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MrRa [10]
3 years ago
12

Michael is biking on a trail and is accelerating at a rate of 1.2 m/s/s for 15 seconds. He began this part of his ride with a ve

locity of 1.62 m/s. Determine Michael's final velocity?
Physics
1 answer:
Serhud [2]3 years ago
7 0

Answer:

Michael's final velocity is 19.62 m/s.    

Explanation:            

We can find the final velocity of Michael by using the following kinematic equation:

v_{f} = v_{0} + at   (1)    

Where:

v_{f}: is the final velocity =?

v_{0}: is the initial velocity = 1.62 m/s

a: is the acceleration = 1.2 m/s²

t: is the time = 15 s

By entering the above values into equation (1) we have:

v_{f} = 1.62 m/s + 1.2 m/s^{2}*15 s

v_{f} = 19.62 m/s

Therefore, Michael's final velocity is 19.62 m/s.

I hope it helps you!                                            

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Two rocks are at the top of a building. Rock 1 is dropped from rest while Rock 2 is thrown horizontaly at a velocity of 5 ms.
LuckyWell [14K]

Answer:

I'm pretty sure the answer is 0 m/s²

Explanation:

The horizontal velocity of the second rock is 5 m/s, so if we pretend air resistance doesn't exist, it will maintain that horizontal velocity, meaning that there is no horizontal acceleration.

7 0
3 years ago
Determine the maximum shearing stress in a solid shaft of 1.5-in. diameter as it transmits 75 hp at a speed of 1800 rpm. (Round
erma4kov [3.2K]

Answer:

\tau=3.96\ ksi

Explanation:

Given that

d= 1.5 in                      ( 1 in = 0.0254 m)

d= 0.0381 m

P= 75 hp                      ( 1 hp = 745.7 W)

P= 55927.5 W

N= 1800 rpm

We know that power P is given as

P=\dfrac{2\pi N\ T}{60}

T=Torque

N=Speed

55927.5=\dfrac{2\times \pi \times 1800\ T}{60}

T=296.85 N.m

The maximum shear stress is given as

\tau=\dfrac{16 T}{\pi d^3}

\tau=\dfrac{16\times 296.85}{\pi \times 0.0381^3}

\tau=27.35\ MPa

We know that 1 MPa =0.145 ksi

\tau=3.96\ ksi

3 0
4 years ago
Which of the following are not vector directions?
Svetradugi [14.3K]

Answer:

here north are not vector option b hope ur help

6 0
3 years ago
Read 2 more answers
A one-piece cylinder has a core section protruding from the larger drum and is free to rotate around its central axis. A rope wr
PilotLPTM [1.2K]

Answer:

Magnitude the net torque about its axis of rotation is 2.41 Nm

Solution:

As per the question:

The radius of the wrapped rope around the drum, r = 1.33 m

Force applied to the right side of the drum, F = 4.35 N

The radius of the rope wrapped around the core, r' = 0.51 m

Force on the cylinder in the downward direction, F' = 6.62 N

Now, the magnitude of the net torque is given by:

\tau_{net} = \tau + \tau'

where

\tau = Torque due to Force, F

\tau' = Torque due to Force, F'

tau = F\times r

tau' = F'\times r'

Now,

\tau_{net} = - F\times r + F'\times r'

\tau_{net} = - 4.35\times 1.33 + 6.62\times 0.51 = - 2.41\ Nm

The net torque comes out to be negative, this shows that rotation of cylinder is in the clockwise direction from its stationary position.

Now, the magnitude of the net torque:

|\tau_{net}| = 2.41\ Nm

 

3 0
3 years ago
Need help please in physics ​
AnnZ [28]

Answer:

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