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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Explanation:
From Newton's second law:
F = ma
Given that m = 4 kg and a = 8 m/s²:
F = (4 kg) (8 m/s²)
F = 32 N
If m is reduced to 1 kg and F stays at 32 N:
32 N = (1 kg) a
a = 32 m/s²
So the acceleration increases by a factor of 4.
Answer:
area = 5733.33 cm²
length = 5.47 ×
cm
Explanation:
Given data
density = 19.32 g/cm³
mass = 33.16 g
thickness = 3.000 µm = 3 ×
cm
radius r = 1.000 µm = 1 × cm
to find out
area of the leaf and length of the fiber
solution
we know volume formula that is
volume = mass / density
volume = 33.16 / 19.32
volume = 1.72 cm³
we know that volume = thickness × area
so area
area = volume / thickness
area = 1.72 / 3 ×
area = 5733.33 cm²
and
we know volume = πr²L
so L = volume / πr²
length = 1.72 / π(1×)²
length = 5.47 × cm
Answer:
which corresponds to the second option shown: "voltage times amperage"
Explanation:
The electric power is the work done to move a charge Q across a given difference of potential V per unit of time.
Since such electrical work is the product of the potential difference V times the charge that moves through that potential, and this work is to be calculated by the unit of time, we need to divide the product by time (t) which leads to the following final simple equation:
Notice that we replaced the quotient representing charge per unit of time (Q/t) by the actual current running through the circuit.
This corresponds to the second option shown in the question: "Voltage times amperage".
To answer this problem, we will use the equations of motions.
Part (a):
For the ball to start falling back to the ground, it has to reach its highest position where its final velocity will be zero.
The equation that we will use here is:
v = u + at where
v is the final velocity = 0 m/sec
u is the initial velocity = 160 m/sec
a is acceleration due to gravity = -9.8 m/sec^2 (the negative sign is because the ball is moving upwards, thus, its moving against gravity)
t is the time that we want to find.
Substitute in the equation to get the time as follows:
v = u + at
0 = 160 - 9.8t
9.8t = 160
t = 160/9.8 = 16.3265 sec
Therefore, the ball would take 16.3265 seconds before it starts falling back to the ground
Part (b):
First, we will get the total distance traveled by the ball as follows:
s = 0.5 (u+v)*t
s = 0.5(160+0)*16.3265
s = 1306.12 meters
The equation that we will use to solve this part is:
v^2 = u^2 + 2as where
v is the final velocity we want to calculate
u is the initial velocity of falling = 0 m/sec (ball starting falling when it reached the highest position, So, the final velocity in part a became the initial velocity here)
a is acceleration due to gravity = 9.8 m/sec^2 (positive as ball is moving downwards)
s is the distance covered = 1306.12 meters
Substitute in the above equation to get the final velocity as follows:
v^2 = u^2 + 2as
v^2 = (0)^2 + 2(9.8)(1306.12)
v^2 = 25599.952 m^2/sec^2
v = 159.99985 m/sec
Therefore, the velocity of the ball would be 159.99985 m/sec when it hits the ground.
