1. Calculate the kinetic energy of a moving object, its mass= 100 kg, and its velocity= 12 m/s.

a street light is mounted at the top of a 15 foot pole. A man 6 ft tall walks away from the pole wit a speed of 7 ft/s along a s

Mariulka [41]

Answer:

16.3 ft/s

Explanation:

Let d=distance

and

x = length of shadow.

Therfore,

x=(d + x)

= 6/15

So,

15x = 6x + 6d

9x = 6d.

x = (2/3)d.

As we know that:

dx=dt

= (2/3) (d/dt)

Also,

Given:

d(d)=dt

= 7 ft/s

Thus,

d(d + x)=dt

= (7/3)d (d/dt)

Substitute, d= 7

d(d + x) = 49/3 ft/s.

Hence,

d(d + x) = 16.3 ft/s.

4 0

3 years ago

Suppose a small planet is discovered that is 16 times as far from the Sun as the Earth's distance is from the Sun. Use Kepler's

mamaluj [8]

Answer:

23376 days

Explanation:

The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.

$T^2\alpha R^3\\T^2=kR^3.......................(1)$

where k is a constant.

From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.

$\frac{T^2}{R^3}=k.......................(2)$

Let the orbital period of the earth be $T_e$ and its mean distance of from the sun be $R_e$ .

Also let the orbital period of the planet be $T_p$ and its mean distance from the sun be $R_p$ .

Equation (2) therefore implies the following;

$\frac{T_e^2}{R_e^3}=\frac{T_p^2}{R_p^3}....................(3)$

We make the period of the planet $T_p$ the subject of formula as follows;

$T_p^2=\frac{T_e^2R_p^3}{R_e^3}\\T_p=\sqrt{\frac{T_e^2R_p^3}{R_e^3}\\}................(4)$

But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore

$R_p=16R_e...............(5)$

Substituting equation (5) into (4), we obtain the following;

$T_p=\sqrt{\frac{T_e^2(16R_e)^3}{(R_e^3}\\}\\T_p=\sqrt{\frac{T_e^24096R_e^3}{R_e^3}\\}$

$R_e^3$ cancels out and we are left with the following;

$T_p=\sqrt{4096T_e^2}\\T_p=64T_e..............(6)$

Recall that the orbital period of the earth is about 365.25 days, hence;

$T_p=64*365.25\\T_p=23376days$

4 0

3 years ago

Which statement best describes the law of conservation of energy?

shtirl [24]

Answer:

b!!!! did it help? :)))

7 0

3 years ago

Read 2 more answers

The sound intensity of a certain type of food processor in normally distributed with standard deviation of 2.9 decibels. If the

Maru [420]

Answer: $(48.41,\ 52.19)$

Explanation:

The confidence interval for population mean is given by :-

$\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

Given : Sample size : $n=9$

Sample mean : $\ovreline{x}=50.3\text{ decibels}$

Standard deviation : $\sigma=2.9\text{ decibels }$

Significance level : $\alpha=1-0.95=0.05$

Critical value : $z_{\alpha/2}=z_{0.025}=1.96$

Now, the 95% confidence interval estimate of the (true, unknown) mean sound intensity of all food processors of this type :-

$50.3\pm (1.96)\dfrac{2.9}{\sqrt{9}}\\\\\approx50.3\pm1.89\\\\=(50.3-1.89,\ 50.3+1.89)=(48.41,\ 52.19)$

6 0

3 years ago

What is the energy of moving electrical charges

Simora [160]

The energy of moving electrical charges is Electrical energy

Hope its the answer you are finding and hope it helps....

3 0

2 years ago