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STALIN [3.7K]
3 years ago
14

A football player throws and accelerates a ball (m = 0.150 kg) from rest to a velocity of 50 m/s in 0.0121 sec. Determine the ac

celeration of the ball first (this is not the answer) and then the force applied by the player's arm to the ball (enter this value). Neglect resistance forces.
Physics
1 answer:
Rasek [7]3 years ago
5 0

Answer:

Acceleration=4m/s²

Force applied =619.8N

Explanation:

Using equation of motion

V=u+at we have: u=o, v=50m/s

50= 0 + a×0.0121

a = 50/0.0121

a= 4m/s²

Neglecting resistance forces

F= ma, where a = v-u/t

F=m×(v-u)/t

F= 0.150 ×(50-0)/0.0121

F=7.5/0.0121

F= 619.8N

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Answer:

132-123+ere

Explanation:

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5 0
3 years ago
Read 2 more answers
Two metal disks, one with radius R1 = 2.45 cm and mass M1 = 0.900 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 k
larisa86 [58]

Answer:

part (a) a_1\ =\ 2.9\ kg

Part (b) a_2\ =\ 6.25\ kg

Explanation:

Given,

  • Mass of the larger disk = M_2\ =\ 1.60\ kg
  • Mass of the smaller disk = M_1\ =\ 0.900\ kg
  • Radius of the larger disk = R_2\ =\ 5.00\ cm\ =\ 0.05\ m
  • Radius of the smaller disk = R_1\ =\ 2.45\ cm\ =\ 0.0245\ m
  • Mass of the block = M = 1.60 kg

Both the disks are welded together, therefore total moment of inertia of the both disks are the summation of the individual moment of inertia of the disks.

\therefore I\ =\ I_1\ +\ I_2\\\Rightarrow I\ =\ \dfrac{1}{2}M_1R_1^2\ +\ \dfrac{1}{2}M_2R_2^2\\\Rightarrow I\ =\ \dfrac{1}{2} (0.9\times 0.0245^2\ +\ 1.60\times 0.05^2)\\\Rightarrow I\ =\ 2.27\times 10^{-3}\ kgm^2

part (a)

Given that a block of mass m which is hanging with the smaller disk,

Let 'T' be 'a' be the tension in the string and acceleration of the block.

From the free body diagram of the smaller block,

mg\ -\ T\ =\ ma\\\Rightarrow T\ =\ mg\ -\ ma\,\,\,\,eqn (1)

From the pulley,

\sum \tau\ =\ I\alpha\\\Rightarow T\times R_1\ =\ I\alpha\ =\ \dfrac{Ia}{R_1}\\\Rightarrow T\ =\ \dfrac{I\alpha}{R_1^2}\,\,\,eqn(2)

From the equation (1) and (2),

mg\ -\ ma\ =\ \dfrac{Ia}{R_1^2}\\\Rightarrow a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}{0.0245^2}}\ +\ 1.60}\\\Rightarow a\ =\ 2.91\ m/s^2

part (b)

Above expression for the acceleration of the block is only depended on the radius of the pulley.

Radius of the larger pulley = R_2\ =\ 0.05\ m

Let a_2 be the acceleration of the block while connecting to the larger pulley.\therefore a\ =\ \dfrac{mg}{\dfrac{I}{R_2^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}{0.05^2}\ +\ 1.60}}\\\Rightarow a\ =\ 6.25\ m/s^2

4 0
3 years ago
A Texas rancher wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A, B, and C in
xz_007 [3.2K]
The answer:
the full question is as follow:
 <span>A Texas rancher wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A, B, and C in Figure below , where A = 4.90 km and θC = 15°. He then correctly calculates the length and orientation of the fourth side D. What is the magnitude and direction of vector D? 

As shown in the figure, 
A + B + C + D = 0, so to find the </span>magnitude and direction of vector D, we should follow the following method:
 D = 0 - (A + B + C) , 
let  W = - (A + B + C), so the magnitude and direction of vector D is the same of the vector W characteristics

Magnitude
 A + B + C = <span> (4.90cos7.5 - 2.48sin16 - 3.02cos15)I</span>
<span>+ (-4.9sin7.5 + 2.48cos16 + 3.02sin15)J
</span>= 1.25I +2.53J
the magnitude of W= abs value of (A + B + C) = sqrt (1.25² + 2.53²)
= 2.82
 
the direction of D can be found by using Dx and Dy value
we know that     tan<span>θo = Dx / Dy = 1.25 / 2.53 =0.49
</span>tanθo =0.49 it implies θo = arctan 0.49 = 26.02°

direction is 26.02°

4 0
3 years ago
Your friend has informed you that his/her uncle has decided not to look after him anymore.Write to your friends uncle giving him
Lorico [155]

Answer:

Cause you are their family and they need you. Do you wanna be someone who abandoned them or be the one they look to and dont dislike into adulthood. Also you signed onto this job no one said it was going to be easy but you making your life easier but harder for someone who is just a kid who still needs you.

7 0
3 years ago
You drop a ball from a window located on an upper floor of a building. It strikes the ground with speed v. You now repeat the dr
bulgar [2K]

Answer:

 y = y₀ (1 - ½ g y₀ / v²)

Explanation:

This is a free fall problem. Let's start with the ball that is released from the window, with initial velocity vo = 0 and a height of the window i

          y = y₀ + v₀ t - ½ g t²

          y = y₀ - ½ g t²

for the ball thrown from the ground with initial velocity v₀₂ = v

         y₂ = y₀₂ + v₀₂ t - ½ g t²

     

in this case y₀ = 0

         y₂2 = v t - ½ g t²

at the point where the two balls meet, they have the same height

         y = y₂

         y₀ - ½ g t² = vt - ½ g t²

         y₀i = v t

         t = y₀ / v

since we have the time it takes to reach the point, we can substitute in either of the two equations to find the height

         y = y₀ - ½ g t²

         y = y₀ - ½ g (y₀ / v)²

         y = y₀ - ½ g y₀² / v²

        y = y₀ (1 - ½ g y₀ / v²)

with this expression we can find the meeting point of the two balls

6 0
4 years ago
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