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3 years ago

14

Suppose you found a rock that has all of the characteristics of a meteorite. You take it to a physicist friend who confirms that

it is a meteorite but says that radioisotope dating indicates an age of only a billion years.
What might be the origin of this meteorite?

1 answer:

Usimov [2.4K]3 years ago

6 0

Answer: The origin might be from MARS

Explanation: Meteorites origin are either from Asteroid, Moon or Mars. Their origin is usually classified according to their age after using radioisotope dating. Meteorites that originate from Mars are between the age of 4.5 billion to 200 million years. Since the meteorite in question is only a billion years old, the origin would be from Mars.

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A car with an initial speed of 23.4 km/h accelerates at a uniform rate of 0.90 m/s2 for 4.0 s.

ss7ja [257]

Final velocity is 10.1 m/s

5 0

3 years ago

19. Find the recoil velocity of a 65 kg ice hockey goalie who

Alexxx [7]

Answer:

c. 0.12 m/s

Explanation:

by using momentum formula $v_2=\frac{m_1\times v_1}{m_1+m_2}$

we get

$v_2=\frac{0.15\times 50}{0.15+65}=0.12$

4 0

4 years ago

A light, rigid rod is 77.0cm long. Its top end is pivoted on a frictionless, horizontal axle. The rod hangs straight down at res

Tju [1.3M]

A minimum speed of 5.49 m/s of is required to make the ball go over the top of the circle.

Length of the rod = 77 cm = 0.77 m

Let the maximum height to which the ball can reach be h.

The kinetic energy of the ball is

$= \frac{1}{2} mv ^{2}$

The potential energy of the ball is

$= mgh$

The velocity of the ball when it is at the top end is 0.

So, the ball's energy from kinetic energy to potential energy is transferred.

Kinetic energy → Potential energy

$\frac{1}{2} mv ^{2} = mgh$

$h = \frac{1 \:mv ^{2}}{2 \: mgh}$

The maximum height to which the ball can reach is equal to twice the length of the rod.

h = 2 L

$h =2 \:L \frac{1 \:mv ^{2}}{2 \: mgh}$

The minimum speed at the bottom required to make the ball go over the top of the circle is,

$v = \sqrt{2g(2L)}$

$v = 4 \times 9.8 \times 0.77$

$v = 5.49 \: m/s$

Therefore, the minimum speed of 5.49 m/s at the bottom is required to make the ball go over the top of the circle.

To know more about potential energy, refer to the below link:

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5 0

2 years ago

Four identical capacitors are connected with a resistor in two different ways. When they are connected as in part a of the drawi

Nina [5.8K]

Answer:

$T_2 = 0.592$

Explanation:

Given

$T_1 = 1.48s$

See attachment for connection

Required

Determine the time constant in (b)

First, we calculate the total capacitance (C1) in (a):

The upper two connections are connected serially:

So, we have:

$\frac{1}{C_{up}} = \frac{1}{C} + \frac{1}{C}$

Take LCM

$\frac{1}{C_{up}} = \frac{1+1}{C}$

$\frac{1}{C_{up}}= \frac{2}{C}$

Cross Multiply

$C_{up} * 2 = C * 1$

$C_{up} * 2 = C$

Make $C_{up}$ the subject

$C_{up} = \frac{1}{2}C$

The bottom two are also connected serially.

In other words, the upper and the bottom have the same capacitance.

So, the total (C) is:

$C_1 = 2 * C_{up}$

$C_1 = 2 * \frac{1}{2}C$

$C_1 = C$

The total capacitance in (b) is calculated as:

First, we calculate the parallel capacitance (Cp) is:

$C_p = C+C$

$C_p = 2C$

So, the total capacitance (C2) is:

$\frac{1}{C_2} = \frac{1}{C_p} + \frac{1}{C} + \frac{1}{C}$

$\frac{1}{C_2} = \frac{1}{2C} + \frac{1}{C} + \frac{1}{C}$

Take LCM

$\frac{1}{C_2} = \frac{1 + 2 + 2}{2C}$

$\frac{1}{C_2} = \frac{5}{2C}$

Inverse both sides

$C_2 = \frac{2}{5}C$

Both (a) and (b) have the same resistance.

So:

We have:

Time constant is directional proportional to capacitance:

So:

$T\ \alpha\ C$

Convert to equation

$T\ =kC$

Make k the subject

$k = \frac{T}{C}$

$k = \frac{T_1}{C_1} = \frac{T_2}{C_2}$

$\frac{T_1}{C_1} = \frac{T_2}{C_2}$

Make T2 the subject

$T_2 = \frac{T_1 * C_2}{C_1}$

Substitute values for T1, C1 and C2

$T_2 = \frac{1.48 * \frac{2}{5}C}{C}$

$T_2 = \frac{1.48 * \frac{2}{5}}{1}$

$T_2 = \frac{0.592}{1}$

$T_2 = 0.592$

Hence, the time constance of (b) is 0.592 s

8 0

3 years ago

a train is moving at a constant speed on a surface and inclined at 10 degrees with the horizontal travels a distance of 400 m in

BabaBlast [244]

Answer:

13.9 m/s

Explanation:

The velocity of the train along the inclined surface is given by the ratio between the distance travelled, d, and the time taken, t:

$v=\frac{d}{t}$

where we have

d = 400 m

t = 5 s

Substituting,

$v=\frac{400 m}{5 s}=80 m/s$

This is the velocity of the train along the inclined plane. Now we can find the vertical component of the velocity by using the formula:

$v_y = v sin \theta$

where $\theta=10^{\circ}$ is the angle of the slope. Substituting, we find

$v_y = (80 m/s)(sin 10^{\circ})=13.9 m/s$

3 0

3 years ago