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3 years ago

Which one of newton's laws does a doll riding a dog represent

1 answer:

4 0

I believe it would best represent Newton’s first law; an object tends to stay at rest and an object tends to stay in motion unless acted upon by an unbalanced force. When the dog stops walking, the doll will continue to go forward because there is no unbalanced force acting in it.

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A student attaches a block of mass M to a vertical spring so that the block-spring system will oscillate if the block-spring sys

Vlada [557]

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3 years ago

During summer break, Jamie's family went camping in the mountains. There were no roads to their camp, 8 miles away, and it took

liraira [26]

The Answer is B

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5 0

3 years ago

Read 2 more answers

How can you tell animal and plant cells apart

Lynna [10]

Well there are a couple ways. 1: The easiest, plant cells have a cell wall or extra protection. You will not find this on an animal cell, as there is only a membrane. 2: Chloroplast. This also will not be found in animal cells as they produce the plant's green color as well as the sugar. Hope this helps!

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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

4 years ago

How can a large construction project impact water quality, supply or flow?

Korolek [52]

Answer:

Construction also impacts the community’s water quality. During the construction phase, pollutants can end up in the water supply. Once construction is over, the water quality can still be a problem due to an increased demand for water, as well as increased pollution due to additional residents.

Explanation:

Hope This Could be of some use

3 0

3 years ago