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mario62 [17]
3 years ago
9

What is the difference between the two graphs

Physics
1 answer:
kupik [55]3 years ago
6 0

Answer:

one of the graph is postion-time graph while the other one is velocity-time graph

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A 3.0-kg mass and a 5.0-kg mass hang vertically at the opposite ends of a rope that goes over an ideal pulley. If the masses are
irga5000 [103]

Answer:

The time taken will be 0.553 seconds.

Explanation:

We should start off by finding the force exerted by the rope on the 3kg weight in this case.

Weight of 5kg mass = 5 * 9.81 = 49.05 N

Weight of 3kg mass = 3 * 9.81 = 29.43 N

The force acting upward on the 3kg mass will equal the weight of the 5kg mass. Thus the resultant force acting on the 3kg mass is:

Total force = 49.05 - 29.43 = 19.62 N (upwards)

We can now find the acceleration:

F = m * a

19.62 = 3 * a

a = 6.54 m/s^2

We now use the following equation of motion to get the time taken to travel 1 meter:

s=u*t+\frac{1}{2} (a*t^2)

1=0*t+\frac{1}{2} (6.54*t^2)

t = 0.553 seconds

4 0
4 years ago
A grinding wheel with a moment of inertia of 2 kg-m2 has a 2.50 N-m torque applied to it. What is its final kinetic energy 10 se
grigory [225]
First you do 2.5 n-m / 2 kg-m2
This equals 1.25 rad/s^2

Then multiply rad/s^2 by 10 because of the amount of seconds
This equals 12.5 rad/s^2

Then do this following equation : (1/2)Iwf^2
Do (0.5) 12.5 * 12.5 * 2
This equals 156.25 J (meaning that the answer is C: 156 J)
6 0
4 years ago
The first game in volleyball is played until a team reaches 25 points.
Zinaida [17]
I don't understand this question
7 1
3 years ago
Read 2 more answers
It took a squirrel 0.50\,\text s0.50s0, point, 50, start text, s, end text to run 5.0\,\text m5.0m5, point, 0, start text, m, en
STALIN [3.7K]

Answer:

-5.0m/s

Explanation:

3 0
4 years ago
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Bill steps off a 3.0-m-high diving board and drops to the water below. At the same time, Ted jumps upward with a speed of 4.2 m/
Pani-rosa [81]

Answer:

equation of  motion for Bill is

y(t) = 4.9t^2

equation of  motion for Ted is

y(t) = 2 + (-4.2)(t) + 4.9t^2

Explanation:

Taking downward position positive and upward position negative

g = 9.8 m/s^2

equation of  motion for Bill is

y(t) = y_0 +v_0 t +\frac{1}{2}gt^2

y(t) = 0 + 0(t) +\frac{1}{2}gt^2

y(t) = \frac{1}{2}\times (9.8t)^2

y(t) = 4.9t^2

equation of  motion for Ted is

y_0 = 2m -1m = 2m

y_0 = -4.2 m/s

y(t) = y_0 +v_0 t +\frac{1}{2}gt^2

y(t) = 2 + (-4.2)(t) +\frac{1}{2}gt^2

y(t) = 2 + (-4.2)(t) +\frac{1}{2}\times (9.8t)^2

y(t) = 2 + (-4.2)(t) + 4.9t^2

8 0
3 years ago
Read 2 more answers
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