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3 years ago

If it takes 0.5 seconds to travel 5 Millimeters? What's the mph?

Physics

1 answer:

Angelina_Jolie [31]3 years ago

3 0

I think it would me 5/0.5 = 10mph

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3 years ago

The force you have to overcome to start an object moving is ____. A. Rolling friction c. Sliding friction b. Static friction d.

finlep [7]

Air resistance Please select the best

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4 years ago

A particle has a charge of +1.5 µC and moves from point A to point B, a distance of 0.15 m. The particle experiences a constant

kirill [66]

Answer:

Part a)

$F = 6 \times 10^{-3} N$

Direction of force is along the motion of charge

Part b)

$E = 4000 N/C$

direction of electric field is along the direction of motion

Explanation:

Part a)

As we know that the change in electric potential energy is equal to the work done by electric field

$W = EPE_A - EPE_B$

$W = 9.0 \times 10^{-4} J$

now from the equation of work done we know that

$W = F.d$

$(9.0 \times 10^{-4}) = F(0.15)$

$F = 6 \times 10^{-3} N$

Direction of force is along the motion of charge

Part b)

As we know the relation between electrostatic force and electric field given as

$F = qE$

$(6 \times 10^{-3}) = 1.5 \times 10^{-6} E$

$E = 4000 N/C$

direction of electric field is along the direction of motion

8 0

3 years ago

What does the dashes line represent

Yuki888 [10]

dashed lines<span> represent bonds away from the viewer.</span>

3 0

3 years ago

A brick falls to the ground. if the time for the collision of the brick and the ground is increased by a factor of 4, the force

melomori [17]

Answer:

By a factor of 1/4.

Explanation:

The impulse force that applies to an object undergoing rapid deceleration just before coming to a stop on the ground is given by the following formula,

$\\\begin{aligned}\\\small F &=\small \frac{\Delta (mV)}{\Delta T}\end{aligned}$

in which $\small \Delta (mV)$ , $\small \Delta t$ represent the change in momentum and the time taken for that change.

If one increases the time that is taken for the momentum change (which remains constant for this situation) by a factor 4 and if that new force is represented by $\small F_1$ , the following manipulation confirms the answer to this question.

$\begin{aligned}\\\small F_1 &=\small \frac{\Delta (mV)}{4\Delta t}\\\\&=\small \frac{1}{4}\times\bigg[\frac{\Delta (mV)}{\Delta t}\bigg]\\\\&=\small \frac{1}{4}F\end{aligned}$

Here $\small F$ is the force that was applied to the object previously.

#SPJ4

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2 years ago