answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
elena-14-01-66 [18.8K]
2 years ago
8

Some children are practicing catching baseballs. The coach tosses the ball into the air and hits the ball with his bat. The ball

travels nearly horizontally, directly at the short stop who manages to catch the line drive. Did the coach, via the bat, do any work on the ball as it was hit
Physics
2 answers:
vova2212 [387]2 years ago
8 0

Answer:

Work is defined as the amount of movement done by a force, and it can be calculated as:

W = f*d

where W is the work, f is the force and d is the distance.

So when the coach hits the ball, he is appliying some force in the ball (via the bat) that impulses the ball briefly (we can considerate that the force is only applied in a differential of time dt)

But we have a force applied and a displacement of the ball, so yes, the bat does work on the ball.

STALIN [3.7K]2 years ago
6 0

Answer:

Yes he did

Explanation:

Horizontally speaking, the ball didn't have any speed prior to the bat hit by the coach. After the hit, the ball has horizontal speed, or kinetic energy. This energy must come from the work performed by the coach during the hit. This work is the product of the hit force and and distance it travels during the brief time the ball makes contact with the bat.

You might be interested in
Heat engines were first envisioned and built during the industrial revolution. Explain the thermodynamics of a heat engine comme
Artyom0805 [142]

Heat engines were developed during industrial revolution.

Generally a heat engine contains three parts i.e source, sink and working substance.

The source of a heat engine is present at a higher temperature as compared to the sink. Due to the temperature difference, the heat will flow from source to sink through working substance.

Let us consider  T_{1}\ and\ T_{2} are the temperature of source and sink.

As the source is at higher temperature as compared to sink, heat will flow from source to sink.

Let\ Q_{1}\ and\ Q_{2} are the heat provided by source and heat rejected to sink.

Hence, the work done by the working substance will be -

                                                W\ =\ Q_{1}-Q_{2}

The efficiency of a heat engine is defined as the ratio of output to the input energy.

Here output = workdone [W]

Hence, the efficiency of a heat engine is calculated as -

                     Efficiency\ [\eta]=\frac{W}{Q_{1}}

                                        \eta\ =\frac{Q_{1}- Q_{2}} {Q_{1}}

                                               =\ 1-\frac{Q_{2}} {Q_{1}}

This is the expression for the efficiency of heat engine.

Here, all the heat absorbed by the working substance can not be converted to desired output. The efficiency of a heat engine can not be 100 percent. Some amount of heat is lost in the form of sound and heat due to the friction which is produced due to the relative motion between various parts of the machine.

6 0
2 years ago
Read 2 more answers
Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant
rosijanka [135]

Answer:

v_f = 13m/s + 0.75 \frac{m}{s^2} * 16 s= 13 m/s +12m/s = 25 m/s

Explanation:

For this case we know that the initial velocity is given by:

v_i = 7 \frac{m}{s}

The final velocity on this case is given by:

v_f = 13 \frac{m}{s}

And we know that it takes 8 seconds to go from 7m/s to 13m/s. We can use the following kinematic formula in order to find the acceleration during the first interval:

v_f = v_i +at

If we solve for the acceleration we got:

a = \frac{v_f -v_i}{t} = \frac{13 m/s -7 m/s}{8 s}= 0.75 \frac{m}{s^2}

So for the other traject we assume that the acceleration is constant and the train travels for 16 s. The initial velocity on this case would be 13m/s from the first interval and we can find the final velocity with the following formula:

v_f = v_ i +a t

And if we replace we got:

v_f = 13m/s + 0.75 \frac{m}{s^2} * 16 s= 13 m/s +12m/s = 25 m/s

5 0
2 years ago
Three couples and two single individuals have been invited to an investment seminar and have agreed to attend. Suppose the proba
lions [1.4K]

Answer:

a) Probability mass function of x

x P(X=x)

0 0.0602

1 0.0908

2 0.1700

3 0.2050

4 0.1800

5 0.1550

6 0.0843

7 0.0390

8 0.0147

b) Cumulative Distribution function of X

x F(x)

0 0.0602

1 0.1510

2 0.3210

3 0.5260

4 0.7060

5 0.8610

6 0.9453

7 0.9843

8 1.0000

The cumulative distribution function gives 1.0000 as it should.

Explanation:

Probability of arriving late = 0.43

Probability of coming late = 0.57

Let's start with the probability P(X=0) that exactly 0 people arrive late, the probability P(X=1) that exactly 1 person arrives late, the probability P(X=2) that exactly 2 people arrive late, and so on up to the probability P(X=8) that 8 people arrive late.

Interpretation(s) of P(X=0)

The two singles must arrive on time, and the three couples also must. It follows that P(X=0) = (0.57)⁵ = 0.0602

Interpretation(s) of P(X=1)

Exactly 1 person, a single, must arrive late, and all the rest must arrive on time. The late single can be chosen in 2 ways. The probabiliy that (s)he arrives late is 0.43.

The probability that the other single and the three couples arrive on time is (0.57)⁴

It follows that

P(X=1) = (2)(0.43)(0.57)⁴ = 0.0908

Interpretation(s) of P(X=2)

Two late can happen in two different ways. Either (i) the two singles are late, and the couples are on time or (ii) the singles are on time but one couple is late.

(i) The probability that the two singles are late, but the couples are not is (0.43)²(0.57)³

(ii) The probability that the two singles are on time is (0.57)²

Given that the singles are on time, the late couple can be chosen in 3 ways. The probability that it is late is 0.43 and the probability the other two couples are on time is (0.57)².

