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3 years ago

A bowling ball is pushed with a force of 22.0N and accerlates at 5.5 m/s square. What is the mass of the bowling ball

Physics

1 answer:

stealth61 [152]3 years ago

7 0

Mass of bowling ball is 4.0kg (A)

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Which is brighter in our sky, a star with apparent magnitude 2 or a star with apparent magnitude 7?

Lina20 [59]

The star with apparent magnitude 2 is more brighter than 7.

To find the answer, we have to know about apparent magnitude.

<h3>What is apparent magnitude?</h3>

100 times as luminous as a star with an apparent brightness of 7 is a star with a magnitude of 2.
The apparent magnitude of bigger stars is always smaller.
The brightest star in the night sky is Sirius.
The brightness of a star or other celestial object perceived from Earth is measured in apparent magnitude (m).
The apparent magnitude of an object is determined by its inherent luminosity, its distance from Earth, and any light extinction brought on by interstellar dust in the path of the observer's line of sight.

Thus, we can conclude that, the star with apparent magnitude 2 is more brighter than 7.

Learn more about the apparent magnitude here:

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7 0

1 year ago

Greg throws a 2.8-kg pumpkin horizontally off the top of the school roof in order to hit Mr. H's car. The car has parked a dista

Igoryamba

Answer:

The horizontal velocity is $v = 9.2 m/s$

Explanation:

From the question we are told that

The mass of the pumpkin is $m = 2.8 \ kg$

The distance of the the car from the building's base is $d = 13.4 \ m$

The height of the roof is $h = 10.4 \ m$

The height is mathematically represented as

$h = \frac{1}{2} gt^2$

Where g is the acceleration due to gravity which has a value of $g =9.8 \ m/s^2$

substituting values

$10.4= 0.5 * 9.8 * t$

making the time taken the subject of the formula

$t = \frac{10.4}{0.5 * 9.8 }$

$t = 1.457 \ s$

The speed at which the pumpkin move horizontally can be represented mathematically as

$v = \frac{d}{t}$

substituting values

$v =\frac{13.4}{1.457}$

$v = 9.2 m/s$

7 0

2 years ago

If a pebble is being transported in a stream by rolling, how does the velocity of it compare to the velocity of the stream?

AleksandrR [38]

Streams carry sediment, like pebbles, in their flows. The pebbles can be in a variety of locations in the flow, depending on it's size, the balance between the upwards velocity on the pebble (drag and lift forces), and it's settling velocity.

3 0

3 years ago

Fernanda is working on a physics project. she is experimenting by rolling a marble down a ramp and then seeing how far it rolls

Ratling [72]

The distance covered on the floor after leaving the ramp is the dependent variable.

As a result of the marble's size, the substance it is constructed of, and the angle at which it is placed onto the ground, the distance it rolls varies.
Therefore, the angle at which the marble is released onto the ground, the type of material used to make the stone, or its size can all be considered independent variables.

<h3>What is Independent variable?</h3>

There are independent and dependent variables in every experiment.
A variable is considered independent if its change is not influenced by the change in another variable or factor.

<h3>What is Dependent variable?</h3>

In any experiment, the dependent variable must be measured or determined, and it must change as the independent variable does.

Learn more about independent and dependent variable here:

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5 0

2 years ago

Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 kg), each having (a) a speed o

Ad libitum [116K]

Answer:

Part A:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 6.05x10^{-14}m$ < $\lambda_{electron} = 1.10x10^{-10}m$

Part B:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 1.29x10^{-13}m$ < $\lambda_{electron} = 5.525x10^{-12}m$

Explanation:

The wavelength of each particle can be determined by means of the De Broglie equation.

$\lambda = \frac{h}{p}$ (1)

Where h is the Planck's constant and p is the momentum.

$\lambda = \frac{h}{mv}$ (2)

Part A

Case for the electron:

$\lambda = \frac{6.624x10^{-34} J.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

But $J = Kg.m^{2}/s^{2}$

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

$\lambda = 1.10x10^{-10}m$

Case for the proton:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(6.55x10^{6}m/s)}$

$\lambda = 6.05x10^{-14}m$

Hence, the proton has a smaller wavelength than the electron.

<em>Part B </em>

For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:

$KE = \frac{1}{2}mv^{2}$

$2KE = mv^{2}$

$v^{2} = \frac{2KE}{m}$

$v = \sqrt{\frac{2KE}{m}}$ (3)

Case for the electron:

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{9.11x10^{-31}Kg}}$

but $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{9.11x10^{-31}Kg}}$

$v = 1.316x10^{8}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(1.316x10^{8}m/s)}$

$\lambda = 5.525x10^{-12}m$

Case for the proton :

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{1.67x10^{-27}Kg}}$

But $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{1.67x10^{-27}Kg}}$

$v = 3.07x10^{6}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(3.07x10^{6}m/s)}$

$\lambda = 1.29x10^{-13}m$

Hence, the proton has a smaller wavelength than the electron.

7 0

3 years ago