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12345 [234]
2 years ago
13

Complete each statement about the sign of the work done on a baseball. Carlton catches a baseball and his hand moves backward as

he brings the ball to a halt, doing work on the baseball. Carlton throws a baseball, doing work on the baseball. Carlton carries a baseball in his hand as he rides his bicycle in a straight line at constant speed, doing work on the baseball. Carlton carries a baseball in his hand as he rides his bicycle in a straight line with increasing speed, doing work on the baseball.
Physics
1 answer:
dezoksy [38]2 years ago
8 0

Answer and Explanation:

  • The work done is negative as the force is applied in the forward direction and the displacement of the hand is in the backward direction.
  • The work done is positive as the work is done on the baseball by Carlton. Moreover, both the force applied and the displacement of Carlton are in a straight line in the forward direction.
  • The work done in this case is zero as Carlton rides with constant speed, the Kinetic energy too remains constant and hence net work done is zero.
  • The work done in this case is positive as the work is done on the baseball by Carlton and as the velocity changes its Kinetic energy also varies wit it.
  • Also  the force applied and the displacement of Carlton in this case are in the directed to the same path.
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A penny falls from a windowsill which is 25.0 m above the sidewalk. How much time does a passerby on the sidewalk below have to
Yakvenalex [24]
T = √(h)/(0.5)(9.81)
t = √(25)/(4.905)
t = √5.1
t = 2.26 seconds

hope this helps and have a great day :)



3 0
2 years ago
The drawing shows three identical springs hanging from the ceiling. Nothing is attached to the first spring, whereas a 4.50-N bl
Pavlova-9 [17]

Answer:

a. 30 N / m

b. 9.0 N

Explanation:

Given that

Unstretched length of the spring, L_o = 20.0cm = 0.2m

a) When the mass of 4.5N is hanging from the second spring, then extended length Is

L_1 = 35.0cm =  0.35m

So, the change in spring length when mass hangs is

x =  L_1 - L_o

= (0.35 - 0.20) m

= 0.15m

As spring are identical

Let us assume that the spring constant be "k", so at equilibrium

Restoring Force on spring = Block weightage

kx =  W =  4.50

k= \frac{4.50}{x} = \frac{4.50}{0.15}

= 30 N / m

b)  Now for the third spring, stretched the length of spring is

L_2 = 50cm = 0.5m

So, the change in spring length is

x'= L_2 - L_o

= (0.5-0.20)m

=  0.30m

At equilibrium,

Restoring Force on spring = Block weightage

Now using all mentioned and computed values in above,

W'= kx'

= 30(0.3)

= 9.0 N

4 0
2 years ago
While playing basketball in PE class, Logan lost his balance after making a lay-up and colliding with the padded wall behind the
olga2289 [7]

Answer:

a.) F = 3515 N

b.) F = 140600 N

Explanation: given that the

Mass M = 74kg

Initial velocity U = 7.6 m/s

Time t = 0.16 s

Force F = change in momentum ÷ time

F = (74×7.6)/0.16

F = 3515 N

b.) If Logan had hit the concrete wall moving at the same speed, his momentum would have been reduced to zero in 0.0080 seconds

Change in momentum = 74×7.6 + 74×7.6

Change in momentum = 562.4 + 562.4 = 1124.8 kgm/s

F = 1124.8/0.0080 = 140600 N

6 0
2 years ago
Assume that the mass has been moving along its circular path for some time. You start timing its motion with a stopwatch when it
lesya692 [45]

Answer:

v = R\omega(-sin\omega t \hat i + cos\omega t \hat j)

Explanation:

As we know that the mass is revolving with constant angular speed in the circle of radius R

So we will have

\theta = \omega t

now the position vector at a given time is

r = Rcos\theta \hat i + R sin\theta \hat j

now the linear velocity is given as

v = \frac{dr}{dt}

v = (-R sin\theta \hat i + R cos\theta \hat j)\frac{d\theta}{dt}

v = R\omega(-sin\omega t \hat i + cos\omega t \hat j)

6 0
1 year ago
The diameters of bolts produced by a certain machine are normally distributed with a mean of 0.30 inches and a standard deviatio
enot [183]

Answer:

2.25 %

Explanation:

65-95-99.7 is a rule to remember the precentages that lies around the mean.

at the range of mean (\mu) plus or minus one standard deviation (\sigma), P([\mu-\sigma \leq X \leq \mu+\sigma])\approx 68.3%

at the range of mean plus or minus two standard deviation, P([\mu -2\sigma \leq X \leq \mu+2\sigma])\approx 95.5%

at the range of mean plus or minus three standard deviation, P([\mu - 3\sigma\leq X \leq \mu+3\sigma])\approx 99.7%

So, note that  they are asking about the probability that it is greater than 0.32, that is the mean (0.3) plus two times the standard deviation (0.1) (P(X \leq \mu+2\sigma))  

So we know that the 95.5% is between \mu - 2\sigma = 0.3 -2*0.1 = 0.28 and \mu + 2\sigma = 0.3 +2*0.1 = 0.32, hence approximately the 4.5% (100%-95.5%) is greater than 0.32 or less than  0.28. But half (4.5%/2=2.25%) is greater than 0.32 and the other half is less than 0.28.

So P(X \leq \mu+2\sigma) \approx 2.25%

8 0
1 year ago
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