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Katen [24]
2 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]2 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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Explanation:

(A) The force of kinetic friction is smaller than that of static friction, but Fg remains the same

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A toaster draws 8 A of current with a voltage of 120 V. Which is the power used by the toaster
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Power = (voltage) x (current) =

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2 years ago
Which of the following types of radiation can penetrate through paper but not through wood?
andrew-mc [135]
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4 0
3 years ago
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A car travels 200km in 3.0 hours. Determine the average velocity of the car
Masja [62]

Answer:

<h2>66.67 km/hr</h2>

Explanation:

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a =  \frac{d}{t }  \\

d is the distance

t is the time taken

From the question we have

a =  \frac{200}{3}  \\  = 66.66666...

We have the final answer as

<h3>66.67 km/hr</h3>

Hope this helps you

4 0
2 years ago
g Two masses are involved in a collision on an axis (one dimensional). One mass is six times the mass of the second. Both masses
statuscvo [17]

Answer:

v₁f = 0.5714 m/s   (→)

v₂f = 2.5714 m/s   (→)

e = 1  

It was a perfectly elastic collision.

Explanation:

m₁ = m

m₂ = 6m₁ = 6m

v₁i = 4 m/s

v₂i = 2 m/s

v₁f = ((m₁ – m₂) / (m₁ + m₂)) v₁i +  ((2m₂) / (m₁ + m₂)) v₂i

v₁f = ((m – 6m) / (m + 6m)) * (4) +  ((2*6m) / (m + 6m)) * (2)  

v₁f = 0.5714 m/s   (→)

v₂f = ((2m₁) / (m₁ + m₂)) v₁i +  ((m₂ – m₁) / (m₁ + m₂)) v₂i

v₂f = ((2m) / (m + 6m)) * (4) + ((6m -m) / (m + 6m)) * (2)

v₂f = 2.5714 m/s   (→)

e = - (v₁f - v₂f) / (v₁i - v₂i)   ⇒   e = - (0.5714 - 2.5714) / (4 - 2) = 1  

It was a perfectly elastic collision.

8 0
2 years ago
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