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2 years ago

Can u find the . in this?

Physics

2 answers:

nikklg [1K]2 years ago

4 0

Answer:

Yeah it's right there from the one next to the exclamation point

Yanka [14]2 years ago

4 0

Answer:

its not there

Explanation:

its literally not there

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I need help please and thanks

asambeis [7]

Answer:

Explanation:

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2 years ago

What is ozone? How and where is it formed in the atmosphere?

mrs_skeptik [129]

Ozone gas is made up of oxygen molecules that have three atoms. it exists in polluted air and ozone layer. Ozone layer is formed in stratosphere(part of atmosphere).

4 0

3 years ago

A 7 kg sled is initially at rest on a horizontal road. the sled is pulled a distance of 2.9 m by a force of 42 n applied to the

Artemon [7]

I have no idea on this question.

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3 years ago

What is Secular Music's instrument's?

nasty-shy [4]

Answer:

Drums, harps, recorders, and bagpipes.

Explanation:

8 0

2 years ago

A 50g ball is released from rest 1.0 above the bottom of thetrack

ludmilkaskok [199]

Answer:

The maximum height of the ball is 2 m.

Explanation:

Given that,

Mass of ball = 50 g

Height = 1.0 m

Angle = 30°

The equation is

$y=\dfrac{1}{4}x^2$

We need to calculate the velocity

Using conservation of energy

$\Delta U_{i}+\Delta K_{i}=\Delta K_{f}+\Delta U_{f}$

Here, ball at rest so initial kinetic energy is zero and at the bottom the potential energy is zero

$\Delta U_{i}=\Delta K_{f}$

Put the value into the formula

$mgh=\dfrac{1}{2}mv^2$

Put the value into the formula

$50\times10^{-3}\times9.8\times1.0=\dfrac{1}{2}\times50\times10^{-3}\times v^2$

$v^2=\dfrac{2\times50\times10^{-3}\times9.8\times1.0}{50\times10^{-3}}$

$v=\sqrt{19.6}$

$v=4.42\ m/s$

We need to calculate the maximum height of the ball

Using again conservation of energy

$\dfrac{1}{2}mv^2=mgh$

Here, h = y highest point

Put the value into the formula

$\dfrac{1}{2}\times50\times10^{-3}\times(4.42)^2=50\times10^{-3}\times9.8\times h$

$y=\dfrac{0.5\times(4.42)^2}{9.8}$

$y=0.996\ m$

Put the value of y in the given equation

$y=\dfrac{1}{4}x^2$

$x^2=4\times0.996$

$x=\sqrt{4\times0.996}$

$x=1.99\ m\ \approx 2 m$

Hence, The maximum height of the ball is 2 m.

4 0

2 years ago