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just olya [345]
3 years ago
6

Suppose the displacement of an object is related to time according to the expression x=By*2, what are the dimensions of B

Physics
1 answer:
laiz [17]3 years ago
5 0

Answer:

3

Explanation:

AP3X

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The correct answer is C.
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look at the circuit in the figure. find the current, voltage, and power in each resistor. please list answer
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fhcjctfkbraf gdovtckcrhha6g

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Marcus drove his Honda Prelude for 4 hours at a rate of 55 miles per hour. How far did he travel?
galben [10]
To solve this, you’d multiply 55 by 4, because he is travelling 55 miles every hour, for four hours, which means 55 miles every hour. The answer would be 220.
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Read 2 more answers
A bullet with mass m = 5.21 g is moving horizontally with a speed v = 443 m/s when it strikes a block of hardened steel with mas
AlladinOne [14]

Answer:

0.312 m/s

Explanation:

Elastic collisions conserve momentum and kinetic energy

The velocity of the center of mass will not change. It continues at

0.00521(443) / 14.80521 = 0.155893... ≈ 0.156 m/s

To conserve kinetic energy we can think of the center of mass (CoM) as an ideal spring returning to each mass that strikes it an identical speed of collision in the opposite direction.

The CoM sees the target approach at - 0.156 and will see it depart at 0.156 m/s

A ground based observer sees the target depart at the velocity of the CoM plus the relative velocity .

v = 0.156 + 0.156  = 0.312 m/s

6 0
2 years ago
Four copper wires of equal length are connected in series. Their cross-sectional areas are 1.6 cm2 , 1.2 cm2 , 4.4 cm2 , and 7 c
EleoNora [17]

Answer:

63.8 V

Explanation:

We are given that

A_1=1.6 cm^2=1.6\times 10^{-4} m^2

1 cm^2=10^{-4} m^2

A_2=1.2 cm^2=1.2\times 10^{-4} m^2

A_3=4.4 cm^2=4.4\times 10^{-4} m^2

A_4=7 cm^2=7\times 10^{-4} m^2

Potential difference,V=140 V

We know that

R=\frac{\rho l}{A}

According to question

l_1=l_2=l_3=l_4=l

In series

R=R_1+R_2+R_3+R_4

R=\rho l(\frac{1}{A_1}+\frac{1}{A_2}+\frac{1}{A_3}+\frac{1}{A_4})

R=\rho l(\frac{1}{1.6\times 10^{-4}}+\frac{1}{1.2\times 10^{-4}}+\frac{1}{4.4\times 10^{-4}}+\frac{1}{7\times 10^{-4}})

R=\rho l(18284.6)

I=\frac{V}{R}=\frac{140}{\rho l\times 18284.6}

Potential across 1.2 square cm=V_1=IR_1=\frac{140}{\rho l\times 18284.6}\times \rho l(\frac{1}{1.2\times 10^{-4}}=63.8 V

Hence, the voltage across the 1.2 square cm wire=63.8 V

3 0
3 years ago
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