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2 years ago

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A projectile launcher has a mass of 3 kg. It fires a projectile of mass 0.08 kg horizontally at a speed of 300 m/s. a. What is t

he total momentum of the system before the projectile is fired?

1 answer:

alexdok [17]2 years ago

6 0

Answer:

0

Explanation:

It’s before the projectile was fired, so nothing has happened yet.

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Suppose a skydiver (mass = 80kg) is falling toward the Earth. Calculate the skydiver’s gravitational potential energy at a point

Vlad [161]

Ep is gravitational potential energy
m is mass (kg)
g is gravitational field strength (N/kg)
h is height (m)

Ep= mgh
= 80kg*9.8N/kg*100m
= 78 400 J

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3 years ago

Lithium was one of the metals studied by the American physicist Robert Millikan in his research on the photoelectric effect. Whe

faltersainse [42]

Answer:

f = 7.9487 10¹³ Hz

Explanation:

The photoelectric effect was correctly explained by Einstein assuming that the radiation is composed of photons, which behave like particles.

hf = K + Ф

It indicates the frequency and the kinetic energy, let's look for the work function

Ф = hf - K

let's reduce the magnitudes to the SI system

K = 0.332 eV (1.6 10⁻¹⁹ J / 1 eV) = 0.5312 10⁻⁻¹⁹ J

let's calculate

Ф = 6.63 10⁻⁻³⁴ 6.64 10¹¹ - 0.5312 10⁻¹⁹

Ф = 4.40 10⁻²² - 0.5312 10⁻¹⁹

Ф = 5.27 10⁻²⁰ J

for the minimum frequency that produces photoelectrons, the kinetic energy is zero

hf = Ф

f = Ф / h

f = 5.27 10⁻²⁰ / 6.63 10⁻³⁴

f = 7.9487 10¹³ Hz

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2 years ago

Three +3.0-μC point charges are at the three corners of a square of side 0.50 m. The last corner is occupied by a −3.0-μC charge

kramer

Answer:

E = 440816.32 N/C

Explanation:

Given data:

Three point charge of charge equal to +3.0 micro coulomb

fourth point charge = - 3.0 micro coulomb

side of square = 0.50 m

$K =1/4 \pi \epsilon_0 = 8.99 \times 10^9$ N.m^2/c^2

Due to having equal charge on center of square, 2 charge produce equal electric field at center and other two also produce electric field at center of same value

So we have

$E_1 + E_3 = 0$

$E =E_2 + E_4$

$E = 2 E_2$

[ $E_2 =\frac{2\times k \times q}{r^2}$

[ $r= \frac{(0.5^2 + 0.5^2)^2}{2} = 0.35 m$ ]

plugging all value

$E = 2 E_2$

$E = 2 E_2 =\frac{2\times k \times q}{r^2}$

$E = \frac{2 \times 8.99 \times 10^93\times 10^{-6}}{0.35^2}$

E = 440816.32 N/C

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2 years ago

Read 2 more answers

If earth's mass were half its actual value but its radius stayed the same, the escape velocity of earth would be:________

siniylev [52]

If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be $V_e = \sqrt{\dfrac{GM}{r}}$ .

<h3>What is an escape velocity?</h3>

The ratio of the object's travel distance over a specific period of time is known as its velocity. As a vector quantity, the velocity requires both the magnitude and the direction. the slowest possible speed at which a body can break out of the gravitational pull of a certain planet or another object.

The formula to calculate the escape velocity of earth is given below:-

$V_e=\sqrt{\dfrac{2GM}{r}}$

Given that earth's mass was half its actual value but its radius stayed the same. The escape velocity will be calculated as below:-

$V_e=\sqrt{\dfrac{2GM}{r\times 2}}$

$V_e = \sqrt{\dfrac{GM}{r}}$ .

Therefore, If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be $V_e = \sqrt{\dfrac{GM}{r}}$ .

To know more about escape velocity follow

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1 year ago

Identify two factors you must know to describe the motion of an object along a straight line

Sveta_85 [38]

In order to describe motion along a straight line, you must state the speed and direction of the motion. Those two quantities, together, comprise what's known as "velocity".

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2 years ago