That's true.
And I'll go ya a better one:
If the object is moving or not moving, at a constant or changing speed, in a straight or curvy line, and the forces on it do not cancel out and add up to zero, the object will accelerate.
The target heart rate for moderate-intensity activity is 80%
Answer:
Hence the weight of the person on the moon is 162.4, and the value of g used is 1.624 m/s²
Explanation:
from the question,
W = mg........................ Equation 1
Where W = weight of the man on Earth, m = mass of the man, g = acceleartion due to gravity of the man
make m the subject of the equation
m = W/g.............. Equation 2
Given: W = 1000 N,
Constant: g = 10 m/s²
Therefore,
m = 1000/10
m = 100 kg
Weight on the moon
W' = mg'
W' = 100(1.624)
W' = 162.4 N.
Hence the weight of tthe person on the moon is 162.4, and the value of g used is 1.624 m/s²
Answer:
h = 67.081 m
Explanation:
Given that,
The time taken by a person to fall down is, t = 2.2 s
The height of the cliff from the ground, h = ?
The distance that the person will fall through the time is given by the formula
S = 1/2 gt² m
Where,
g - acceleration due to gravity
Substituting the values in the above equation
S = 1/2 x 9.8 m/s² x (3.7 s)²
= 67.081 m
Therefore, the height of the cliff from the ground is, h = 67.081 m
For a total charge of Q coulomb is uniformly distributed along a rod 40cm in length, the electric field intensity 20cm away from the rod is mathematically given as
E1=1.598*10^11v/m
<h3>What is the e
lectric field intensity 20cm away from the
rod along its perpendicular
bisector?</h3>
Generally, the equation for the initial electric field intensity is mathematically given as
Therefore
Hence
E1=B*9*10^{13})/(10*110)*
E1=1.598*10^11v/m
In conclusion, the electric field intensity
E1=1.598*10^11v/m
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