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
poizon [28]
3 years ago
8

What is the weight of a 75kg. Mass?

Physics
2 answers:
AysviL [449]3 years ago
7 0
W=(75kg)(9.8m/s^2

W = 735N

wariber [46]3 years ago
3 0

Weight = (mass) x (gravity wherever the mass is located)

The weight of a 75 kg mass is about  122 newtons on the Moon,
242 newtons on Mars, and 735.5 newtons on Earth. 
You might be interested in
Out of hypothesis, theories and laws “can help predict future events”
Amanda [17]

Answer:

true

Explanation:

7 0
3 years ago
Asexual reproduction involves only one parent. The offspring of this type of reproduction have - F DNA identical to the DNA of t
Sedbober [7]

Answer:

DNA identical to the DNA of the parent

Explanation:

5 0
3 years ago
2) [25 pts] The bob of mass m=5 kg shown in the figure is being held by a force F. If the angle is 0⃗
Colt1911 [192]

Complete Question

The complete question is shown on the first uploaded image  

Answer:

a

   The tension is  T =  48.255 \ N

b

   The time taken is  t =  0.226 \ s

c

   The position for maximum velocity is  

        S = 0  

d

     The maximum velocity is  V_{max} =0.384 \ m/s

Explanation:

The free body for this question is shown on the second uploaded image

From the question we are told that

    The mass of the bob  is m_b  =  5 \ kg

     The angle is  \theta = 10^o

     The length of the string is  L =  0.5 \ m

The tension on the string is mathematically represented as

              T  = mg cos \theta

substituting values

             T =  5 * 9.8 cos(10)

             T =  48.255 \ N

The motion of the bob is mathematically represented as

             S =  A sin (w t  + \frac{\pi }{2} )

     =>   S =  A sin (wt)

Where  w is the angular speed

and  \frac{\pi}{2} is the phase change

At initial position S =  0

   So   wt  = cos^{-1} (0)

          wt  = 1

Generally w can be mathematically represented as

          w =  \frac{2 \pi }{T}

Where T is the period of oscillation which i mathematically represented as

          T  =  2 \pi \sqrt{\frac{L}{g} }

So  

      t   = \frac{1}{w}

       t   = \frac{T}{2 \pi}

       t =  \sqrt{\frac{L}{g} }

substituting values

       t =  \sqrt{\frac{0.5}{9.8} }

       t =  0.226 \ s

Looking at the equation

         wt =  1

We see that maximum velocity of the bob will be at  S = 0  

i. e     w =  \frac{1}{t}

The maximum velocity is mathematically represented as

          V_{max}   =  w A

Where A is the amplitude which is mathematically represented as

         A = L sin \theta

So

      V_{max} = \frac{2 \pi }{T }  L sin \theta

      V_{max} = \frac{2 \pi }{2 \pi } \sqrt{\frac{g}{L} }   L sin \theta  Recall   T  =  2 \pi \sqrt{\frac{L}{g} }

     V_{max} = \sqrt{gL} sin \theta

substituting values        

       V_{max} = \sqrt{9.8 * 0.5 } sin (10)

       V_{max} =0.384 \ m/s

6 0
2 years ago
A 2.0-kg projectile is fired with initial velocity components v0x = 30 m/s and v0y = 40 m/s from a point on the Earth's surface.
EleoNora [17]

(a) The kinetic energy of the projectile when it reaches the highest point in its trajectory is 900 J.

(b) The work done  in firing the projectile is 2,500 J.

<h3>Kinetic energy of the projectile at maximum height</h3>

The kinetic energy of the projectile when it reaches the highest point in its trajectory is calculated as follows;

K.E = ¹/₂mv₀ₓ²

where;

  • m is mass of the projectile
  • v₀ₓ is the initial horizontal component of the velocity at maximum height

<u>Note:</u> At maximum height the final vertical velocity is zero and the final horizontal velocity is equal to the initial horizontal velocity.

K.E = (0.5)(2)(30²)

K.E = 900 J

<h3>Work done in firing the projectile</h3>

Based on the principle of conservation of energy, the work done in firing the projectile is equal to the initial kinetic energy of the projectile.

W = K.E(i) = ¹/₂mv²

where;

  • v is the resultant velocity

v = √(30² + 40²)

v = 50 m/s

W = (0.5)(2)(50²)

W = 2,500 J

Thus, the kinetic energy of the projectile when it reaches the highest point in its trajectory is 900 J.

The work done  in firing the projectile is 2,500 J.

Learn more about kinetic energy here: brainly.com/question/25959744

#SPJ1

5 0
1 year ago
Consider a Cassegrain-focus, reflecting telescope. Images recorded at Cassegrain-focus will be:
Contact [7]
B. Flipped compared to what is in the sky
3 0
3 years ago
Other questions:
  • Plz I need help !!!!!!!
    7·2 answers
  • Which of the following is true?
    10·1 answer
  • Does adding electrons change the mass
    14·1 answer
  • show that the value of g is independent to the mass of the free falling body. The weigh of the person on the earth is 72 kg wt,
    10·1 answer
  • 1. You are driving along a highway at 35.0 m/s when you hear the siren of a police car approaching you at 44.2 m/s from behind a
    11·1 answer
  • What element has conductivity,shiney luster,magnitic <br> -science
    8·1 answer
  • What is the gravity of sombrero galaxy compared to earth?
    5·2 answers
  • The image below The image below The image below The image below The image below The image below The image below The image below
    13·1 answer
  • 4 kg of water was
    6·1 answer
  • a___of water a. of bread a. of soap a. of juice a. salt. a. sand a. of glass a. of corn fill in the blanks I follow him but righ
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!