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3 years ago

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Objects have the same velocity only if they are moving the same speed and in the same direction. objects moving at different spe

eds, in different directions or both have different velocities.

1 answer:

EastWind [94]3 years ago

7 0

Answer:

Hey

Yes i think you are correct.

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Look at the circuit diagram. Which of these components is part of the circuit?

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I think the answer is c AC power source
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3 years ago

Two bicycle tires are set rolling with the same initial speed of 3.30 m/s along a long, straight road, and the distance each tra

vredina [299]

Answer:

At low pressure- $\mu_{k}=0.02315$

At high pressure- $\mu_{k}=0.00445$

Explanation:

Initial speed, $V_{i}=3.3 m/s$

Final speed, $V_{f}=3.3/2= 1.65 m/s$

Net horizontal force due to rolling friction $F_{net}=\mu_{k}$ mg where m is mass, g is acceleration due to gravity, $\mu_{k}$ is coefficient of rolling friction

From kinematic relation, $V_{f}^{2}- V_{i}^{2}=2ad$

For each tire,

$V_{f}^{2}- V_{i}^{2}=2\mu_{k}gd$

Making $\mu_{k}$ the subject

$\mu_{k}=\frac {V_{f}^{2}- V_{i}^{2}}{2gd}$

Under low pressure of 40 Psi, d=18 m

$\mu_{k}=\frac {1.65^{2}- 3.3^{2}}{2*9.8*18}=-0.02315$

Therefore, $\mu_{k}=0.02315$

At a pressure of 105 Psi, d=93.7

$\mu_{k}=\frac {1.65^{2}- 3.3^{2}}{2*9.8*93.7}=-0.00445$

Therefore, $\mu_{k}=0.00445$

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3 years ago

The relationship between distance from the sun and orbital period is that as the distance from the sun increases, the orbital pe

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Second law is it:))))))))

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3 years ago

Using your Periodic Table, which element below has the smallest atomic radius? A.) Sodium, B.) Chlorine, C.) Phosphorus, D.) Iro

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Explanation:

Chlorine is the smallest atomic radius

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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