Answer:
For a velocity versus time graph how do you know what the velocity is at a certain time?
Ans: By drawing a line parallel to the y axis (Velocity axis) and perpendicular to the co-ordinate of the Time on the x axis (Time Axis). The point on the slope of the graph where this line intersects, will be the desired velocity at the certain time.
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How do you know the acceleration at a certain time?
Hence,
By dividing the difference of the Final and Initial Velocity by the Time Taken, we could find the acceleration.
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How do you know the Displacement at a certain time?
Ans: As Displacement equals to the area enclosed by the slope of the Velocity-Time Graph, By finding the area under the slope till the perpendicular at the desired time, we find the Displacement.
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To solve this problem it is necessary to apply the equations related to the conservation of momentum. Mathematically this can be expressed as
Where,
= Mass of each object
= Initial velocity of each object
= Final Velocity
Since the receiver's body is static for the initial velocity we have that the equation would become
Therefore the velocity right after catching the ball is 0.0975m/s
Answer:
The magnitude of the charge on each sphere is 0.135 μC
Explanation:
Given that,
Mass = 1.0
Distance = 2.0 cm
Acceleration = 414 m/s²
We need to calculate the magnitude of charge
Using newton's second law
Put the value of F
Put the value into the formula
Hence, The magnitude of the charge on each sphere is 0.135μC.
Answer:
3.0883 x 10^10mg
Explanation:
1 kilogram = 1000 000 milligrams
So, 30 883 x 1000 000 = 30 883 000 000mg
The vertical weight carried by the builder at the rear end is F = 308.1 N
<h3>Calculations and Parameters</h3>
Given that:
The weight is carried up along the plane in rotational equilibrium condition
The torque equilibrium condition can be used to solve
We can note that the torque due to the force of the rear person about the position of the front person = Torque due to the weight of the block about the position of the front person
This would lead to:
F(W*cosθ) = mgsinθ(L/2) + mgcosθ(W/2)
F(1cos20)= 197/2(3.10sin20 + 2 cos 20)
Fcos20= 289.55
F= 308.1N
Read more about vertical weight here:
brainly.com/question/15244771
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