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kykrilka [37]
3 years ago
13

The kinetic energy of a particle of mass 500g is 4.8j. Determine the velocity of the particle

Physics
2 answers:
musickatia [10]3 years ago
7 0

Answer:

4.38 m/s

Explanation:

Otrada [13]3 years ago
5 0
The answer is 4.38 m/s

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A rubber ball and a steel ball are dropped from the same height onto a concrete floor. They have the same mass, and lets assume
Mkey [24]

Answer:

The average forces would be the same

Explanation:

Both have the same velocity on impact as they fell from the same height.

Both have the same velocity after the bounce because they reach the same height.

Both have the same mass

Both will thus experience the same impulse because both have the same change in momentum.

Therefore both experience the same average force.

5 0
3 years ago
I need this done pls
Aleonysh [2.5K]

Answer:

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5 0
2 years ago
A 500kg car is driven forward with a thrust force of 1500N. Air resistance and friction acts against the motion of the car with
Wittaler [7]
2m/s^2, this is because F=ma, meaning a is also equal to F/m. The car applies 1500N in one direction and outside sources apply a total of -500N, meaning the 500kg car is moving forward with a total of 1000N of force. Taking the total 1000N and dividing it by 500kg gives you and acceleration of 2m/s^2. Hope this helps!
8 0
2 years ago
Which of the following is NOT proof of a chemical change
Blababa [14]

Answer:

E

Explanation:

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4 0
3 years ago
Read 2 more answers
The electric field strength in the space between two closely spaced parallel disks is 1.0 10^5 N/C. This field is the result of
alex41 [277]

To solve this problem it is necessary to apply the concepts related to the capacitance in the disks, the difference of the potential and the load in the disc.

The capacitance can be expressed in terms of the Area, the permeability constant and the diameter:

C = \frac{\epsilon_0 A}{d}

Where,

\epsilon_0 = Permeability constant

A = Cross-sectional Area

d = Diameter

Potential difference between the two disks,

V = Ed

Where,

E = Electric field

d = diameter

Q = Charge on the disk equal to \rightarrow Q=ne=(3.9*10^9)(1.6*10^{-19})= 6.24*10^{-10}C

Through the value found and the expression given for capacitance and potential, we can define the electric charge as

Q = CV

Q = \frac{\epsilon A}{d}(Ed)

Q = \epsilon_0 AE

Q = \epsilon_0 \pi(\frac{d}{2})^2E

Q = \frac{\epsilon \pi d^2E}{4}

Re-arranging the equation to find the diameter of the disks, the equation will be:

d = \sqrt{\frac{4D}{\epsilon_0 \pi E}}

Replacing,

d = \sqrt{\frac{4(6.24*10^{-10})}{(8.85*10^{-12})\pi(1*10^{5})}}

d = 0.0299m

Therefore the diameter of the disks is 0.03m

8 0
3 years ago
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