1answer.
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
nekit [7.7K]
3 years ago
6

A baseball has a mass of about 155 g. What is the magnitude of the momentum of a baseball thrown at a speed of 87 miles per hour

? (Note that you need to convert mass to kilograms and speed to meters/second. A mile is 1.6 kilometers or 1600 meters.)
Physics
1 answer:
n200080 [17]3 years ago
6 0

Answer:

5.994 kgm/s

Explanation:

Momentum: This can be defined as the product of the mass of a body to the velocity of that body. The S.I unit of momentum is kgm/s.

Mathematically, momentum can be expressed as,

M = mv.................................... Equation 1

Where M = momentum of the baseball, m = mass of the baseball, v = velocity of the baseball.

Given: m = 155 g = (155/1000) kg = 0.155 kg. v = 87 miles per hour = 87(1600/3600) m/s = 38.67 m/s.

Substituting into equation 1

M = 0.155(38.67)

M = 5.994 kgm/s.

Thus the momentum of the baseball = 5.994 kgm/s

You might be interested in
Due Ma<br> duart<br> ded<br> out<br> 25 N<br> 35 N<br> 1-03
Anuta_ua [19.1K]

Answer:

what's that all about

hehehwhe

Explanation:

dgbjjjedgkigdssfhkkoyddwrhkoyeqaxghjjhasghffhjiopjtewqetujjgda

8 0
2 years ago
I need help with question 2 and 2 please
vitfil [10]
Alright here the answer to number 2

5 0
3 years ago
an athlete is running at a constant velocity with a javelin held in his right hand.the force he is applying on thejavelin as he
mestny [16]
50 Nm
work=force*change in position
7 0
2 years ago
31. If you threw a baseball straight out at 45 m/s from a height of 1.5 meters (A) how long would it be in the air? B) How far o
coldgirl [10]

Answer:

A) t = 0.55 s

B) x = 24.8 m

Explanation:

A) We can find the time at which the ball will be in the air using the following equation:

y_{f} = y_{0} + v_{0y}t - \frac{1}{2}gt^{2}    

Where:

y_{f} is the final height= 0  

y_{0} is the initial height= 1.5 m

v_{0y} is the component of the initial speed in the vertical direction = 0 m/s        

t: is the time =?      

g: is the gravity = 9.81 m/s²

0 = 1.5 m - \frac{1}{2}9.81 m/s^{2}t^{2}

By solving the above equation for t we have:

t = \sqrt{\frac{2*1.5 m}{9.81 m/s^{2}}} = 0.55 s  

Hence, the ball will stay 0.55 seconds in the air.

                             

B) We can find the distance traveled by the ball as follows:

x_{f} = x_{0} + v_{0x}t + \frac{1}{2}at^{2}

Where:  

a: is the acceleration in the horizontal direction = 0  

x_{f} is the final position =?  

x_{0} is the initial position = 0      

v_{0x} is the component of the initial speed in the horizontal direction = 45 m/s                                                                                            

x_{f} = x_{0} + v_{0x}t + \frac{1}{2}at^{2}

x_{f} = 0 + 45 m/s*0.55 s + 0 = 24.8 m

Therefore, the ball will travel 24.8 meters.

I hope it helps you!

3 0
2 years ago
Two manned satellites approach one another at a relative velocity of v=0.190 m/s, intending to dock. The first has a mass of m1=
drek231 [11]

Answer:

Their final relative velocity is 0.190 m/s

Explanation:

The relative velocity of the satellites, v = 0.190 m/s

The mass of the first satellite, m₁ = 4.00 × 10³ kg

The mass of the second satellite, m₂ = 7.50 × 10³ kg

Given that the satellites have elastic collision, we have;

v_2 = \dfrac{2 \cdot m_1}{m_1 + m_2} \cdot u_1 - \dfrac{m_1 - m_2}{m_1 + m_2} \cdot u_2

v_2 = \dfrac{ m_1 - m_2}{m_1 + m_2} \cdot u_1 + \dfrac{2 \cdot m_2}{m_1 + m_2} \cdot u_2

Given that the initial velocities are equal in magnitude, we have;

u₁ = u₂ = v/2

u₁ = u₂ = 0.190 m/s/2 = 0.095 m/s

v₁ and v₂ = The final velocities of the satellites

We get;

v_1 = \dfrac{2 \times 4.0 \times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 - \dfrac{4.0 \times 10^3- 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095

v_2 = \dfrac{ 4.0 \times 10^3 - 7.50\times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 + \dfrac{2 \times 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095

The final relative velocity of the satellite, v_f = v₁ + v₂

∴ v_f = 0.095 + 0.095 = 0.190

The final relative velocity of the satellite, v_f = 0.190 m/s

4 0
2 years ago
Other questions:
  • Which statement describes one feature of a closed circuit? Charges do not flow. Bulbs will not shine. The circuit is broken. The
    13·2 answers
  • A wave is traveling through a medium ad it travels it displaces particles of matter in the same direction as the wave is traveli
    14·2 answers
  • 9<br>How can you prove that air has weight​
    12·2 answers
  • Why was the st valentine's day massacre important? And why are students studying this case?
    14·1 answer
  • What causes interstellar dust and clouds to condense in the planets and stars
    5·1 answer
  • Three parallel wires of length l each carry current Iin the same direction. They’re positioned at the vertices of an equilateral
    8·2 answers
  • What is the main idea of the paragraph in one sentence?
    6·2 answers
  • If a line segment on a distance-time graph is fairly flat, the speed is relatively slow.
    14·1 answer
  • A bullet was fired from a gun. The bullet had a mass of 50g and the kinetic energy of the bullet was 25 kJ. How fast did it go?
    13·1 answer
  • What are the 3 basic parts of an atom and what are their charges?
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!