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3 years ago

An object moving with a speed or 67m/s and has kinetic energy of 500 J what is the mass of the object

1 answer:

8 0

To solve this equation, simply plug the values into the equation for calculating kinetic energy.

KE = 1/2mv^2
500 = 1/2(m)(67^2)
500 =2244.5m
m = 500/2244.5 = 0.222 kg.

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You drop a small ball, and then a second small ball. When you drop the second ball, the distance between them is 3 cm. What stat

Alja [10]

Answer:

c) The distance between the balls increases.

Explanation:

If you drop the balls at the same time, regardless of their masses they accelerate equally, since they will be in free fall.

However, if you drop one of the balls earlier, then that ball will gain velocity, whereas the second ball has zero initial velocity. At the time the second ball is dropped, both balls have the same acceleration but different initial velocities.

According to the below kinematics equation:

$x = v_0t + \frac{1}{2}at^2$

The initial velocity of the first ball will make the difference, and the first ball will travel a greater distance than the second ball. Hence, their distance increases.

3 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

A life preserver is thrown from an helicopter straight down to a person in distress. The initial velocity of the life preserver

Leno4ka [110]

Answer:29.627 m

Explanation:

Given

Initial velocity of life preserver(u) is 1.6 m/s

it takes 2.3 s to reach the water

using equation of motion

v=u+at

$v=1.6+9.81\times 2.3$

v=24.163 m/s

Let s be the height of life preserver

$v^2-u^2=2gs$

$24.163^2-1.6^2=2\times 9.81\times s$

$s=\frac{581.29}{2\times 9.81}$

s=29.627 m

6 0

3 years ago

Determine the slope of end a of the cantilevered beam. E = 200 gpa and i = 65. 0(106) mm4

DENIUS [597]

For E = 200 gpa and i = 65. 0(106) mm4, the slope of end a of the cantilevered beam is mathematically given as

A=0.0048rads

<h3>What is the slope of end a of the cantilevered beam?</h3>

Generally, the equation for the is mathematically given as

$A=\frac{PL^2}{2EI}+\frac{ML}{EI}$

Therefore

A=\frac{10+10^2+3^2}{2*240*10^9*65*10^6}+\frac{10+10^3*3}{240*10^9*65*10^{-6}}

A=0.00288+0.00192=0.0048rads

A=0.0048rads

In conclusion, the slope is

A=0.0048rads