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likoan [24]
3 years ago
9

You can only score 1 point when shooting a free throw. Group of answer choices True False

Physics
2 answers:
Arte-miy333 [17]3 years ago
8 0

Answer:

true

Explanation:

When shooting a free throw you can only receive 1 point for the team

emmainna [20.7K]3 years ago
4 0

Answer:

2 points

Explanation:

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True or false Digitized information can be used by computers.
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True! ^-^ It can be used!
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A uniform solid cylinder with a radius of 10 cm and a mass of 3.0 kg is rotating about its center with an angular speed of 33.4
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Is the equation for Ec=1/2 m(Dv)^2 where Dv is the difference between the angular speed & the areolar speed?

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A soccer ball is traveling at a velocity of 50 m/s. The kinetic energy of the ball is 500 J. What is the mass of the soccer ball
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A soccer ball is traveling at a velocity of 50 m/s. The kinetic energy of the ball is 500 J.The mass of the soccer ball is 0.4 kgs. This answer is derived from the formula K=1/2 MV^2.So velocity and kinetic energy are given from that mass of the ball is calculated.By substituting the values 500=1/2*M*50*50 which gives M=0.4 Kgs.<span>
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4 0
4 years ago
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What is your weight on Venus, assuming that the acceleration due to gravity on Venus is 8.875 m/s^2 and your mass is 50 kg.
Volgvan

Answer:

Weight on Venus = 443.75 N

Explanation:

Weight of a body is the product of mass and acceleration due to gravity.

So we have

       Weight =  Mass x Acceleration due to gravity

        W = mg

        Mass, m = 50 kg

        Acceleration due to gravity, g = 8.875 m/s²

        W = 50 x 8.875 = 443.75 N

Weight on Venus = 443.75 N

3 0
3 years ago
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A luggage handler pulls a 20.0 kg suitcase up a ramp inclined at 25 degrees above the horizontal by a force F of magnitude 145 N
Nonamiya [84]

Answer:

A) 667 J

B) 381.4 J

C) 0 J

D) 245.4 J

E) 40.2J

F) 2 m/s

Explanation:

Let g = 9.81 m/s2

A) The work done on the suitcase is the product of the force applied and the distance travelled:

w = Fs = 145 * 4.6 = 667 J

B) The work done by gravitational force the dot product between the gravity vector and the distance vector

W_g = \vec{P}\vec{s} = mgs sin\alpha = 20*9.81*4.6*sin25^o = 381.4 J

C) As the normal force vector is perpendicular to the distance vector, the work done by the normal force is 0

D) The work done on the suitcase by friction force is the product of the force applied and the distance travelled, whereas friction force is the product of normal force and coefficient

W_f = F_fs = \mu N s = \mu s mgcos\alpha = 0.3* 4.6 * 20*9.81*cos25^o = 245.4 J

E) The total workdone on the suite case would be the pulling work subtracted by gravity work and friction work

W = w – W_g – W_f = 667 – 381.4 – 245.4 = 40.2 J

F) As the suit case has 0 kinetic and potential energy at the bottom, and the total work done is converted to kinetic energy at 4.6 m along the ramp, we can conclude that:

E_k = W = 40.2 j

mv^2/2 = 40.2

20v^2/2 = 40.2

10v^2 = 40.2

v^2 = 4.02

v = \sqrt{4.02} = 2 m/s  

3 0
3 years ago
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