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2 years ago

6

You are learning about energy transformations in science class. Mel and Sam's built this set-up to see if light energy could be

transformed into _________ energy.

1 answer:

Allisa [31]2 years ago

5 0

I think it might be heat energy. light transforms into heat energy

You might be interested in

As a 5.0 x 10^2 newton basketball player jumps from the floor up toward the basket, the magnitude of the force of her feet on th

Yuliya22 [10]

I believe the answer is (4) The reason that is, is because if the exponents are the same like 10^2 and 10^3, you can add them. Then you would get 10^5. You can go ahead though and multiply 5.0 and 1.0. Now remember that with decimals you don't need the zeros behind the decimal point. So that simplifies it with just 5 x 1. Leaving you with 5.0 x 10^5.

4 0

2 years ago

Read 2 more answers

A 0.500-kg ball traveling horizontally on a frictionless surface approaches a very massive stone at 20.0 m/s perpendicular to wa

gregori [183]

The magnitude of the change in momentum of the stone is about 18.4 kg.m/s

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<h3>Further explanation</h3>

Let's recall Impulse formula as follows:

$\boxed {I = \Sigma F \times t}$

<em>where:</em>

<em>I = impulse on the object ( kg m/s )</em>

<em>∑F = net force acting on object ( kg m /s² = Newton )</em>

<em>t = elapsed time ( s )</em>

Let us now tackle the problem!

$\texttt{ }$

<u>Given:</u>

mass of ball = m = 0.500 kg

initial speed of ball = vo = 20.0 m/s

final kinetic energy = Ek = 70% Eko

<u>Asked:</u>

magnitude of the change of momentum of the stone = Δp = ?

<u>Solution:</u>

<em>Firstly, we will calculate the final speed of the ball as follows:</em>

$Ek = 70\% \ Ek_o$

$\frac{1}{2} m v^2 = 70\% \ ( \frac{1}{2} m (v_o)^2 )$

$v^2 = 70 \% \ (v_o)^2$

$v = - v_o \sqrt{70 \%}$ → <em>negative sign due to ball rebounds</em>

$v = - v_o \sqrt{0.7} \texttt{ m/s}$

$\texttt{ }$

<em>Next, we could find the magnitude of the change of momentum of the stone as follows:</em>

$\Delta p_{stone} = - \Delta p_{ball}$

$\Delta p_{stone} = - [ mv - mv_o ]$

$\Delta p_{stone} = m[ v_o - v ]$

$\Delta p_{stone} = m[ v_o + v_o\sqrt{0.7} ]$

$\Delta p_{stone} = mv_o [ 1 + \sqrt{0.7} ]$

$\Delta p_{stone} = 0.500 ( 20.0 ) [ 1 + \sqrt{0.7} ]$

$\Delta p_{stone} \approx 18.4 \texttt{ kg.m/s}$

$\texttt{ }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437

Kinetic Energy : brainly.com/question/692781

Acceleration : brainly.com/question/2283922

The Speed of Car : brainly.com/question/568302
Average Speed of Plane : brainly.com/question/12826372
Impulse : brainly.com/question/12855855
Gravity : brainly.com/question/1724648

$\texttt{ }$

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

8 0

2 years ago

An airplane is delivering food to a small island. It flies 100 m above the ground at a speed of 150 m/s .

miss Akunina [59]

Answer:

The airplane should release the parcel $6.7*10^2$ m before reaching the island

Explanation:

The height of the plane is $y_0=100m$ , and its speed is v=150 m/s

When an object moves horizontally in free air (no friction), the equation for the y measured with respect to ground is

$y=y_0 - \frac{gt^2}{2}$ [1]

And the distance X is

x = V.t [2]

Being t the time elapsed since the release of the parcel

If we isolate t from the equation [1] and replace it in equation [2] we get

$X = V . \sqrt{\frac{2y_0}{g}}$

Using the given values:

$x = 150 m/s \sqrt{\frac{2\times 100m}{9.8 m/sec^2}}$

x = $6.7*10^2$ m

4 0

2 years ago

A metallic sphere of radius 2.0 cm is charged with +5.0-μC+5.0-μC charge, which spreads on the surface of the sphere uniformly.

sladkih [1.3K]

Answer:

Explanation:

Potential due to a charged metallic sphere having charge Q and radius r on its surface will be

v = k Q / r . On the surface and inside the metallic sphere , potential is the same . Outside the sphere , at a distance R from the centre potential is

v = k Q / R

a ) On the surface of the shell , potential due to positive charge is

V₁ = $\frac{9\times10^9\times5\times10^{-6}}{6\times10^{-2}}$

On the surface of the shell , potential due to negative charge is

V₁ = $\frac{- 9\times10^9\times5\times10^{-6}}{6\times10^{-2}}$

Total potential will be zero . they will cancel each other.

b ) On the surface of the sphere potential

= $\frac{9\times10^9\times5\times10^{-6}}{2\times10^{-2}}$

= 22.5 x 10⁵ V

On the surface of the sphere potential due to outer shell

= $\frac{9\times10^9\times5\times10^{-6}}{5\times10^{-2}}$

= -9 x 10⁵

Total potential

=( 22.5 - 9 ) x 10⁵

= 13.5 x 10⁵ V

c ) In the space between the two , potential will depend upon the distance of the point from the common centre .

d ) Inside the sphere , potential will be same as that on the surface that is

13.5 x 10⁵ V.

e ) Outside the shell , potential due to both positive and negative charge will cancel each other so it will be zero.

5 0

2 years ago

Compared to the resistivity of a 0.4-meter length of 1-millimeter-diameter copper wire at 0 degrees Celsius, the resistivity of

Mazyrski [523]

Answer:

Resistivity of both wires are same

Explanation:

Length of one wire, $l_1=0.4 m$

Diameter, $d_1=1mm$

Radius, $r_1=\frac{d_1}{2}=\frac{1}{2}mm=0.5\times 10^{-3} m$

$1mm=10^{-3} m$

$l_2=0.8 m$

$d_2=1mm$

$r_2=0.5\times 10^{-3} m$

Temperature in each case is same.

Area of each wire, $A_1=A_2=A=\pi r^2=\pi (0.5\times 10^{-3})^2m^2$

Resistivity is the property of material due to which it offers resistance to the flow of current.

Resistivity of material depends upon the temperature and material by which it is made.

It does not depends upon the length of object.

Therefore, the resistivity of both wires of different length are same.

3 0

2 years ago

Read 2 more answers