So the probability of (ii) is (0.57)²(3)(0.43)(0.57)² which looks better as (3)(0.43)(0.57)⁴ It follows that

P(X=2) = (0.43)²(0.57)³ + (3)(0.43)(0.57)⁴ = 0.0342 + 0.136 = 0.1700

Interpretations of P(X=3).

Here a single must arrive late, and also a couple. The late single can be chosen in 2 ways. The probability the person is late but the other single is not is (0.43)(0.57).

The late couple can be chosen in 3 ways. The probability one couple is late and the other two couples are not is (0.43)(0.57)². Putting things together, we find that

P(X=3) = (2)(3)(0.43)²(0.57)³ = 0.2050

Interpretation(s) P(X=4)

Since we either (i) have the two singles and one couple late, or (ii) two couples late. So the calculation will break up into two cases.

(i) Two singles and one couple late

Two singles' probability of being late = (0.43)² and One couple being late can be done in 3 ways, so its probability = 3(0.43)(0.57)²

(ii) Two couples late, one couple and two singles early

This can be done in only 3 ways, and its probability is 2(0.57)³(0.43)²

P(X=4) = (3)(0.43)³(0.57)² + (3)(0.57)³(0.43)² = 0.0775 + 0.103 = 0.1800

Interpretations of P(X=5)

For 5 people to be late, it has to be two couples and 1 single person.

For couples, The two late couples can be picked in 3 ways. Probability is 3(0.43)²(0.57)

The late single person can be picked in two ways too, 2(0.43)(0.57)

P(X=5) = 2(3)(0.43)³(0.57)² = 0.1550

Interpretations of P(X=6)

For 6 people to be late, we have either (i) the three couples are late or (ii) two couples and the two singles.

(i) Three couples late with two singles on time = (0.43)³(0.57)²

(ii) Two couples and two singles late

Two couples can be selected in 3 ways, so probability = 3(0.43)²(0.57)(0.43)²

P(X=6) = (0.43)³(0.57)² + 3(0.43)⁴(0.57) = 0.0258 + 0.0585 = 0.0843

Interpretation(s) of P(X=7)

For 7 people to be late, it has to be all three couples and only one single (which can be picked in two ways)

P(X=7) = 2(0.57)(0.43)⁴ = 0.0390

Interpretations of P(X=8)

Everybody had to be late

P(X=8) = (0.43)⁵ = 0.0147

6 0
2 years ago
ASK YOUR TEACHER A 2.0-kg mass swings at the end of a light string with the length of 3.0 m. Its speed at the lowest point on it
Nadya [2.5K]

Answer:

  K_b = 78 J

Explanation:

For this exercise we can use the conservation of energy relations

starting point. Lowest of the trajectory

        Em₀ = K = ½ mv²

final point. When it is at tea = 50º

        Em_f = K + U

        Em_f = ½ m v_b² + m g h

where h is the height from the lowest point

        h = L - L cos 50

        Em_f = ½ m v_b² + mg L (1 - cos50)

energy be conserve

        Em₀ = Em_f

         ½ mv² = ½ m v_b² + mg L (1 - cos50)

         K_b = ½ m v_b² + mg L (1 - cos50)

let's calculate

          K_b = ½ 2.0 6.0² + 2.0 9.8 6.0 (1 - cos50)

          K_b = 36 +42.0

          K_b = 78 J

4 0
2 years ago
How could you test the hypothesis that elephants interpreted the ground signal as being farther away than the air signal?
bekas [8.4K]

Answer:

One way to test the hypothesis is to create two waves, one in the air and one on the ground at the same time. One of them for the elephant to get closer and another for the elephants to move away. Observe the reaction of the animal and with this we know which sound came first.

Explanation:

This hypothesis is based on the fact that the speed of sound in air is v = 343 m / s with a small variation with temperature.

The speed of sound in solid soil is an average of the speed of its constituent media, giving values ​​between

 wood      3900 m / s

 concrete 4000 m / s

 fabrics     1540 m / s

 earth       5000 m / s wave S

 ground    7000 m / s P wave

 

we can see that the speed on solid earth is an order of magnitude greater than in air.

One way to test the hypothesis is to create two waves, one in the air and one on the ground at the same time. One of them for the elephant to get closer and another for the elephants to move away. Observe the reaction of the animal and with this we know which sound came first.

From the initial information, the wave going through the ground should arrive first.

3 0
2 years ago
Other questions:
  • A stone is dropped from a tower 100 meters above the ground. The stone falls past ground level and into a well. It hits the wate
    9·1 answer
  • Which one of the following devices converts radioactive emissions to light for detection?
    8·1 answer
  • If the 80 ohm resistor fails, will the 50 ohm and 100 ohm resistor continue to operate? Why or why not?
    5·1 answer
  • Two swimmers begin a race and Swimmer A completes each length of the pool in 30 seconds, while Swimmer B completes each length i
    6·1 answer
  • A ledge on a building is 18 m above the ground. A taut rope attached to a 4.0-kg can of paint sitting on the ledge passes up ove
    15·1 answer
  • Male Rana catesbeiana bullfrogs are known for their loud mating call. The call is emitted not by the frog's mouth but by its ear
    14·1 answer
  • The Orion nebula is one of the brightest diffuse nebulae in the sky (look for it in the winter, just below the three bright star
    7·1 answer
  • Calculate the time taken by the light to pass through a nucleus of diameter 1.56 10 -16 m. (speed of light is 3 10 8 m/s)
    10·1 answer
  • Which optical device can focus light to a point through reflection?
    14·1 answer
  • A hockey puck with a mass of 0.16 kg travels at a velocity of 40 m/s toward a goalkeeper. The goalkeeper has a mass of 120 kg an
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!