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Tanya [424]
3 years ago
15

Asap PLEASE HELP!!!! Match these vocabulary words with their definitions.

Physics
1 answer:
lubasha [3.4K]3 years ago
3 0
1. Isotherm
2. Labrador current
3. Longitude
4. Gulf stream
5. Horse latitudes
6. Maritime
7. Middle latitude climates
8. North Atlantic drift

hope this helps

mark me brainliest please!
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Answer:0.27*10^{6} (m)

Explanation:

with a wave we know that  speed= wavelenght* frecuency   so

wavelenght= 3*10^{7} / 109  = 0.27*10^{6} (m)

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How is the average kinetic energy of particles related to temperature of a substance?
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As kinetic energy increases, substance temperature increases
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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
3 years ago
You get a new pair of sunglasses as a birthday gift from a friend. When wrapping the present, your friend has removed the price
Gwar [14]

Answer:

Explanation:By using the sunglasses while looking at If your sunglasses are polarized, instead of only seeing the surface of a lake or river, you will suddenly be able to see through the glare and into the water below.

Another one is by comparing two sunglasses, hold your glasses and another one simultaneously and look through both pairs at the same time. Then, rotate one pair of sunglasses about 60 degrees. If both pairs of glasses are polarized, the overlapping area will darken as they filter out excess light. If your pair isn't polarized, however, you won't notice any difference.

Also it can be determined by simple computer test If you rotate your glasses sideways while looking at a computer monitor through polarized glasses, portions of your screen will become blank or go dark. The same also applicable to LCD display screens such as the ones on a gas pump.

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3 years ago
A 4 kg block is pulled with 5.6 N of force to the right. The block experiences 1.1 N of friction. What is the acceleration of th
BARSIC [14]

Answer:

The acceleration of the box is 1.125 m/s² towards right.

Explanation:

Mass of the box, m=4 kg

Force acting towards right, F=5.6 N

Frictional force acting towards left, f=1.1 N

Let the acceleration be a m/s².

Now, net force acting on the box towards right is given as:

F_{net}=F-f=5.6-1.1=4.5\textrm{ N}

From Newton's second law of motion,

F_{net}=ma\\4.5=4a\\a=\frac{4.5}{4}=1.125\textrm{ }m/s^2

Therefore, the acceleration of the box is 1.125 m/s² towards right.

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