What is the power generated by a 3.7 V cell phone battery that provides 2.0 A of current?

The space station is 4.41 x 10^5 kg and orbits the earth 6.78 x 10^6 m from the center of earth. The mass of earth is 5.97 x 10^

allochka39001 [22]

Answer:

3 820 885 N

Explanation:

Gravitational equation

F = G m1 m2 / r^2

G = gravitational constant = 6.6713 x 10^-11 m^3/kg-s^2

F = 6.6713 x 10^-11 * 4.41 x 10^5 * 5.97 x 10^24 / ( 6.78x 10^6)^2

= 3820885 .3 N

6 0

2 years ago

A player kicks a football from ground level with a velocity of 26.2m/s at an angle of 34.2° above the horizontal. How far back f

Amanda [17]

For the ball to go straight into the goal, the kicker needs to be no more than 6.54 meters away from the goal.

For the ball to arc into the goal, the kicker needs to be between 58.5 and 65.1 meters away from the goal.

<h3>Explanation</h3>

How long does it take for the ball to reach the goal?

Let the distance between the kicker and the goal be $x$ meters.

Horizontal velocity of the ball will always be $26.2\times\cos{34.2\textdegree}$ until it lands if there's no air resistance.

The ball will arrive at the goal in $\displaystyle \frac{x}{26.2\times\cos{34.2\textdegree}}$ seconds after it leaves the kicker.

What will be the height of the ball when it reaches the goal?

Consider the equation

$\displaystyle h(t) = -\frac{1}{2}\cdot g\cdot t^{2} + v_{0,\;\text{vertical}} \cdot t + h_0$ .

For this soccer ball:

$g = 9.81\;\text{m}\cdot\text{s}^{-2}$ ,
$v_{0,\;\text{vertical}} = 26.2\times \sin{34.2\textdegree{}}\;\text{m}\cdot\text{s}^{-2}$ ,
$h_0 = 0$ since the player kicks the ball "from ground level."

$\displaystyle t=\frac{x}{26.2\times\cos{34.2\textdegree}}$

when the ball reaches the goal.

$\displaystyle h= - 9.81 \times \frac{x^2}{(26.2\times\cos{34.2\textdegree})^2} + (26.2 \times \sin{34.2\textdegree})\times\frac{x}{26.2\times\cos{34.2\textdegree}} \\\phantom{h} = -\frac{9.81}{(26.2\times\cos{34.2\textdegree})^2}\cdot x^{2} + \frac{\sin{34.2\textdegree}}{\cos{34.2\textdegree}}\cdot x$ .

Solve this quadratic equation for $x$ , $x > 0$ .

$x = 65.1$ meters when $h = 0$ meters.
$x = 6.54$ or $58.5$ meters when $h = 4$ meters.

In other words,

For the ball to go straight into the goal, the kicker needs to be no more than 6.54 meters away from the goal.
For the ball to arc into the goal, the kicker needs to be between 58.5 and 65.1 meters away from the goal.

3 0

3 years ago

Calculate the potential difference across a 25- resistor if a 0.3-A current is flowing through it.

alekssr [168]

Voltage = (current) x (resistance)

Voltage = (0.3 A) x (25 ohms)

Voltage = (0.3 x 25) (A-Ω)

<em>Voltage = 7.5 volts</em>

8 0

3 years ago

Read 2 more answers

A force F is used to raise a 4-kg mass M from the ground to a height of 5 m. What is the work done by the force F? (Note: sin 60

miskamm [114]

Answer:

Answer:196 Joules

Explanation:

Hello

Note: I think the text in parentheses corresponds to another exercise, or this is incomplete, I will solve it with the first part of the problem

the work is the product of a force applied to a body and the displacement of the body in the direction of this force

assuming that the force goes in the same direction of the displacement, that is upwards

W=F*D (work, force,displacement)

the force necessary to move the object will be

$F=mg(mass *gravity)\\F=4kgm*9.8\frac{m}{s^{2} }\\ F=39.2 Newtons\\replace\\\\W=39.2 N*5m\\W=196\ Joules$

Answer:196 Joules

I hope it helps

5 0

2 years ago

A beam of electrons passes through a single slit, and a beam of protons passes through a second, but identical, slit. The electr

ryzh [129]

Answer:

(b) The electrons, because they have the smaller momentum and, hence, the larger de Broglie wavelength

Explanation:

de Broglie wavelength λ = h / m v

Since both electrons and protons have same velocity , momentum mv will be less for electrons because mass of electron is less .

for electron , momentum is less so . Therefore de Broglie wavelength λ will be more for electrons .

Amount of diffraction that is angle of diffraction is proportional to λ

Therefore electrons having greater de Broglie wavelength will show greater diffraction.

7 0

2 years ago