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2 years ago

Question 1 of 10

Physics

1 answer:

Dahasolnce [82]2 years ago

3 0

Answer:

They have the same rate of motion.

Explanation:

Electromagnetic radiation is a type of energy that is commonly known as light. Generally, light travels in waves, and all electromagnetic radiation travels at the same speed, which is about 3.0 × 108 ms− 1 through a vacuum.

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Calculate the radius of each ball to the nearest tenth. Use the formula r = d/2 where r is the radius and d is the diameter. Rec

erma4kov [3.2K]

Answer:

What is the radius of the table tennis ball?

⇒ 2.1 cm

What is the radius of the golf ball?

2.0 cm

Explanation:

divide the radius and round it to the nearest 10th place..... but hope that help ;)

8 0

3 years ago

Read 2 more answers

A woman pushes a 27 kg lawnmower at a steady speed. She exerts a 120 N force in a direction 35◦ below the horizontal. The accele

nikklg [1K]

The vertical force exerted on the lawn is 68.8 N downward

Explanation:

The vertical force exerted by the lawnmower on the lawn is equal to the vertical component of the force applied, therefore:

$F_y = F sin \theta$

where

F is the magnitude of the force applied

$\theta$ is the angle between the direction of the force and the horizontal

In this problem:

F = 120 N

$\theta=-35^{\circ}$

Substituting,

$F_y = (120)(sin 30)=-68.8 N$

where the negative sign means the direction of the force is downward.

Learn more about vector components and forces here:

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3 0

3 years ago

Based on the mass and speed data, which object will have the greatest amount of kinetic energy

Alexus [3.1K]

The objects will have to be listed

5 0

2 years ago

Read 2 more answers

A 175-kg roller coaster car starts from rest at the top of an 18.0-m hill and rolls down the hill, then up a second hill that ha

Anni [7]

Answer:

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem we present the equations that describe the situation of the roller coaster car on each top of the hill. Let consider that bottom has a height of zero meters.

From top of the first hill to the bottom

$m\cdot g \cdot h_{1} = \frac{1}{2}\cdot m\cdot v_{1}^{2} +W_{1, loss}$ (1)

From the bottom to the top of the second hill

$\frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2}+W_{2,loss}$ (2)

Where:

$m$ - Mass of the roller coaster car, in kilograms.

$v_{1}$ - Speed of the roller coaster car at the bottom between the two hills, in meters per second.

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

$h_{1}$ - Height of the first top of the hill with respect to the bottom, in meters.

$W_{1, loss}$ - Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.

$v_{2}$ - Speed of the roller coaster car at the top of the second hill, in meters per seconds.

$h_{2}$ - Height of the second top of the hill with respect to the bottom, in meters.

$W_{2, loss}$ - Work done by non-conservative forces on the car bewteen the bottom between the two hills and the top of the second hill, in joules.

By using (1) and (2), we reduce the system of equation into a sole expression:

$m\cdot g\cdot h_{1} = m\cdot g\cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2} + W_{loss}$ (3)

Where $W_{loss}$ is the work done by non-conservative forces on the car from the top of the first hill to the top of the second hill, in joules.

If we know that $m = 175\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $h_{1} = 18\,m$ , $h_{2} = 8\,m$ and $v_{2} = 11\,\frac{m}{s}$ , then the work done by non-conservative force is:

$W_{loss} = m\cdot\left[ g\cdot \left(h_{1}-h_{2}\right)-\frac{1}{2}\cdot v_{2}^{2} \right]$

$W_{loss} = 6574.75\,J$

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

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3 years ago

When observing a specimen in a microscope you are told that the total magnification of the specimen is 630x assuming you are usi

Norma-Jean [14]

<span>Objective Lenses: Usually you will find 3 or 4 objective lenses on a microscope. They almost always consist of 4X, 10X, 40X and 100X powers. When coupled with a10X (most common) eyepiece lens, we get total magnifications of 40X (4X times10X), 100X , 400X and 1000X.</span>

5 0

3 years ago