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____ [38]
3 years ago
10

Each statement below is a definition or an example of force, work, potential energy, or kinetic energy. Drag each into the box t

hat it matches best. when an object is displaced in the direction of the net force acting on it the energy present when you run when pushing a box across a floor, it is the pushing force times the distance between the starting and ending point gravity and friction energy of motion energy that is stored something that pushes or pulls an object the energy present when you stand on a high dive force work potential Energy kinetic Energy
Physics
1 answer:
marshall27 [118]3 years ago
6 0

Answer:

yes

Explanation:

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Instructions: Select the correct answer.Why does air at 75% relative humidity and 30eC make your skin feel stickier than air at
antoniya [11.8K]
B............,.,,.............
4 0
3 years ago
A 5.50 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 3.20 m by a force of 25.
kiruha [24]

(a) 69.3 J

The work done by the applied force is given by:

W=Fd cos \theta

where:

F = 25.0 N is the magnitude of the applied force

d = 3.20 m is the displacement of the sled

\theta=30^{\circ} is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J

(b) 0

The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

\Delta K = W

where

\Delta K is the change in kinetic energy

W is the work done

Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

\Delta K=69.3 J

(d) 4.9 m/s

The change in kinetic energy of the sled can be rewritten as:

\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2 (1)

where

Kf is the final kinetic energy

Ki is the initial kinetic energy

m = 5.50 kg is the mass of the sled

u = 0 is the initial speed of the sled

v = ? is the final speed of the sled

We can calculate the variation of kinetic energy of the sled, \Delta K, after it has travelled for d=3 m. Using the work-energy theorem again, we find

\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J

And substituting into (1) and re-arrangin the equation, we find

v=\sqrt{\frac{2 \Delta K}{m}}=\sqrt{\frac{2(65.0 J)}{5.50 kg}}=4.9 m/s

6 0
3 years ago
What are autotrophs? Name three types of organisms that are autotrophs.
ValentinkaMS [17]
Autotrophs are organisms that can make its own food by synthesizing organic nutrients from inorganic materials. Three types include: photoautotrophs, chemoautotrophs, and plants.
4 0
3 years ago
Which of the following describes the way a population can be dispersed?
zubka84 [21]

Answer:

option b. <em>uniform</em><em> </em><em>population</em>

5 0
3 years ago
Given a force of 56 N and an acceleration of 7 m/s2, what is the mass?
love history [14]

Answer:

<h2>8 kg</h2>

Explanation:

The mass of an object can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

From the question we have

m =  \frac{56}{7}  \\

We have the final answer as

<h3>8 kg</h3>

Hope this helps you

6 0
3 years ago
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