Two steamrollers begin 100 m apart and head toward each other, each at a constant speed of 1.00 m/s . At the same instant, a fly
1 answer:
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1) The motion of the baseball consists of two separate motions along the horizontal (x) and vertical (y) axis. The position on the two directions at time t is given by:
where the motion on the x-axis is a uniform motion with constant speed
, while the motion on the y-axis is a uniformly accelerated motion with constant acceleration 
, and initial height 
.
First of all, we can find the time t at which the ball reaches the ground by requiring
:
and if now we substitute this time into the equation for x(t), we find the horizontal distance covered during this time interval, which is the horizontal distance covered by the ball before hitting the ground:
2) Speed of the ball as it hits the ground
We need to find the components of the velocity on the two directions, at the istant when the ball hits the ground.
On the x-axis, the velocity is the same as the initial one:
On the y-axis, the velocity is given by
If we substitute t=0.61 s (the time at which the ball reaches the ground), we find
And the speed of the ball is the magnitude of the resultant of the two components:
where the motion on the x-axis is a uniform motion with constant speed
First of all, we can find the time t at which the ball reaches the ground by requiring
and if now we substitute this time into the equation for x(t), we find the horizontal distance covered during this time interval, which is the horizontal distance covered by the ball before hitting the ground:
2) Speed of the ball as it hits the ground
We need to find the components of the velocity on the two directions, at the istant when the ball hits the ground.
On the x-axis, the velocity is the same as the initial one:
On the y-axis, the velocity is given by
If we substitute t=0.61 s (the time at which the ball reaches the ground), we find
And the speed of the ball is the magnitude of the resultant of the two components:
Answer:
E = 1.19 N/C
Explanation:
Let's first determine the length of the arc which can be given as:
L= Rθ
where:
L = length of the arc
R = radius of curvature
θ = angle in radius
L = (9.09×10⁻²m)(2.59)
L = (0.0909)(2.59)
L = 0.235431 m
Then, the magnitude of electric field that Q produces at the center of curvature can be calculated by using the formula:
Since
where;
L = length
Q = charge
λ = density of the charge;
then substituting for λ, we have :
substituting our given parameter; we have:
E = 1.1889 N/C
E = 1.19 N/C
∴ the magnitude of the electric field that Q produces at the center of curvature = 1.19 N/C
The work done by gas is 0.753 J and change in internal energy is 4.247J
So we are given that mass is 2kg , radius 1 cm and the amount of heat is 5 cm
The piston raised by 2.4cm
As we know that Work done is PΔV
Where ΔV is change in volume
Therefore ΔV = πr^2 h = π x (.01)^2 x .024 =7.53×10^(-6)m^3
Here pressure is 10^5 pa
So W =
Therefore W = 0.753 J
Now coming to change in internal energy
Change in Internal Energy = Heat Added - Energy lost in work
∴ 5J - 0.753 J = 4.247J
Hence the change in internal energy is 4.247 J
Learn more about Work done here
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