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Katen [24]
3 years ago
8

A robot on a basketball court needs to throw the ball in the hoop. The ball have a velocity of 10 m/s and an an angle of 60 degr

ees. The high of hoop is 0.75 meters and the horizontal distance between the robot and hoop is 5 meters. A) draw a digram.
B)What is the high that the robot needs to throw the ball from.

Physics
1 answer:
IRISSAK [1]3 years ago
5 0

Answer:

Initial velocity = 10 m/s

θ = 60°

This is the case of projectile motion

So the horizontal component of velocity 10 m/s  =  10 cosθ

u = 10 cosθ

u = 10 cos 60°

u=5 m/s

x= 5 m

So in the horizontal direction

x = u .t

5 = 5 .t

t = 1 sec  The vertical component of velocity 10 m/s = 10 sinθ

Vo= 10 sinθ

Vo= 10 sin 60°

Vo = 8.66 m/s

V^2=V_o^2+2gh

0=V_o^2-2gh

0=8.66^2-2\times 10\times h

h=3.75 m

So height of robot = 3.75 - 0.75 m

   height of robot =3 m

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Answer:

Choice a. 1 kg, assuming that all other forces on the object (if any) are balanced.

Explanation:

By Newton's Second Law,

\displaystyle a = \frac{\Sigma F}{m},

where

  • a is the acceleration of the object in \text{m}\cdot\text{s}^{-2},
  • \Sigma F is the net force on the object in Newtons, and
  • m is the mass of the object in kilograms.

As a result,

\displaystyle m = \frac{\Sigma F}{a}.

Assume that all other forces on this object are balanced. The net force on the object will be 100\;\text{N}. The net force is constant. Acceleration should also be constant and the same as the average acceleration in the two seconds.

<h3>What is the average acceleration of this object?</h3>

\displaystyle \begin{aligned}\text{Acceleration} &= \text{Average Acceleration}=\frac{\text{Change in Velocity}}{\text{Time Taken}}\end{aligned}.

\displaystyle {a} = \frac{200\;\text{m}\cdot\text{s}^{-1}}{2\;\text{s}}=100\;\text{m}\cdot\text{s}^{-2}.

<h3>Apply Newton's Second Law to find the mass of the object.</h3>

\displaystyle m = \frac{\Sigma F}{a} = \frac{100\;\text{N}}{100\;\text{m}\cdot\text{s}^{-2}} = 1\;\text{kg}.

6 0
3 years ago
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Of waterfalls with a height of more than 50 m , Niagara Falls in Canada has the highest flow rate of any waterfall in the world.
Vinil7 [7]

Answer:

Power output: W=1426.9MW

Explanation:

The power output of the falls is given mainly by its change in potential energy:

Q=-P_{tot}=-(P_{2}-P_{1})

The potential energy for any point can be calculated as:

P=m*g*h

If we consider the base of the falls to be the reference height, at point 2 h=0, so P2=0, and height at point 1 equals 52m:

Q=P_{1}=m*g*h

If we replace m with the mass rate M we obtain the rate of change in potential energy over time, so the power generated:

W=M*g*h=2.8*10^{3}m^{3}/s*1*10^{3}kg/m^{3}*9.8m/s^{2}*52m =1426.9MW

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2 years ago
A 100-W lightbulb is placed in a cylinder equipped with a moveable piston. The lightbulb is turned on for 0.010 hour, and the as
Taya2010 [7]

Answer:

w =  - 508.53 joules

q = - 3091.47 joules

Explanation:

Let us convert the time in hours into seconds

0.010* 3600\\= 36

Change in internal energy

\delta E = p * \delta t

where E is the internal energy in Joules

p is the power in watts

and t is the time in seconds

\delta E = - 100 * 36\\

\delta E = - 3600 Joules

Amount of work done by the system

w = - P * \delta V

where P is the pressure and V is the volume

Substituting the given values in above equation, we get -

w = - 1 * ( 5.92 -0.90)\\

w = -5.02 liter-atmospheres

Work done in Joules

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q = \delta E - w\\

Substituting the given values we get -

q = - 3600 - (-508.53)\\q = - 3091.47

Thus

w =  - 508.53 joules

q = - 3091.47 joules

7 0
3 years ago
What is one way in which radioactive decay is used to benefit society?
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Answer:

the same    588 N

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