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2 years ago

Investigations allow for the control of variables

2 answers:

8 0

I believe that the statement is true. <span>Investigations allow for the control of variables. Changing variables will lead you to observations that may prove your hypothesis. Hope this answers the question. Have a nice day. Please feel free to ask more questions.</span>

Bogdan [553]2 years ago

7 0

<h3>I believe that the statement is true. Investigations allow for the control of variables. Changing variables will lead you to observations that may prove your hypothesis</h3>

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Will the velocity of the book change as it moves across the surface with NO friction? Explain your answer.

MakcuM [25]

No velocity will not be changed

Why?

According to Newtons 1st law the velocity of a moving object remains unchanged unless a external force affect that.

6 0

2 years ago

Read 2 more answers

An object accelerates in a direction that is always perpendicular to its motion.what is the effect if any of the acceleration on

mars1129 [50]

If its accelerating it will increase velocity in the direction of the acceleration which is perpendicular to the velocity.

3 0

2 years ago

Which of the following is equal to impulse?<br> A. Ft<br> B. m/s<br> C. pt<br> D. mv

tankabanditka [31]

Answer: A (Ft)

Explanation: The impulse experienced by the object equals the change in momentum of the object. In equation form, F • t = m • Δ v

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2 years ago

A sewing machine uses 1.6kWh of electricity in one day. If electricity costs 9p per unit, what is the total cost in pence of usi

atroni [7]

Answer:

$\huge\boxed{\sf 1.6\ units = 14.4p}$

Explanation:

Since,

So,

1.6 kWh = 1.6 units

If,

1.6 units = 9p × 1.6

1.6 units = 14.4p

$\rule[225]{225}{2}$

8 0

1 year ago

A spherical asteroid of average density would have a mass of 8.7×1013kg if its radius were 2.0 km. 1. If you and your spacesuit

Law Incorporation [45]

1. 0.16 N

The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:

$F=G\frac{Mm}{r^2}$

where

G is the gravitational constant

$M=8.7\cdot 10^{13}kg$ is the mass of the asteroid

m = 100 kg is the mass of the man

r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid

Substituting, we find

$F=(6.67\cdot 10^{-11}m^3 kg^{-1} s^{-2})\frac{(8.7\cdot 10^{13} kg)(110 kg)}{(2000 m)^2}=0.16 N$

2. 1.7 m/s

In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force

$m\frac{v^2}{r}=G\frac{Mm}{r^2}$

where v is the minimum speed required to stay in orbit.

Re-arranging the equation and solving for v, we find:

$v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{(6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2})(8.7\cdot 10^{13} kg)}{2000 m}}=1.7 m/s$

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3 years ago