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HACTEHA [7]
2 years ago
5

A 356.3 G rock is traveling at 14.2 m/s. How much kinetic energy in joules does it have?

Physics
1 answer:
geniusboy [140]2 years ago
3 0

Answer:

Ek=35.922J

Explanation:

Ek=0.5mv^2

Ek=0.5(0.3563kg)(14.2)^2

Ek=35.922J

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At a race car driving event, a staff member notices that the skid marks left by the race car are 9.06 m long. The very experienc
vaieri [72.5K]

Answer:

27.1 m/s

Explanation:

Given that at a race car driving event, a staff member notices that the skid marks left by the race car are 9.06 m long. The very experienced staff member knows that the deceleration of a car when skidding is -40.52 m/s2.

Using third equation of motion,

V^2 = U^2 + 2aS

Since the car is decelerating, the final velocity V = 0

Substitute all the parameter into the equation above,

0 = U^2 - 2 * 40.52 * 9.06

U^2 = 734.22

U = \sqrt{734.22}

U = 27.096

U = 27.1 m/s  approximately

Therefore, the staff member can estimate for the original speed of the race car to be 27.1 m/s if it came to a stop during the skid

5 0
3 years ago
What is the momentum of a 12 kg condor flying at 6 m/s?
rusak2 [61]

Answer:

72

Explanation:

Formula

p=mv\\=12*6\\=72

5 0
3 years ago
What is the frequency of a sound wave with a wavelength of 0.04 meter in air? What type of wave is
Viktor [21]
8500 Hz and Longitudinal


Speed = frequency x wavelength

Speed of sound at 20 degrees Celsius is approximately 340 m/s
4 0
3 years ago
Read 2 more answers
A particle of charge 2.0 x 10^-8C experiences an upward force of magnitude 4.0 x10^-6 when it is placed in a particular point in
koban [17]

Answer:

a) The electric field at that point is 2.0\times 10^{2} newtons per coulomb.

b) The electric force is 2.0\times 10^{-6} newtons.

Explanation:

a) Let suppose that electric field is uniform, then the following electric field can be applied:

E = \frac{F_{e}}{q} (1)

Where:

E - Electric field, measured in newtons per coulomb.

F_{e} - Electric force, measured in newtons.

q - Electric charge, measured in coulombs.

If we know that F_{e} = 4.0\times 10^{-6}\,N and q = 2.0\times 10^{-8}\,C, then the electric field at that point is:

E = \frac{4.0\times 10^{-6}\,N}{2.0\times 10^{-8}\,C}

E = 2.0\times 10^{2}\,\frac{N}{C}

The electric field at that point is 2.0\times 10^{2} newtons per coulomb.

b) If we know that E = 2.0\times 10^{2}\,\frac{N}{C} and q = 1.0\times 10^{-8}\,C, then the electric force is:

F_{e} = E\cdot q

F_{e} = \left(2.0\times 10^{2}\,\frac{N}{C} \right)\cdot (1.0\times 10^{-8}\,C)

F_{e} = 2.0\times 10^{-6}\,N

The electric force is 2.0\times 10^{-6} newtons.

7 0
3 years ago
When a warm air mass is trapped between two cooler air masses, it is called a/an
Semenov [28]
The answer is C. An occluded front.
8 0
3 years ago
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