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GREYUIT [131]
3 years ago
8

Billy pulls his friend Sue on a sled at a constant velocity with a horizontal force of 200 N. The weight of the sled and Sue is

400 N. Which way will the sled move if Billy suddenly starts to pull with a force of 600 N?
Physics
1 answer:
Nezavi [6.7K]3 years ago
6 0
Pulls would be the logical answer.
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An elephant travels 16metes to the left in 20s, what is the average velocity?​
SCORPION-xisa [38]

Answer:

S=16metes

t=20sec

V=?

V=S/t

V=16/20

V=0.8m/s

4 0
3 years ago
An 19-cm-long bicycle crank arm, with a pedal at one end, is attached to a 23-cm-diameter sprocket, the toothed disk around whic
Cloud [144]

Answer:

The tangential acceleration of the pedal is 0.0301 m/s².

Explanation:

Given that,

Length = 19 cm

Diameter = 23 cm

Time = 10 sec

Initial angular velocity = 65 rpm

Final velocity = 90 rpm

Suppose we need to find the tangential acceleration of the pedal

We need to calculate the tangential acceleration of the pedal

Using formula of tangential acceleration

a_{t}=r\alpha

a_{t}=\dfrac{23\times10^{-2}}{2}\times\dfrac{\omega_{2}-\omega_{1}}{t}

a_{t}=\dfrac{23\times10^{-2}}{2}\times\dfrac{90\times\dfrac{2]pi}{60}-65\times\dfrac{2\pi}{60}}{10}

a_{t}=0.0301\ m/s^2

Hence, The tangential acceleration of the pedal is 0.0301 m/s².

4 0
3 years ago
A shunting engine in a rail yard is working on a straight east-west section of line. First it travels west for 150 s with an ave
Ymorist [56]

Displacement-1

\\ \rm\longmapsto Velocity(Time)

\\ \rm\longmapsto 150(4)

\\ \rm\longmapsto 600m

Displcaement-2

\\ \rm\longmapsto 200(3)

\\ \rm\longmapsto 600m

Total displacement=600+600=1200m

7 0
3 years ago
Read 2 more answers
Am i pertty and who do think is going to win the presidential election who do u want to win and why
rodikova [14]
Honestly it doesn’t matter to me
8 0
3 years ago
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A student connects a 1 hp motor to a bicycle. How much time will it take for the bicycle to accelerate from rest to a speed of 5
bija089 [108]

Answer:

t=2s

Explanation:

The definition of power is:

P=\frac{W}{t}

And the work-energy theorem states that:

W=\Delta K

Since the movement starts from rest, we have that:

\Delta K=\frac{mv_f^2}{2}-\frac{mv_0^2}{2}=\frac{mv_f^2}{2}

And putting all together:

P=\frac{mv_f^2}{2t}

Since we want the time taken:

t=\frac{mv_f^2}{2P}

Which for our values is:

t=\frac{(120kg)(5m/s)^2}{2(746W)}=2s

7 0
3 years ago
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