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kogti [31]
3 years ago
6

A block is projected up a frictionless inclined plane with initial speed v0 = 1.72 m/s. The angle of incline is θ = 44.8°. (a) H

ow far up the plane does it go? (b) How long does it take to get there? (c) What is its speed when it gets back to the bottom?
Physics
1 answer:
Snowcat [4.5K]3 years ago
3 0

Explanation:

Given

initial velocity(v_0)=1.72 m/s

\theta =44.8{\circ}

using v^2-u^2=2as

Where v=final velocity (Here v=0)

u=initial velocity(1.72 m/s)

a=acceleration   (gsin\theta )

s=distance traveled

0-(1.72)^2=2(-9.81\times sin(44.8))s

s=0.214 m

(b)time taken to travel 0.214 m

v=u+at

0=1.72-gsin(44.8)\times t

t=\frac{1.72}{9.8\times sin(44.8)}

t=0.249 s

(c)Speed of the block at bottom

v^2-u^2=2as

Here u=0 as it started coming downward

v^2=2\times gsin(44.8)\times 0.214

v=\sqrt{2.985}

v=1.72 m/s

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10.4 N

Given

m = 1.10 kg

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g = 9.81 m/s2

Solution

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cosθ = mgcosθ

Fn = (1.10 kg)(9.81 m/s2

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3 0
3 years ago
the velocity of a car traveling in the positive direction decreases from 32 m/s to 24 m/s in 4 seconds. what is the average acce
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Answer:

a=-2m/sec^2

Negative sign shows that velocity of the car is decreases at a constant rate

Explanation:

We have given velocity of the car is decreases from 32 m /sec to 24 m/sec in 4 second

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And finally car reaches to a velocity of 24 m/sec

Time taken to change in velocity = 4 sec

So final velocity v = 24 m/sec

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So 24=32+a\times 4

a=-2m/sec^2

Negative sign shows that velocity of the car is decreases at a constant rate

6 0
3 years ago
At the local playground, a 21-kg child sits on the right end of a horizontal teeter-totter, 1.8 m from the pivot point. On the l
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Answer:

By convention a negative torque leads to clockwise rotation and a positive torque leads to counterclockwise rotation.

here weight of the child =21kgx9.8m/s2 = 205.8N

the torque exerted by the child Tc = - (1.8)(205.8) = -370.44N-m ,negative sign is inserted because this torque is clockwise and is therefore negative by convention.

torque exerted by adult Ta = 3(151) = 453N , counterclockwise torque.

net torque Tnet = -370.44+453 =82.56N , which is positive means counterclockwise rotation.

b) Ta = 2.5x151 = 377.5N-m

Tnet = -370.44+377.5 = 7.06N-m , positive ,counterclockwise rotation.

c)Ta = 2x151 = 302N-m

Tnet = -370.44+302 = -68.44N-m, negative,clockwise rotation.

5 0
3 years ago
A particle with a charge of 2e moves between two points which have a potential difference of 75V. What is the change in potentia
Sonbull [250]
Electric potential energy is defined as Ep=Q*V where Q is the magnitude of the charge and V is the potential difference. So when a charge moves between the points that have a potential difference, it's energy changes. 

In our case: 

Q=2e=2*(-1.6*10^-19) C
V=75 V

Ep=(-3.2*10^-19)*75

Ep=-2.4*10^-17 J

The change in potential energy of the charge is -2.4*10^-17 J 
5 0
3 years ago
A bucket that has a mass of 20 kg when filled with sand needs to be lifted to the top of a 15 meter tall building. You have a ro
prohojiy [21]

Answer:

work done lifting the bucket (sand and rope) to the top of the building,

W=67.46 Nm

Explanation:

in this question we have given

mass of bucket=20kg

mass of rope=.2\frac{kg}{m}

height of building= 15 meter

We have to find the work done lifting the bucket (sand and rope) to the building =work done in lifting the rope + work done in lifting the sand

work done in lifting the rope is given as,W_{1}=Force \times displacement

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work done in lifting the sand is given as,W_{2}=Force \times displacement

W_{2}=\int\limits^{15}_0 F \, dx.................(2)

Here,

F=mx+c

here,

c=20-18

c=2

m=\frac{20-18}{15-0}

m=.133

Therefore,

F=.133x+2

Put value of F in equation 2

W_{2}=\int\limits^{15}_0 (.133x+2) \, dx

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W=22.5 Nm+44.96 Nm

W=67.46 Nm

4 0
3 years ago
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