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RideAnS [48]
3 years ago
9

Suppose a treadmill has an average acceleration of 4.7x10^-3 m/s. a)how much does its speed change after 5min? b)if the treadmil

l's speed 1.7 m/s, what will its final speed be?
Physics
1 answer:
Mars2501 [29]3 years ago
8 0

Answer:

a)Change in the speed  = 1.41 m/s

b)The final speed will be 3.11 m/s

Explanation:

Given that

Acceleration ,a= 4.7 x 10⁻³ m/s²

a)

We know that

v= u + a t

v=final speed ,u=initial speed

t= time ,a= acceleration

Change in the speed

v- u = a t

t= 5 min  = 5 x 60 s = 300 s

v- u = 4.7 x 10⁻³ x 5 x 60 m/s

v-u = 1.41 m/s

Change in the speed  = 1.41 m/s

b)

Given that

u= 1.7 m/s

v-u = 1.41 m/s

v= 1.7 + 1.41 m/s

v=3.11 m/s

The final speed will be 3.11 m/s

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A 2000 kg car slams on the brakes and slows down at a rate of -10 m/s2. How much force are the brakes applying?
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2 years ago
jet is flying at 500 mph east relative to the ground. A Cessna is flying at 150 mph 60° north of west relative to the ground. Wh
Greeley [361]

Answer:

C. 590 mph

\vert v_{cj}\vert=589.49\ mph

Explanation:

Given:

  • velocity of jet, v_j=500\ mph
  • direction of velocity of jet, east relative to the ground
  • velocity of Cessna, v_c=150\ mph
  • direction of velocity of Cessna, 60° north of west

Taking the x-axis alignment towards east and hence we have the velocity vector of the jet as reference.

Refer the attached schematic.

So,

\vec v_j=500\ \hat i\ mph

&

\vec v_c=150\times (\cos120\ \hat i+\sin120\ \hat j)

\vec v_c=-75\ \hat i+75\sqrt{3}\ \hat j\ mph

Now the vector of relative velocity of Cessna with respect to jet:

\vec v_{cj}=\vec v_j-\vec v_c

\vec v_{cj}=500\ \hat i-(-75\ \hat i+75\sqrt{3}\ \hat j )

\vec v_{cj}=575\ \hat i-75\sqrt{3}\ \hat j\ mph

Now the magnitude of this velocity:

\vert v_{cj}\vert=\sqrt{(575)^2+(75\sqrt{3} )^2}

\vert v_{cj}\vert=589.49\ mph is the relative velocity of Cessna with respect to the jet.

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