3 years ago

PLEASE ASAP ILL GIVE BRAINLIEST.

Physics

1 answer:

Fiesta28 [93]3 years ago

5 0

Acceleration no longer exist as the car stops.

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The frequency is slowly increased. Once it passes the critical value fb, the student hears the ball bounce. There is now enough

vitfil [10]

Answer: g = acceleration = A*w^2 = A*(2*pi*fb)^2.

Explanation:

The ball bounces when the acceleration of the ball exceeds that of gravity. If A and fb are measured at that point, g = acceleration = A*w^2 = A*(2*pi*fb)^2.

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3 years ago

What is the heat needed to raise the temperature of 24.7 kg silver from 14.0 degrees celsius to 30.0 degrees celsius? specific h

Citrus2011 [14]

The amount of heat needed to increase the temperature of a substance by $\Delta T$ is given by
$Q=m C_s \Delta T$
where
m is the mass of the substance
$C_s$ is its specific heat capacity
$\Delta T$ is the increase of temperature

The sample of silver of our problem has a mass of $m=24.7 kg$ . Its specific heat capacity is $C_s = 236 J/g^{\circ}C$ and the increase in temperature is
$\Delta T=30.0^{\circ}-14.0^{\circ}C=16.0^{\circ}C$

Therefore, the amount of heat needed is
$Q=mC_s \Delta T=(24.7 kg)(236 J/g^{\circ}C)(16.0^{\circ}C)=9.32 \cdot 10^4 J$

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4 years ago

How are the frequency and period of a wave related to each other?

AVprozaik [17]

Hope this answer helps

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3 years ago

The density of the object on the left is 1.5 g/cm3 and the density of the fluid is 1.0 g/cm3. Which has greater density? The obj

gizmo_the_mogwai [7]

the first one is object and the second one is sink

4 0

4 years ago

Read 2 more answers

a 5.0 kg block is at rest on a horizontal floor. if you push horizontally on the 5.0 kg block with a force of 15.0 n, it just st

zheka24 [161]

For the 5.0Kg block with the force of 15 Newton, the coefficient of static friction (μ) comes out to be 0.306. Applying the formula F = μmg.

Static friction acts on stationary body/body at rest. Let μ be the coefficient of static friction between the block and the horizontal floor.

Using the formula: F = μmg

Where,

F = Force , m= mass of the block and g = gravity.

and values of: m= 5.0Kg, F= 15.0N, g= 9.8m/s²

We can get: μ = F/(mg)

μ = 15/49

μ =0.306

Therefore, coefficient of static friction (μ) = 0.306.

To learn more about static friction visit the link- brainly.com/question/13680415

#SPJ4

4 0

2 years ago