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Travka [436]
3 years ago
6

A child rides her bike at a rate of 12.0 km/hr down the street. A squirrel suddenly runs in front of her so she applies the brak

es and slows to 8.0 km/hr in 0.25 sec. What is the acceleration of the bike rider during this period?
Physics
2 answers:
olya-2409 [2.1K]3 years ago
6 0

it should be 16 kilometers per second

Sedbober [7]3 years ago
3 0
By v = u - at 
<span>=>8 = 12 - a x 0.25 </span>
<span>=>a = 4/0.25 km/hr/sec </span>
<span>=>a = 16km/hr/sec

I hope this helped!</span>
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Exercises
Crank

\\ \rm\Rrightarrow \dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}

\\ \rm\Rrightarrow \dfrac{1}{u}=\dfrac{1}{-10}+\dfrac{1}{38}

\\ \rm\Rrightarrow \dfrac{1}{u}=\dfrac{-19+5}{190}

\\ \rm\Rrightarrow \dfrac{1}{u}=\dfrac{-14}{190}

\\ \rm\Rrightarrow u=\dfrac{190}{-14}

\\ \rm\Rrightarrow u=13.6cm

Real

5 0
2 years ago
A particle with positive charge q = 9.61 10-19 C moves with a velocity v = (5î + 4ĵ − k) m/s through a region where both a unifo
user100 [1]

Answer:

\vec{F}=(1.537*10^{-17}\hat{i}-9.61*10^{-19}\hat{j}-2.883*10^{-18}\hat{k})N

Explanation:

The total force on the particle is given by

\vec{F}=q\vec{v}\ X\ \vec{B}+q\vec{E}

Then, by replacing we have:

q\vec{v}\ X \vec{B}=q[7\hat{k}-9\hat{j}-\hat{k}]\\\\q\vec{E}=q[5\hat{i}-\hat{j}-2\hat{k}]\\\\\vec{F}=(9.61*10^{-19}C)[(7+9)\hat{i}+(-9-1)\hat{j}+(-1-2)\hat{k}]\\\\\vec{F}=(1.537*10^{-17}\hat{i}-9.61*10^{-19}\hat{j}-2.883*10^{-18}\hat{k})N

where the cross product can be made with the determinant method.

Hope this helps!!

3 0
3 years ago
Read 2 more answers
How do you find the water pressure at the bottom of the 55-m-high water tower?
svp [43]
-- What's the volume of a cylinder with radius=1m and height=55m ?

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3 years ago
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pickupchik [31]

Answer:

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Explanation:

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Look at the rock sitting on the hill in the picture above. Gravity should make the rock slide down the hill. What force is acting to balance gravity,keeping the rock in place? - D. friction
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5 0
3 years ago
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