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pishuonlain [190]
3 years ago
9

A lab cart is loaded with different masses and moved at various velocities

Physics
1 answer:
Sedaia [141]3 years ago
6 0
A lab cart is loaded with different masses and moved at various constant velocities? the anser should be

1.0m/s → 4kg
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Earth’s surface is subjected to unusually high rates of weathering. The Moon and some of the inner planets of our solar system a
Komok [63]
<h2>Answer:</h2>

<em><u>Running water on the Surface of Earth.</u></em>

<h2>Explanation:</h2>

Weathering can be defined as the phenomenon of breaking or disintegration of the rocks, minerals, soil etc. due to coming in contact with the Earth's atmosphere such as the Water, Wind, Organisms etc.

In this process the rocks, minerals disintegrate into lot of small particles.

But the rate of weathering on Earth is at a very unusual or unique rate because Earth's surface has the presence of the running water that leads to the high rate of surface weathering.

<em><u>Therefore, the reason of Earth's rapid weathering rates is Running water on the Surface of Earth.</u></em>

<em><u></u></em>

3 0
3 years ago
3. the definition of the FREQUENCY of a sound is​
sergiy2304 [10]
Definition: Frequency is the measurement of the number of times that a repeated event occurs per unit of time. Sound moves through a medium such as air or water as waves. It is measured in terms of frequency and amplitude. Frequency, sometimes referred to as pitch, is the number of times per second that a sound pressure wave repeats itself.
8 0
3 years ago
Layer of the atmosphere where jets fly through and contains ozone layer is called ________
yaroslaw [1]

Answer:

The stratosphere

3 0
3 years ago
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What was one effect of plate movement on the pacific region?
djyliett [7]
The formation of new volcanoes and the eruption of some existing ones already. 
5 0
2 years ago
The magnetic field at the center of a 1.40-cm-diameter loop is 2.50 mT . PART A) What is the current in the loop?
NISA [10]

Explanation:

It is given that,

Diameter of loop, d = 1.4 cm

Radius of loop, r = 0.7 cm = 0.007 m

Magnetic field, B=2.5\ mT=2.5\times 10^{-3}\ T

(A) Magnetic field of a current loop is given by :

B=\dfrac{\mu_oI}{2r}

I is the current in the loop

I=\dfrac{2Br}{\mu_o}

I=\dfrac{2\times 2.5\times 10^{-3}\times 0.007}{4\pi \times 10^{-7}}

I = 27.85 A

(B) Magnetic field at a distance r from a wire is given by :

B=\dfrac{\mu_o I}{2\pi r}

r=\dfrac{\mu_o I}{2\pi B}

r=\dfrac{4\pi \times 10^{-7}\times 27.85}{2\pi \times 2.5\times 10^{-3}}

r = 0.00222 m

r=2.2\times 10^{-3}\ m

Hence, this is the required solution.

3 0
3 years ago
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