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3 years ago

50 POINTS

Physics

2 answers:

strojnjashka [21]3 years ago

8 0

Answer: A Answers. Assuming that the terminal velocity doesn't change during the fall, then the kinetic energy would remain constant. However the terminal velocity decreases during the fall since the air becomes denser at lower altitudes.

Explanation:

What happens to the KE of an object when it slows down and heats up? - Quora. The kinetic energy goes down and the loss of the kinetic energy is through the production of heat energy. In real world this is due to friction, or an opposing force that decelerates the object, or a combination of both.

Rina8888 [55]3 years ago

5 0

Answer:

Kinetic energy stays the same

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Consider a model of a hydrogen atom in which an electron is in a circular orbit of radius r = 5.92×10−11 m around a stationary p

DaniilM [7]

Answer:

2.068 x 10^6 m / s

Explanation:

radius, r = 5.92 x 10^-11 m

mass of electron, m = 9.1 x 10^-31 kg

charge of electron, q = 1.6 x 10^-19 C

As the electron is revolving in a circular path, it experiences a centripetal force which is balanced by the electrostatic force between the electron and the nucleus.

centripetal force = $\frac{mv^{2}}{r}$

Electrostatic force = $\frac{kq^{2}}{r^{2}}$

where, k be the Coulombic constant, k = 9 x 10^9 Nm^2 / C^2

So, balancing both the forces we get

$\frac{kq^{2}}{r^{2}}=\frac{mv^{2}}{r}$

$v=\sqrt{\frac{kq^{2}}{mr}}$

$v=\sqrt{\frac{9\times 10^{9}\times1.6\times 10^{-19}\times 1.6\times 10^{-19}}{9.1\times 10^{-31}\times 5.92\times10^{-11}}}$

v = 2.068 x 10^6 m / s

Thus, the speed of the electron is give by 2.068 x 10^6 m / s.

6 0

3 years ago

If two equal forces act on an object in opposite directions, what is the net

labwork [276]

Answer:

Net Force = 0

Explanation:

Causes objects to accelerate. Balanced Forces. Two equal forces push in opposite direction causing no change in motion causing net force = 0.

4 0

3 years ago

3) How far will 20 N of force stretch a spring with a spring constant of 140 N/m?

muminat

Answer:

7 meters, 2.8 meters

Explanation:

work done (nm) = force (n) * distance (m)

140= 20 * m

140/20 = m

m=7 meters

140= 50 * m

140/50 = m

m= 2.8 meters

4 0

2 years ago

What are scientists who study space called?

Brilliant_brown [7]

Answer:

Astronomer

Explanation:

A scientist who studies the objects in the sky, including planets, galaxies, black holes, and stars, is called an astronomer. These days, the terms astronomer and astrophysicist are used interchangeably, to talk about any physicist who specializes in celestial bodies and the forces that affect them.

4 0

3 years ago

Read 2 more answers

. Consider the equation =0+0+02/2+03/6+04/24+5/120, where s is a length and t is a time. What are the dimensions and SI units of

Olegator [25]

Answer:

See Explanation

Explanation:

Given

$s=s_0+v_0t+\frac{a_0t^2}{2}+ \frac{j_0t^3}{6}+\frac{S_0t^4}{24}+\frac{ct^5}{120}$

Solving (a): Units and dimension of $s_0$

From the question, we understand that:

$s \to L$ --- length

$t \to T$ --- time

Remove the other terms of the equation, we have:

$s=s_0$

Rewrite as:

$s_0=s$

This implies that $s_0$ has the same unit and dimension as $s$

Hence:

$s_0 \to L$ --- dimension

$s_o \to$ Length (meters, kilometers, etc.)

Solving (b): Units and dimension of $v_0$

Remove the other terms of the equation, we have:

$s=v_0t$

Rewrite as:

$v_0t = s$

Make $v_0$ the subject

$v_0 = \frac{s}{t}$

Replace s and t with their units

$v_0 = \frac{L}{T}$

$v_0 = LT^{-1}$

Hence:

$v_0 \to LT^{-1}$ --- dimension

$v_0 \to$ $m/s$ --- unit

Solving (c): Units and dimension of $a_0$

Remove the other terms of the equation, we have:

$s=\frac{a_0t^2}{2}$

Rewrite as:

$\frac{a_0t^2}{2} = s_0$

Make $a_0$ the subject

$a_0 = \frac{2s_0}{t^2}$

Replace s and t with their units [ignore all constants]

$a_0 = \frac{L}{T^2}\\$

$a_0 = LT^{-2$

Hence:

$a_0 = LT^{-2$ --- dimension

$a_0 \to$ $m/s^2$ --- acceleration

Solving (d): Units and dimension of $j_0$

Remove the other terms of the equation, we have:

$s=\frac{j_0t^3}{6}$

Rewrite as:

$\frac{j_0t^3}{6} = s$

Make $j_0$ the subject

$j_0 = \frac{6s}{t^3}$

Replace s and t with their units [Ignore all constants]

$j_0 = \frac{L}{T^3}$

$j_0 = LT^{-3}$

Hence:

$j_0 = LT^{-3}$ --- dimension

$j_0 \to$ $m/s^3$ --- unit

Solving (e): Units and dimension of $s_0$

Remove the other terms of the equation, we have:

$s=\frac{S_0t^4}{24}$

Rewrite as:

$\frac{S_0t^4}{24} = s$

Make $S_0$ the subject

$S_0 = \frac{24s}{t^4}$

Replace s and t with their units [ignore all constants]

$S_0 = \frac{L}{T^4}$

$S_0 = LT^{-4$

Hence:

$S_0 = LT^{-4$ --- dimension

$S_0 \to$ $m/s^4$ --- unit

Solving (e): Units and dimension of $c$

Ignore other terms of the equation, we have:

$s=\frac{ct^5}{120}$

Rewrite as:

$\frac{ct^5}{120} = s$

Make $c$ the subject

$c = \frac{120s}{t^5}$

Replace s and t with their units [Ignore all constants]

$c = \frac{L}{T^5}$

$c = LT^{-5}$

Hence:

$c \to LT^{-5}$ --- dimension

$c \to m/s^5$ --- units

4 0

3 years ago