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3 years ago

What happens when velocity and acceleration are at right angles to each other

1 answer:

7 0

You get circular motion, where the acceleration is pointing towards the center of the circle, as long as they are constant, and not fluctuating.

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What is a difference between a law and a hypothesis?

Len [333]

A hypothesis is an educated guess. It refers to a theory you have based on some sort of evidence you collected or noticed. It is not fact and can sometimes be proven wrong. A law would refer to a scientific fact that has been proven correct and is a fact about the subject.

5 0

4 years ago

Two blocks of masses m1 and m2 are connected to each other on an Atwood machine (the blocks are connected by a string going over

ddd [48]

Answer:211.11 gm

Explanation:

Given

mass of lighter block $m_1=100 gm$

mass of heavier block is $m_2$

acceleration of system $a=3.5 m/s^2$

From diagram

for lighter block

$T-m_1g=m_1a$

$T=m_1(g+a)$

For heavier Block

$m_2g-T=m_2a$

$T=m_2(g-a)$

Equating Tension T

$m_1(g+a)=m_2(g-a)$

$m_2=m_1\cdot \frac{g+a}{g-a}$

$m_2=m_1\cdot \frac{9.8+3.5}{9.8-3.5}$

$m_2=0.1\cdot 2.11$

$m_2=0.2111 kg$

$m_2=211.11 gm$

8 0

3 years ago

A horizontal spring with a constant, k = 291 N/m, is stretched 11 cm with a 2 kg block attached to it. What is the maximum speed

Nuetrik [128]

Maximum speed acquired by the block is 1.32 m/s

<u>Explanation:</u>

Given-

spring constant, k = 291 N/m

distance, x = 11cm = 0.11m

Mass of the block, m = 2 kg

Maximum speed of the block, Vmax = ?

We know,

Period of its motion, ω = √k/m

ω = √291/2 N/mkg

ω = 12.06 rad/s

Maximum speed, Vmax = ωx

Vmax = 12.06 rad/s X 0.11m

Vmax = 1.32 m/s

Therefore, maximum speed acquired by the block is 1.32 m/s

3 0

3 years ago

If the distance between two masses is tripled, the gravitational force between changes by a factor of

Kruka [31]

Changes by a factor of 7n.

6 0

4 years ago

Thermopane window is constructed, using two layers of glass 4.0 mm thick, separated by an air space of 5.0 mm.

Bond [772]

To solve this problem it is necessary to apply the concepts related to rate of thermal conduction

$\frac{Q}{t} = \frac{kA\Delta T}{d}$

The letter Q represents the amount of heat transferred in a time t, k is the thermal conductivity constant for the material, A is the cross sectional area of the material transferring heat, $\Delta T$ , T is the difference in temperature between one side of the material and the other, and d is the thickness of the material.