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Mazyrski [523]
2 years ago
12

The weight of an object is the product of its mass m, and acceleration of gravity , g . If an objects mass is m=10 . Kg what is

its weight?
Physics
1 answer:
zzz [600]2 years ago
8 0

Its weight is 10g .  The 'g' is the acceleration of gravity
wherever the object happens to be at the time.

On Mars, g=3.71 m/s², the object weighs 37.1 newtons.

On the Moon, g=1.62 m/s², the object weights16.2 newtons.

On Earth, g=9.81 m/s², the object weighs 98.1 newtons.
.
.
etc.

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a particle of mass m sits at rest at x = 0. At time t = 0 a force given by F = Fe^(-t/T) is applied in the +x direction; F and T
wlad13 [49]

Explanation:

Given:F=m\ddot{x}=Fe^{-\frac{t}{T}}

Solving for \ddot{x}:

\ddot{x}=\frac{F}{m}e^{-\sqrt{\frac{F}{m} } t}

where:

T=\sqrt{\frac{m}{F}}

Integrating to get \dot{x} with initial conditions \dot{x}(0)=0:

\dot{x}=\sqrt{\frac{F}{m}}-\sqrt{\frac{F}{m}} e^{-\sqrt{\frac{F}{m}} t}

Integrating to get x with initial conditions x(0) = 0:

x=-1+\sqrt{\frac{F}{m}} t+e^{-\sqrt{\frac{F}{m}}t}

When t=T:

x=-1+\sqrt{\frac{F}{m}}\sqrt{\frac{m}{F}}+e^{-\sqrt{\frac{F}{m}}\sqrt{\frac{m}{F}}}=\frac{1}{e}

\dot{x}=\sqrt{\frac{F}{m}}-\sqrt{\frac{F}{m}} e^{-\sqrt{\frac{F}{m}}\sqrt{\frac{m}{F}}}=\sqrt{\frac{F}{m}}(1-\frac{1}{e})

4 0
2 years ago
A vertical, solid steel post 25 cm in diameter and 2.50m long is required to support a load of 8000kg. You can ignore the weight
Gwar [14]

(a) The stress in the post is 1,568,000 N/m²

(b) The strain in the post is  7.61 x 10⁻⁶  

(c) The change in the post’s length when the load is applied is 1.9 x 10⁻⁵ m.

<h3>Area of the steel post</h3>

A = πd²/4

where;

d is the diameter

A = π(0.25²)/4 = 0.05 m²

<h3>Stress on the steel post</h3>

σ = F/A

σ = mg/A

where;

  • m is mass supported by the steel
  • g is acceleration due to gravity
  • A is the area of the steel post

σ = (8000 x 9.8)/(0.05)

σ = 1,568,000 N/m²

<h3>Strain of the post</h3>

E = stress / strain

where;

  • E is Young's modulus of steel = 206 Gpa

strain = stress/E

strain = (1,568,000) / (206 x 10⁹)

strain = 7.61 x 10⁻⁶

<h3>Change in length of the steel post</h3>

strain = ΔL/L

where;

  • ΔL is change in length
  • L is original length

ΔL = 7.61 x 10⁻⁶ x 2.5

ΔL = 1.9 x 10⁻⁵ m

Learn more about Young's modulus of steel here: brainly.com/question/14772333

#SPJ1

7 0
1 year ago
A motorcycle at a stoplight takes 4.22 seconds to accelerate 28.0 m/s after the light turns green. what is the acceleration?
stira [4]
Acceleration=velocity/time
acceleration=28/4.22
therefore, acceleration=6.64
4 0
2 years ago
What force is required to accelerate a 385 kg couch at 0.2 m/s^2 ?
Juliette [100K]

Answer:

It takes 77 N

Explanation:

Using Newton's second law of motion, F=ma (Force equals mass times acceleration. Since the mass of the couch is 385 kg and the target acceleration is 0.2 m/s, you simply multiply mass times acceleration (ma) to get the total force, or 77 N.

6 0
2 years ago
A mango is dropped and fall freely from rest. What are its position and velocity after 1.0secs,2.0secs,and 3.0secs
irakobra [83]

Answer:

sxsdfsd

Explanation:

4 0
3 years ago
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