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2 years ago

What is static and sliding friction

1 answer:

5 0

Static friction is the friction that exists between a stationary object and the surface on which it's resting.
frictional force occurs when you try to push an object alongside a surface.

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To make a given sound seem twice as loud, how should a musician change the intensity of the sound?

Serhud [2]

Answer:

C. Quadruple the intensity

Explanation:

The intensity of the sound is proportional to square of amplitude of the sound.

I ∝ A²

$\frac{I_1}{A_1^2} = \frac{I_2}{A_2^2}\\\\I_2 = \frac{I_1A_2^2}{A_1^2}$

When the given sound is twice loud as the initial value, then the new amplitude is twice the former.

A₂ = 2A₁

$I_2 = \frac{I_1A_2^2}{A_1^2} \\\\I_2 = \frac{I_1(2A_1)^2}{A_1^2} \\\\I_2 = \frac{4I_1A_1^2}{A_1^2}\\\\ I_2 = 4I_1$

Thus, to make a given sound seem twice as loud, the musician should Quadruple the intensity

3 0

2 years ago

What type of fat may help lower the risk of heart disease

enyata [817]

Polyunsaturated fatty acids which is Omega-3 fatty acids.

8 0

2 years ago

Suppose a proton ( = 1. 67×10^−27 kg) is confined to a box of width = 1. 00×10^−14 m (a typical nuclear radius).

dimulka [17.4K]

Answer:

22e837281949222324

Explanation:

3 0

2 years ago

A vertical spring stretches 4.0 cm when a 12-g object is hung from it. The object is replaced with a block of mass 28 g that osc

marin [14]

Answer:

Time period of the osculation will be 0.0671 sec

Explanation:

It is given a vertical spring is stretched by 4 cm

So change in length of the spring x = 4 cm = 0.04 m

Mass which is hung from it m = 12 gram = 0.012 kg

Sprig force will be equal to weight of the mass

So $kx=mg$

$k\times 0.04=0.012\times 9.8$

k = 244.7 N/m

Now new mass is m = 28 gram = 0.028 kg

So time period with new mass will be

$T=2\pi \sqrt{\frac{m}{k}}$

$=2\times 3.14 \sqrt{\frac{0.028}{244.7}}=0.0671sec$

4 0

2 years ago

As part of a safety investigation, two 1900 kg cars traveling at 20 m/s are crashed into different barriers. Part A Find the ave

DedPeter [7]

Answer:

- $29.2\times 10^{3} N$

Explanation:

We are given that

Mass of cars= m=1900 kg

Initial speed of car=u=20 m/s

Final speed of car=v=0

Time= $\Delta t$ =1.3 s

We have to find the average force exerted on the car.

Average force= $\frac{change\;in\;momentum}{\Delta t}$

$F_{avg}=\frac{mv-mu}{1.3}$

$F_{avg}=\frac{1900(0)-1900(20)}{1.3}$

$F_{avg}=\frac{-38000}{1.3}=-29.2\times 10^{3} N$

Hence, the average force exerted on the car that hits a line of water barrels=- $29.2\times 10^{3} N$

8 0

3 years ago