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Alex787 [66]
3 years ago
11

A runner traveled 7.7 miles an hour and a half what was their average speed

Physics
1 answer:
Alenkasestr [34]3 years ago
6 0

Answer:5.13333333...

Explanation: 7.7 is the distance and an hour and a half is the average speed. You would have to divide the total distance by the total time. So it would be 7.7 divided by 1.5 which would equal 5.13333333333...

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If an astronaut had a mass of 30 kg on the moon what would his mass be on earth?
Novay_Z [31]

Explanation:

A one-kilogram mass is still a one-kilogram(as mass is an intrinsic property of the object) but the downward force due to gravity, and therefore it's weight, is only one-sixth of what the object would have on the Earth. So man of mass 180 pounds weights only about 30 pounds-force when visiting the moon

hope it help..... pls add me as brainlist.

Have a nice day

7 0
3 years ago
Atoms of the same element that have the same number of protons but different numbers of neutrons are called_____
saveliy_v [14]
The answer is:  "isotopes" .
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7 0
3 years ago
What is the difference between regional metamorphism and contact metamorphism?
-Dominant- [34]
Contact metamorphism<span> is a type of </span>metamorphism<span> where rock minerals and texture are changed, mainly by heat, due to </span>contact<span> with magma. </span>Regional metamorphism<span> is a type of </span>metamorphism<span> where rock minerals and texture are changed by heat and pressure over a wide area or region.</span>
4 0
3 years ago
A 170 kg astronaut (including space suit) acquires a speed of 2.25 m/s by pushing off with his legs from a 2600 kg space capsule
saw5 [17]

Explanation:

Mass of the astronaut, m₁ = 170 kg

Speed of astronaut, v₁ = 2.25 m/s

mass of space capsule, m₂ = 2600 kg

Let v₂ is the speed of the space capsule. It can be calculated using the conservation of momentum as :

initial momentum = final momentum

Since, initial momentum is zero. So,

m_1v_1+m_2v_2=0

170\ kg\times 2.25\ m/s+2600\ kg\times v_2=0

v_2=-0.17\ m/s

So, the change in speed of the space capsule is 0.17 m/s. Hence, this is the required solution.

8 0
3 years ago
A simple and common technique for accelerating electrons is shown in the figure, where there is a uniform electric field between
Sunny_sXe [5.5K]

Answer : 4.483\times 10^{15}\ m/s^2.

Explanation:

It is given that,

Electric field strength, E=2.55\times 10^{4}\ N/C

We know that,

Charge of electron, q=1.6\times 10^{-19}\ C

Mass of electron, m=9.1\times 10^{-31}\ kg

From the definition of electric field, F=qE...............(1)

According to Newton's second law, F = ma..........(2)

From equation (1) and (2)

ma=qE

a=\dfrac{qE}{m}

a=\dfrac{1.6\times 10^{-19}\ C\times 2.55\times 10^{4}\ N/C }{9.1\times 10^{-31}\ kg}

a=0.4483\times 10^{16}\ m/s^2

or

a=4.483\times 10^{15}\ m/s^2

So, the horizontal component of acceleration of an electron is 4.483\times 10^{15}\ m/s^2.

Hence, it is the required solution.

7 0
3 years ago
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