Answer:
F = 120 N
Explanation:
Given that,
The mass of a runner, m = 60 kg
Acceleration of the runner, a = 2 m/s²
We need to find the force provided from her legs. The formula for force is given by :
F = ma
Substitute all the values,
F = 60 kg × 2 m/s²
= 120 N
So, the required force is equal to 120 N.
Answer:
ω = 3.66 rad/s
Explanation:
Given that
L= 1 m
M = 270 g = 0.27 kg
m = 3 g
u= 250 m/s
v= 140 m/s
We know that moment of inertia of the rod about it center
I=0.0225 kg.m²
The initial angular momentum
L₁ = m u r
Final angular momentum
L₂= I ω+m v r
Here r= L/4
r= 0.25 m
There is no any external torque ,it means that angular momentum will be conserve
L₁ = L₂
m u r = I ω+m v r
m r (u - v)=I ω
Now by putting the values
m r (u - v)=I ω
0.003 x 0.25 x (250- 140)=0.0225 x ω
ω = 3.66 rad/s
<h2>
Answer: Both objects have charges of the same sign</h2>
When two electrically charged bodies come closer, appears a force that attracts or repels them, depending on the sign of the charges of this two bodies.
This is stated by Coulomb's Law:
"The electrostatic force between two point charges and is proportional to the product of the charges and inversely proportional to the square of the distance that separates them, and has the direction of the line that joins them"
Mathematically this law is written as:
Where is a proportionality constant.
Now, if and have the same sign charge (both positive or both negative), a repulsive force will act on these charges.
What is the work done in stretching a spring by a distance of 0.5m if the restoring force is 48N is 12 J.
Hooke's Law states that the more you deform a spring, the more force it will take to deform it further.
Notations: F= Restoring force
k= spring constant
x = spring displacement
Given: F = 48N,
x= 0.5m
By using Formula; F= kx
F = 48 = kx, k = F/x
k = 48/0.5 = 96 N/m
work done by spring = 1/2 (kx)² = 1/2(96x0.5)² = 12 J
work done by spring is 12J.
Learn more about the Hooke's law with the help of the following link:
brainly.com/question/2449067
#SPJ4