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enyata [817]
3 years ago
9

For this problem, we assume that we are on planet-i. the radius of this planet is r =4200 km, the gravitational acceleration at

the surface is gi =9.1 m/s 2 , and the gravitational constant g = 6.67 × x10−11 n m2 /kg2 in si units. the mass of planet-i is not given. not all the quantities given here will be used. suppose a cannon ball of mass m = 5130 kg is projected vertically upward from the surface of this planet. it rises to a maximum height h =12306 km above the surface of the planet. caution: here the gravitational acceleration decreases as the cannon ball travels away from planet-i. r h determine the kinetic energy (in joules) of the cannon ball immediately after it is fired off. answer in units of j.
Physics
1 answer:
Minchanka [31]3 years ago
3 0
The expression commonly used for potential gravitational energy is just simplification. It is actually just the first term in Taylor expansion of the real expression. 
In general, the potential energy of gravitational field is defined as:
U=-G \frac{mM}{r}
Where G is universal gravitational constant, and r is the distance between the objects centers of mass. Negative sign represents the bound state.
Since we are not given the mass of the planet we have to calculate it.
F_g=G\frac{mM}{r_p^2}\\ mg=G\frac{mM}{r_p^2}\\ g=G\frac{M}{r_p^2}
This formula can be used for any planet. It gives you the gravitational acceleration on the planet's surface. We can use it to calculate the planet's mass:
g=G\frac{M}{r_p^2}\\ M=\frac{gr_p^2}{G}=2.41\cdot 10^{24}kg
Now we can calculate the potential energy of that cannonball when it reaches its maximum height.
U=-G \frac{mM}{r}\\ U=-G \frac{mM}{r_p+h}
When we plug in the numbers we get:
U=-4.99\cdot 10^{10} J
The potential energy has to be equal to the kinetic energy.
E_k=4.99\cdot 10^{10} J

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A 1.50 cm high diamond ring is placed 20.0 cm from a concave mirror with radius of curvature 30.00 cm. The magnification is ____
Rama09 [41]

Answer:

Magnification, m = -0.42

Explanation:

It is given that,

Height of diamond ring, h = 1.5 cm

Object distance, u = -20 cm

Radius of curvature of concave mirror, R = 30 cm

Focal length of mirror, f = R/2 = -15 cm (focal length is negative for concave mirror)

Using mirror's formula :

\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}, f = focal length of the mirror

\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}

\dfrac{1}{v}=\dfrac{1}{-15}+\dfrac{1}{-20}

v = -8.57 cm

The magnification of a mirror is given by,

m=\dfrac{-v}{u}

m=\dfrac{-(-8.57)}{-20}

m = -0.42

So, the magnification of the concave mirror is 0.42. Thew negative sign shows that the image is inverted.

5 0
3 years ago
People cannot see certain types of light waves because:
enot [183]

Answer:

Light comes in different colors like radio, ultra violet, gamma-ray, etc, and they are invisible to the bare eye

Explanation:

5 0
2 years ago
Suppose that you connect the terminals of two batteries of different emfs positive to positive and negative to negative (opposin
gulaghasi [49]

Answer:

Answer is explained in the explanation section below.

Explanation:

This question is very basic and easy. The answer to this question is:

Answer: If both batteries are connected we would get less amount of charge as compared to connected a single battery.

Reasoning:

If both batteries are connected in a manner of positive terminal to positive terminal and negative terminal to negative terminal then a capacitor is added to charge it from the batteries then, total electromotive force (emf) would decrease.

As a result, the capacitor added would get less amount of charge stored. But capacitor added will get more amount of charge stored when a single battery is connected.

7 0
2 years ago
A piece of aluminum has a volume of 1.50 10-3 m3. the coefficient of volume expansion for aluminum is β = 69 ✕ 10-6 (°c)-1. the
Alex17521 [72]

Answer:

W = 3.12 J

Explanation:

Given the volume is 1.50*10^-3  m^3 and the coefficient of volume for aluminum is β = 69*10^-6 (°C)^-1. The temperature rises from 22°C to 320°C. The difference in temperature is 320 - 22 = 298°C, so ΔT = 298°C. To reiterate our known values we have:

β = 69*10^-6 (°C)^-1       V = 1.50*10^-3  m^3       ΔT = 298°C

So we can plug into the thermal expansion equation to find ΔV which is how much the volume expanded (I'll use d instead of Δ because of format):

dV = \beta V_{0} dT\\dV = (69*10^{-6})( C)^{-1} * (1.50*10^{-3})m^{3} * (298)C\\dV = 3.0843*10^-5

So ΔV = 3.0843*10^-5 m^3

Now we have ΔV, next we have to solve for the work done by thermal expansion. The air pressure is 1.01 * 10^5 Pa

To get work, multiply the air pressure and the volume change.

W = P * dV = (1.01 * 10^5)Pa * (3.0843*10^{-5})m^3\\W = 3.115143J

W = 3.12 J

Hope this helps!

4 0
3 years ago
What are the <br>difference between clinical and laboratory thermometer?<br><br><br>​
Mamont248 [21]

Answer:

Explanation:

Clinical Thermometer is meant for clinical purposes. It is developed for measuring the human body temperature. A laboratory thermometer, which is colloquially known as the lab thermometer, is used for measuring temperatures other than the human body temperature.

6 0
3 years ago
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