m

The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.

Now, by using lens maker formula:

\frac{1}{f} = \frac{1}{u} + \frac{1}{u'}

where

u = object distance = 25 cm = 0.25 m = near point of a normal eye

u' = image distance

Now,

\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}

\frac{1}{u'} = \frac{1}{0.5} - \frac{1}{0.25}

\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}

Solving the above eqn, we get:

u' = - 0.5 m = - 50 cm

7 0
3 years ago
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