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3 years ago

8

The speed of a boat undergoing constant acceleration increases from 5 meters per second to 15 meters per second over a period of

20 seconds. How far does the boat travel during this time?

1 answer:

defon3 years ago

4 0

Should be 250m. check with your teacher and let me know

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Now that you are familiar with MRI's, nanotechnology and micro-bots, use your imagination to brainstorm other probable invention

dlinn [17]

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<u>Explanation:</u>

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3 years ago

Please Help Quick ASAP Hurry This is Physical Science

krok68 [10]

Answer:

Think it is C

Explanation:

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3 years ago

A car traveling at 20 m/s when the driver sees a child standing in the road. He takes 0.80 s to react, then steps on the brakes

mr Goodwill [35]

When driver see the child standing on road his speed is 20 m/s

So here at that instant his reaction time is 0.80 s

He will cover a total distance given by product of speed and time

$d_1 = v* t$

$d_1 = 20 * 0.8$

$d_1 = 16 m$

now after this he will apply brakes with acceleration a = 7 m/s^2

so the distance covered before it stop is given by

$v_f^2 - v_i^2 = 2 a d$

$0 - 20^2 = 2*(-7)*d_2$

$d_2 = 28.6 m$

so the total distance covered by it

$d = d_1 + d_2$

$d = 16 + 28.6 = 44.6 m$

<em>so it will cover a total distance of 44.6 m</em>

3 0

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, $\mu_{k}=0.00445$

4 0

3 years ago

A car traveling in a straight line has a velocity of 6m/s at some instant. After 6.32s its velocity is 13.2m/s . What is the ave

NeX [460]

Average acceleration = (change in speed) / (time for the change) .

Average acceleration = (13.2 - 6) / (6.32) = 7.2 / 6.32 = about <em>1.139... m/s²</em> .

8 0

3 years ago