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Delicious77 [7]
3 years ago
6

A car with a mass of 600 kg is traveling at a velocity of 10 m/s. How much kinetic energy does it have? The car has J of kinetic

energy.
Physics
1 answer:
konstantin123 [22]3 years ago
5 0
The kinetic energy of an object is given by:
K= \frac{1}{2}mv^2
where m is the mass of the object and v its velocity. 

The car in this problem has a mass of m=600 kg and a velocity of v=10 m/s, therefore if we put these numbers into the equation, we find the kinetic energy of the car:
K= \frac{1}{2}mv^2= \frac{1}{2}(600 kg)(10 m/s)^2=30000 J
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Answer:

no it's opaque

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Explanation:

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3 years ago
A force of 350 N causes a body to move with an acceleration of 10 m/s2. What’s the mass of the body?
saveliy_v [14]
Hi Maria.

The formula for finding the mass of the object is M = F/A, or the force divided by the acceleration. Since we have our force and acceleration, we are ready to solve!

M = 350 / 10.

M = 35.

The unit of measurement for mass is kg, so your answer is going to be
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I hope this helps!
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3 years ago
Would you rather be hit by a 1-kg mass traveling at 10m/s or a 2-kg mass traveling at 5m/s? Why?
yawa3891 [41]

Answer:

a 2 kg mass traveling at 5ms

Explanation:

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3 years ago
Help me with the question b.​
Morgarella [4.7K]

Answer:

a) The specific heat capacity means the amount of heat needed by a unit mass of a material to increase its temperature in one unit.

b) Liquid P - Q = 3840\,J, Liquid Q - Q = 5500\,J, Liquid R - Q = 7800\,J, Liquid S - Q = 2856\,J

Explanation:

a) The specific heat capacity means the amount of heat needed by a unit mass of a material to increase its temperature in one unit.

b) Let suppose that heat transfer rates between liquids and surroundings are stable. The quantity of the heat released is determined by the following expression:

Q = m\cdot c\cdot (T_{r} - T_{f}) (1)

Where:

m - Mass of the liquid, in kilograms.

c - Specific heat capacity, in joules per kilogram-degree Celsius.

T_{r} - Initial temperature of the sample, in degrees Celsius.

T_{f} - Freezing point, in degrees Celsius.

Liquid P (m = 1\,kg, c = 160\,\frac{J}{kg\cdot ^{\circ}C}, T_{r} = 30\,^{\circ}C, T_{f} = 6\,^{\circ}C)

Q = (1\,kg)\cdot \left(160\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (30\,^{\circ}C - 6\,^{\circ}C)

Q = 3840\,J

Liquid Q (m = 1\,kg, c = 220\,\frac{J}{kg\cdot ^{\circ}C}, T_{r} = 30\,^{\circ}C, T_{f} = 5\,^{\circ}C)

Q = (1\,kg)\cdot \left(220\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (30\,^{\circ}C - 5\,^{\circ}C)

Q = 5500\,J

Liquid R (m = 1\,kg, c = 300\,\frac{J}{kg\cdot ^{\circ}C}, T_{r} = 30\,^{\circ}C, T_{f} = 4\,^{\circ}C)

Q = (1\,kg)\cdot \left(300\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (30\,^{\circ}C - 4\,^{\circ}C)

Q = 7800\,J

Liquid S (m = 1\,kg, c = 102\,\frac{J}{kg\cdot ^{\circ}C}, T_{r} = 30\,^{\circ}C, T_{f} = 2\,^{\circ}C)

Q = (1\,kg)\cdot \left(102\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (30\,^{\circ}C - 2\,^{\circ}C)

Q = 2856\,J

6 0
3 years ago
What is friction and the types withe examples.​
Blizzard [7]

Explanation:

The answer is In the picture. Thanks.

7 0
3 years ago
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