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3 years ago

7

A boy is going down a slide. As he reaches the bottom, friction causes him to slow down and stop. Which one of Newton's laws is

this?
2nd law
3rd law
1st law
4th law

1 answer:

lara31 [8.8K]3 years ago

8 0

I think it’s the second law

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If we interpret the large, angular rocks to have originated from the outcrop at the top of the hill, we are using __________ rea

IrinaK [193]

The reasoning which is in use when large, angular rocks are interpreted to have originated from the outcrop at the top of the hill is; Fossil succession

<h3>Fossil succession of rocks</h3>

The principle of fossil succession in characterized by the fact that fossil entities succeed one another upward through rock layers in a definite and determinable order.

On this note, any time period can be dated by its fossil content.

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2 years ago

Which of the following tools is used to observe objects that are very far away? A. a anemometer B. a hand lens C. a microscope D

inn [45]

Answer:

D. a telescope

Explanation:

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3 years ago

Read 2 more answers

The central star of a planetary nebula emits ultraviolet light with wavelength 104nm. This light passes through a diffraction gr

Gala2k [10]

Answer: 31.33 degrees

Explanation:

The diffraction angles $\theta_{n}$ when we have a slit divided into $n$ parts are obtained by the following equation:

$dsin\theta_{n}=n\lambda$ (1)

Where:

$d$ is the width of the slit

$\lambda$ is the wavelength of the light

$n$ is an integer different from zero.

Now, the first-order diffraction angle is given when $n=1$ , hence equation (1) becomes:

$dsin\theta_{1}=\lambda$ (2)

Now we have to find the value of $\theta_{1}$ :

$sin\theta_{1}=\frac{\lambda}{d}$

$\theta_{1}=arcsin(\frac{\lambda}{d})$ (3)

We know:

$\lambda=104nm=104(10)^{-9}m$

In addition we are told the diffraction grating has 5000 slits per mm, this means:

$d=\frac{1mm}{5000}=\frac{1(10)^{-3}m}{5000}$

Substituting the known values in (3):

$\theta_{1}=arcsin(\frac{104(10)^{-9}m}{\frac{1(10)^{-3}m}{5000}})$

$\theta_{1}=arcsin(0.52)$

<u>Finally:</u>

$\theta_{1}=31.33\º$ >>>This is the first-order diffraction angle

4 0

3 years ago

A small rubber ball is launched by a compressed-air cannon from ground level with an initial speed of 11.8 m/s directly upward.

Fantom [35]

Answer:

7.09683 m

1.20285 s

2.4057 s

11.8 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² (negative up, positive down)

From equation of motion we have

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-11.8^2}{2\times -9.81}\\\Rightarrow s=7.09683\ m$

The maximum height above the ground that the ball reaches is 7.09683 m

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-11.8}{-9.81}\\\Rightarrow t=1.20285\ s$

Time taken to go up is 1.20285 s it will take the same time to come down so total time taken to reach the ground after it is shot is 1.20285+1.20285 = 2.4057 s

$v=u+at\\\Rightarrow v=0+9.81\times 1.20285\\\Rightarrow v=11.8\ m/s$

The velocity just before it hits the ground is 11.8 m/s

6 0

3 years ago

Given a wire with a cross section of .45cm^2, a length of 3cm, and an internal resistance of 3 ohms, show your 5 steps to solve

Gre4nikov [31]

Resistance ∞ (proportional) length
resistance ∞ 1/ area

therefore,
(the constant that we take is known as the resistivity)

resistance = (resistivity*length )/ area
resistivity = (resistance * area ) / length
= (3 * 45) / 3 = 135/3 = 45 Ωm

in short your answer is 45 Ωm

4 0

3 years ago