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2 years ago

15

represents the space-time, speed-time and acceleration-time graphs for a ball pulled upwards with a speed of 10 m / s from 1 met

er of ground

1 answer:

White raven [17]2 years ago

5 0

Represent the sleaze time speed time and acceleration time graphs

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Please tell me the order the answers go. Best and most correct answer gets Brainliest.

iogann1982 [59]

See the attached picture:

8 0

3 years ago

A ball having a weight of 1.5 N is dropped from a height of 4 meters. (Neglect air friction.) How much mechanical energy is "los

NikAS [45]

As this mechanical energy is associated with height, it would be "Potential Energy" in particular.

U = mgh
U = F.h
U = 1.5 * 4
U = 6 Joules

So, 6 J of energy is lost before it hits the ground.

Hope this helps!

4 0

3 years ago

Read 2 more answers

A block of ice(m = 14.0 kg) with an attached rope is at rest on a frictionless surface. You pull the block with a horizontal for

nadezda [96]

Answer:

a) The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) The final speed of the block of ice is 9.8 meters per second.

Explanation:

a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.

The weight of the block ( $W$ ), measured in newtons, is:

$W = m\cdot g$ (1)

Where:

$m$ - Mass of the block of ice, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $m = 14\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , the magnitudes of the weight and normal force of the block of ice are, respectively:

$N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$N = W = 137.298\,N$

And the pull force is:

$F_{pull} = 98\,N$

The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:

$m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f}$ (2)

Where:

$v_{o}$ , $v_{f}$ - Initial and final speeds of the block, measured in meters per second.

$\Sigma F$ - Horizontal net force, measured in newtons.

$\Delta t$ - Impact time, measured in seconds.

Now we clear the final speed in (2):

$v_{f} = v_{o}+\frac{\Sigma F\cdot \Delta t}{m}$

If we know that $v_{o} = 0\,\frac{m}{s}$ , $m = 14\,kg$ , $\Sigma F = 98\,N$ and $\Delta t = 1.40\,s$ , then final speed of the ice block is:

$v_{f} = 0\,\frac{m}{s}+\frac{(98\,N)\cdot (1.40\,s)}{14\,kg}$

$v_{f} = 9.8\,\frac{m}{s}$

The final speed of the block of ice is 9.8 meters per second.

6 0

2 years ago

Sam, whose mass is 60 Kg, is riding on a 5.0 kg sled initially traveling at 8.0 m/s. He

umka2103 [35]

<h3>Answer: 130 newtons</h3>

===============================================================

Explanation:

We'll need the acceleration first.

The initial speed (let's call that Vi) is 8.0 m/s
The final speed (Vf) is 0 m/s since Sam comes to a complete stop at the end.
This happens over a duration of t = 4.0 seconds

The acceleration is equal to the change in speed over change in time

a = acceleration

a = (change in speed)/(change in time)

a = (Vf - Vi)/(4 seconds)

a = (0 - 8.0)/4

a = -8/4

a = -2

The acceleration is -2 m/s^2, meaning that Sam slows down by 2 m/s every second. Negative accelerations are often associated with slowing down. The term "deceleration" can be used here.

Here's a further break down of Sam's speeds at the four points of interest

At 0 seconds, he's going 8 m/s
At the 1 second mark, he's slowing down to 8-2 = 6 m/s
At the 2 second mark, he's now at 6-2 = 4 m/s
At the 3 second mark, he's at 4-2 = 2 m/s
Finally, at the 4 second mark, he's at 2-2 = 0 m/s

Next, we'll apply Newton's Second Law of motion

F = m*a

where,

F = force applied
m = mass
a = acceleration

We just found the acceleration, and the mass is fairly easy as all we need to do is add Sam's mass with the sled's mass to get 60+5.0 = 65 kg

So the force applied must be:

F = m*a

F = 65*(-2)

F = -130 newtons

This force is negative to indicate it's pushing against the sled's momentum to slow Sam down.

The magnitude of this force is |F| = |-130| = 130 newtons

8 0

3 years ago

A kite is 100m above the ground. if there are 200m of string out what is the angle between the string and the horizontal? (assum

jeka57 [31]

Answer : The angle between the string and the horizontal is 30 degrees

Explanation: Imagine this a a triangle where the length of the string (200m) is the hypotenuse and the height of the kite is the opposite side (100m) .

Let the angle between the string and the horizontal be theta.

Now sin (Theta) = opposite side/hypotenuse

= 100/200 = 1/2

Therefore Theta = Sin ⁻¹ ( 1/2 )

Theta = 30 degrees

4 0

3 years ago