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egoroff_w [7]
3 years ago
6

A straight line with a negative slope on a velocity-time graph indicates which of the following?

Physics
1 answer:
podryga [215]3 years ago
5 0

Answer:

A

Explanation:

Straight line with a negative slope

On a velocity_time graph

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A 70 kg student stands on top of a 5.0 m platform diving board . how much gravitational potential energy does the student have?
Marianna [84]

Answer:

a. P.E = 3430Joules.

b. Workdone = 3430Nm

Explanation:

<u>Given the following data;</u>

Mass = 70kg

Distance = 5m

We know that acceleration due to gravity is equal to 9.8m/s²

To find the potential energy;

Potential energy = mgh

P.E = 70*9.8*5

<em>P.E = 3430J</em>

b. To find the workdone;

Workdone = force * distance

But force = mass * acceleration

Force = 70*9.8

Force = 686 Newton.

Workdone = 686 * 5

<em>Workdone = 3430Nm</em>

6 0
3 years ago
All objects emit ______ radiation?<br> A-electromagnetic<br> B-kinetic<br> D-solar
natulia [17]
A, electromagnetic radiation
6 0
3 years ago
What force is required to accelerate a 385 kg couch at 0.2 m/s^2 ?
Juliette [100K]

Answer:

It takes 77 N

Explanation:

Using Newton's second law of motion, F=ma (Force equals mass times acceleration. Since the mass of the couch is 385 kg and the target acceleration is 0.2 m/s, you simply multiply mass times acceleration (ma) to get the total force, or 77 N.

6 0
3 years ago
3. Maverick and Goose are flying a training mission in their F-14. They are
Elanso [62]

Answer:

A. The bomb will take <em>17.5 seconds </em>to hit the ground

B. The bomb will land <em>12040 meters </em>on the ground ahead from where they released it

Explanation:

Maverick and Goose are flying at an initial height of y_0=1500m, and their speed is v=688 m/s

When they release the bomb, it will initially have the same height and speed as the plane. Then it will describe a free fall horizontal movement

The equation for the height y with respect to ground in a horizontal movement (no friction) is

y=y_0 - \frac{gt^2}{2}    [1]

With g equal to the acceleration of gravity of our planet and t the time measured with respect to the moment the bomb was released

The height will be zero when the bomb lands on ground, so if we set y=0 we can find the flight time

The range (horizontal displacement) of the bomb x is

x = v.t     [2]

Since the bomb won't have any friction, its horizontal component of the speed won't change. We need to find t from the equation [1] and replace it in equation [2]:

Setting y=0 and isolating t we get

t=\sqrt{\frac{2y_0}{g}}

Since we have y_0=1500m

t=\sqrt{\frac{2(1500)}{9.8}}

t=17.5 sec

Replacing in [2]

x = 688\ m/sec \ (17.5sec)

x = 12040\ m

A. The bomb will take 17.5 seconds to hit the ground

B. The bomb will land 12040 meters on the ground ahead from where they released it

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3 years ago
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Rzqust [24]
The maximum force acts between B and C as the graph is steepest showing maximum deceleration
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