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AlekseyPX
3 years ago
10

A professional boxer hits his opponent with a 1035 N horizontal blow that lasts 0.175 s. The opponent's total body mass is 120 k

g and the blow strikes him near his center of mass and while he is motionless in midair. Determine the following.(a) The opponent's final velocity after the blow(b) Calculate the recoil velocity of the opponent's 5.0-kg head if hit in this manner, assuming the head does not initially transfer significant momentum to the boxer's body.
Physics
1 answer:
12345 [234]3 years ago
7 0

Answer:

(a) vf = 1.51 m/s

(b) vf = 36.22 m/s

Explanation:

The rate of change of momentum is equal to the force:

F = \frac{mv_f-mv_i}{t}

Ft = m(v_f-v_i)

where,

F = Force = 1035 N

t = time = 0.175 s

vi = initial speed = 0 m /s

vf = final speed = ?

(a)

m = mass of body = 120 kg

Therefore,

(1035\ N)(0.175\ s)=(120\ kg)(v_f - 0\ m/s)\\\\v_f = \frac{181.125\ Ns}{120\ kg} \\\\

<u>vf = 1.51 m/s</u>

<u></u>

(b)

m = mass of head = 5 kg

Therefore,

(1035\ N)(0.175\ s)=(5\ kg)(v_f - 0\ m/s)\\\\v_f = \frac{181.125\ Ns}{5\ kg} \\\\

<u>vf = 36.22 m/s</u>

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a jet plane traveling 1890 km/h pulls out of a dive by moving in an arc of radius 5.20km. what is planes acceleration in g's
Andrew [12]

Answer:

53 m/s^2

Explanation:

Unit conversions:

1890 km/h = 1890 km/h * 1000m/km * 1/3600 h/s = 525 m/s

5.2 km = 5200 m

Assume that the jets is traveling in perfect circular motion, we can calculate the centripetal acceleration of the motion:

a = \frac{v^2}{r}

where v = 525m/s is the velocity of the jet and r = 5200 is the radius of the arc

a = \frac{525^2}{5200} = 53m/s^2

7 0
3 years ago
Two strings with linear densities of 5 g/m are stretched over pulleys, adjusted to have vibrating lengths of 0.50 m, and attache
HACTEHA [7]

Answer:

2.18 kg

Explanation:

The frequency of a wave in a stretched string f = n/2L√(T/μ) where n = harmonic number, L = length of string, T = tension = mg where m = mass of object on string and g = acceleration due to gravity = 9.8 m/s² and μ = linear density of string.

For string 1, its fundamental frequency f  is when n = 1. So,

f = 1/2L√(T/μ) =  1/2L√(mg/μ)

Now for string 1, L = 0.50 m, m = 20 kg and μ = 5 g/m = 0.005 kg/m

substituting the values of the variables into f, we have

f = 1/2L√(mg/μ)

f = 1/2 × 0.50 m√(20 kg × 9.8 m/s²/0.005 kg/m)

f = 1/1 m√(196 kgm/s²/0.005 kg/m)

f = 1/1 m√(39200 m²/s²)

f = 1/1 m × 197.99 m/s

f = 197.99 /s

f = 197.99 Hz

f ≅ 198 Hz

For string 2, at its third harmonic frequency f'  is when n = 3. So,

f' = 3/2L√(T/μ) =  3/2L√(mg/μ)

Now for string 2, L = 0.50 m, m = M kg and μ = 5 g/m = 0.005 kg/m

substituting the values of the variables into f, we have

f' = 3/2L√(Mg/μ)

f' = 3/2 × 0.50 m√(M × 9.8 m/s²/0.005 kg/m)

f' = 3/1 m√(M1960 m²/s²kg)

f' = 3/1 m√M√(1960 m²/s²kg)

f' = 3/1 m √M × 44.27 m/s√kg

f' = 132.81√M/s√kg

f' = 132.81√M Hz/√kg

Since the frequency of the beat heard is 2 Hz,

f - f' = 2 Hz

So, 198 Hz - 132.81√M Hz/√kg = 2 Hz

132.81√M Hz/√kg = 198 Hz - 2 Hz

132.81√M Hz/√kg = 196 Hz

√M Hz/√kg = 196 Hz/138.81 Hz

√M/√kg = 1.476

squaring both sides,

[√M/√kg] = (1.476)²

M/kg = 2.178

M = 2.178 kg

M ≅ 2.18 kg

8 0
2 years ago
A car has a momentum of 20,000 kg • m/s. What would the car’s momentum be if its velocity doubles?
AnnyKZ [126]

To answer this question, it helps enormously if you know
the formula for momentum:

           Momentum = (mass) x (speed) .

Looking at the formula, you can see that momentum is directly
proportional to speed.  So if speed doubles, so does momentum.

If the car's momentum is 20,000 kg-m/s now, then after its speed
doubles, its momentum has also doubled, to 40,000 kg-m/s.

6 0
2 years ago
Read 2 more answers
How does the kinetic energy from the forward motion of a car traveling at 16 m/s
elena55 [62]

The kinetic energy in the first case is 4 times more than the second case.

Hence, option D)It is 4 times greater is the correct answer.

<h3>What is Kinetic Energy?</h3>

Kinetic energy is simply a form of energy a particle or object possesses due to its motion.

It is expressed as;

K = (1/2)mv²

Where m is mass of the object and v is its velocity.

Given that;

  • For the first case, velocity v = 16m/s
  • For the second case, velocity = 8m/s
  • Let the mass of the car be m

For the first case, kinetic energy of the car will be;

K = (1/2)mv²

K = (1/2) × m × (16m/s)²

K = (1/2) × m × 256m²/s²

K = mass × 128m²/s²

For the second case, kinetic energy of the car will be;

K = (1/2)mv²

K = (1/2) × m × (8m/s)²

K = (1/2) × m × 64m²/s²

K = mass × 32m²/s²

Comparing the kinetic energy of the car with the same mass but different velocity, we can see that the kinetic energy in the first case is 4 times more than the second case.

Hence, option D)It is 4 times greater is the correct answer.

Learn more about kinetic energy here: brainly.com/question/12669551

#SPJ1

7 0
1 year ago
When a fixed amount of ideal gas goes through an isobaric expansion A) its internal (thermal) energy does not change.B) the gas
Bingel [31]
<h2>Answer: its temperature must increase.</h2>

Explanation:

In an isobaric process the pressure remains constant, which means the initial pressure and the final pressure will be the same.

In addition, during this thermodynamic process, the volume of the ideal gas expands or contracts in such a way that the variation of pressure \Delta P is neutralized.

Now, according to the First law of Thermodynamics that establishes the conservation of energy:

\Delta U=\Delta Q-\Delta W   (1)

Where:

\Delta U is the internal energy

\Delta Q is the heat transferred

\Delta W is the work

Now, for an isobaric process:

\Delta W=P\Delta V    (2)

Where:

P is the pressure (<u>always positive</u>)

\Delta V is the volume variation of the gas

<u />

<u>Here we have two possible results:</u>

-If the gas expands (positive \Delta V), the work is positive.

-If the gas compresses (negative \Delta V), the work is negative.

In this case we are talking about the first result (work is positive).

Then, according to the above, equation (1) can be written as follows:

\Delta U=\Delta Q - P\Delta V   (3)

Clearing \Delta Q:

\Delta Q=\Delta U+P \Delta V    (4)

Then, for an ideal gas in an isobaric process, part of the heat (Q) added to the system will be used to do work (positive in this case) and the other part <u>will increase the internal energy</u>, hence <u>the temperature will increase as well.</u>

7 0
3 years ago
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