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1 year ago

This lab is investigating the relationship between mass, ________, and momentum.

Physics

1 answer:

Sav [38]1 year ago

7 0

This lab is investigating the relationship between mass, <u>Speed </u>, and momentum.

Momentum is manufactured from the mass and speed of an object. it's miles a vector quantity, owning a significance and a direction. If m is an object's mass and v is its speed, then the object's momentum is p.

Momentum in an easy way is a quantity of movement. right here amount is measurable because if an item is moving and has mass, then it has momentum. If an object no longer flows then it has no momentum. however, in regular existence, it has an important but many people didn't understand it.

Momentum gives the connection between the mass, pace, and direction of an object. Any exchange in momentum results in pressure. So, an exchange in momentum is used to determine the force appearing upon the item.

Learn more about momentum here:-brainly.com/question/1042017

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What is the wavelength of an electromagnetic wave that travels at 3 10 m/s and has a frequency of 60 mhz? (1 mhz = 1,000,000 hz)

Dahasolnce [82]

V = 310 m/s
f = 60 MHz = 60 × 10^6 Hz
v = xf
x = v/f
x = 310/(60 × 10^6) m
x = 5.166667 × 10^(−6) m

8 0

3 years ago

Read 2 more answers

A solar cell generates a potential difference of 0.25 V when a 550 Ω resistor is connected across it, and a potential difference

Andre45 [30]

a) $400 \Omega$

b) 0.43 V

c) 0.44 %

Explanation:

For a battery with internal resistance, the relationship between emf of the battery and the terminal voltage (the voltage provided) is

$V=E-Ir$ (1)

where

V is the terminal voltage

E is the emf of the battery

I is the current

r is the internal resistance

In this problem, we have two situations:

1) when $R_1=550 \Omega$ , $V_1=0.25 V$

Using Ohm's Law, the current is:

$I_1=\frac{V_1}{R_1}=\frac{0.25}{550}=4.5\cdot 10^{-4} A$

2) when $R_2=1000 \Omega$ , $V_2=0.31 V$

Using Ohm's Law, the current is:

$I_2=\frac{V_2}{R_2}=\frac{0.31}{1000}=3.1\cdot 10^{-4} A$

Now we can rewrite eq.(1) in two forms:

$V_1 = E-I_1 r$

$V_2=E-I_2 r$

And we can solve this system of equations to find r, the internal resistance. We do it by substracting eq.(2) from eq(1), we find:

$V_1-V_2=r(I_2-I_1)\\r=\frac{V_1-V_2}{I_2-I_1}=\frac{0.25-0.31}{3.1\cdot 10^{-4}-4.5\cdot 10^{-4}}=400 \Omega$

To find the electromotive force (emf) of the solar cell, we simply use the equation used in part a)

$V=E-Ir$

where

V is the terminal voltage

E is the emf of the battery

I is the current

r is the internal resistance

Using the first set of data,

$V=0.25 V$ is the voltage

$I=4.5\cdot 10^{-4}A$ is the current

$r=400\Omega$ is the internal resistance

Solving for E,

$E=V+Ir=0.25+(4.5\cdot 10^{-4})(400)=0.43 V$

In this part, we are told that the area of the cell is

$A=4.0 cm^2$

While the intensity of incoming radiation (the energy received per unit area) is

$Int.=5.5 mW/cm^2$

This means that the power of the incoming radiation is:

$P=Int.\cdot A=(5.5)(4.0)=22 mW = 0.022 W$

This is the power in input to the resistor.

The power in output to the resistor can be found by using

$P'=I^2R$

where:

$R=1000 \Omega$ is the resistance of the resistor

$I=3.1\cdot 10^{-4} A$ is the current on the resistor (found in part A)

Susbtituting,

$P'=(3.1\cdot 10^{-4})^2(1000)=9.61\cdot 10^{-5} W$

Therefore, the efficiency of the cell in converting light energy to thermal energy is:

$\epsilon = \frac{P'}{P}\cdot 100 = \frac{9.6\cdot 10^{-5}}{0.022}=0.0044\cdot 100 = 0.44\%$

7 0

3 years ago

What are some ways that electric fields are similar to magnetic fields?

sattari [20]

Answer:

The correct answers are the proportionality of the fields concerning distance, vector fields, and forces at a distance.

Explanation:

The similarities between magnetic fields and electric fields are that electric fields are produced by two charges that can be positive and negative. Magnetic fields are associated with two magnetic poles, although they are also produced by moving charges. Both fields are inversely proportional to the square of the distance between the sources, both fields are vectorial and both act by distant forces.

Have a nice day!

3 0

3 years ago

A positive point charge q is placed at the center of an uncharged metal sphere insulated from the ground. The outside of the sph

Olenka [21]

Answer:

B.The charge on A is -q; there is no charge on B.

Explanation:

We are given that

Charge=+q

We have to find the correct statement.

When positive charge is placed at center of uncharged metal sphere

insulated from the ground then negative charge(-q) induced on inner

surface A of sphere and the outer surface B is grounded then the surface is neutral .

It means there is no charge on surface B.

Hence, option B is true .

B.The charge on A is -q; there is no charge on B.

4 0

2 years ago

State the 3 laws of motion

Georgia [21]

Answer:

Explanation:

Newton's first law of motion:

An object in motion stays in motion, and an object at rest stays at rest, until acted upon by an unbalanced force.

Newton's second law:

The net force on an object is equal to its mass times its acceleration.

Newton's third law:

For every action, there is an opposite and equal reaction.

6 0

3 years ago