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san4es73 [151]
3 years ago
5

If an object has a kinetic energy of 30 j and mass of 34kg how fast is the object moving ?

Physics
1 answer:
cricket20 [7]3 years ago
3 0

Ek = (m*V^2) / 2 where m is mass and V is speed, then we can take this equation and manipulate it a little to isolate the speed.

Ek = mv^2 / 2 — multiply both sides by 2

2Ek = mv^2 — divide both sides by m

2Ek / m = V^2 — switch sides

V^2 = 2Ek / m — plug in values

V^2 = 2*30J / 34kg

V^2 = 60J/34kg

V^2 = 1.76 m/s — sqrt of both sides

V = sqrt(1.76)

V = 1.32m/s (roughly)

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scZoUnD [109]

Answer: Acceleration

4 0
3 years ago
A whale swims due east for a distance of 6.9 km, turns around and goes due west for 1.8 km, and finally turns around again and h
xeze [42]

The question is incomplete. Here is the complete question:

A whale swims due east for a  distance of 6.9 km, turns around and  goes due west for 1.8 km and finally  turns around again and heads 3.7 km  due east.  (a) What is the total distance  traveled by the whale? (b) What are the  magnitude and direction of the displacement of the whale?

Answer:

(a) Distance = 12.4 km

(b) Displacement = 8.8 km due east

Explanation:

Consider east direction as positive and west direction as negative.

Given:

The motion is along the east-west line.

The whale first swims 6.9 km due east, then 1.8 km due west and again 3.7 km due east.

(a)

Distance traveled by the whale is equal to the sum of the lengths of all the distances traveled. Therefore,

Distance traveled = 6.9 km + 1.8 km + 3.7 km = 12.4 km

Therefore, the distance traveled by the whale is 12.4 km.

(b)

Displacement of the whale is given by considering the sign of each of the individual displacements. Therefore,

Displacement of the whale = (+6.9 km) + (-1.8 km) + (+3.7 km)

Displacement of the whale = 6.9 km - 1.8 km + 3.7 km = 8.8 km

The answer is positive. So, the direction is due east.

Therefore, the displacement of the whale is <u>8.8 km due east.</u>

6 0
3 years ago
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vitfil [10]

Answer:

12.31 m/s

Explanation:

If we recall from the previous knowledge we had about speed,

we will know that:

speed = distance/ time.

As such:

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Mathematically, it can be computed as:

v_{avg} = \dfrac{d+d}{\dfrac{d}{v_1}+ \dfrac{d}{v_2}}

v_{avg} = \dfrac{2d}{\dfrac{d}{10 \ m/s}+ \dfrac{d}{16 \ m/s}}

v_{avg} = \dfrac{2}{\dfrac{1}{10 \ m/s}+ \dfrac{1}{16 \ m/s}}

v_{avg} = \dfrac{2}{\dfrac{13}{80 \ m/s}}

\mathbf{v_{avg} =12.31 \ m/s}

8 0
2 years ago
The -is the process by which gases in the atmosphere absorb and reradiate heat?
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Water vapor and carbon dioxide

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The correct answer is

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