A compact car moving at 60 m/s has 320,000 J of kinetic energy. What is its mass?

A spherical ball is dropped through a liquid, explain why it reaches terminal velocity.

Alekssandra [29.7K]

Probably because of the drag coefficient and the density of the liquid.

8 0

3 years ago

a projectile is shot horizontally from the edge of a cliff, 230m above the ground. the projectile lands 300m from base of the cl

bekas [8.4K]

Answer:

The time taken by the projectile to hit the ground is 6.85 sec.

Explanation:

Given that,

Vertical height of cliff = 230 m

Distance = 300 m

Suppose, determine the time taken by the projectile to hit the ground.

We need to calculate the time

Using second equation of motion

$s=ut+\dfrac{1}{2}gt^2$

Where, s = vertical height of cliff

u = initial vertical velocity

g = acceleration due to gravity

Put the value in the equation

$230=0+\dfrac{1}{2}\times9.8\times t^2$

$t=\sqrt{\dfrac{230\times2}{9.8}}$

$t=6.85 sec$

Hence, The time taken by the projectile to hit the ground is 6.85 sec.

7 0

3 years ago

75 pts!

jasenka [17]

again CORRECT ans is 4) light strikes the grooves → different wavelengths of light bend at different angles → diffracted wavelengths reach the eyes → the eyes see different colors.

moderator - plz review the ans as u deleted my right ans n approved the wrong ans :(

6 0

3 years ago

Read 2 more answers

An astronaut lands on a new, recently discovered planet in a different star system. The astronaut measures the acceleration due

Nezavi [6.7K]

Answer:

The radius of the new planet is ~2.04 * 10⁶ m, or 2,041,752 m.

Explanation:

We can use Newton's Law of Universal Gravitation:

$\displaystyle F_g=G\frac{Mm}{r^2}$

Let's look at Newton's 2nd Law:

$F=ma$

We can set these equations equal to each other:

$\displaystyle G\frac{Mm}{r^2} =ma$

The mass of the second mass (astronaut) cancels out. We are left with:

$\displaystyle G\frac{M}{r^2} =a$

We are solving for the radius of the new planet, so we can rearrange the equation:

$\displaystyle r=\sqrt{\frac{GM}{a} }$

Substitute in our known values given in the problem (G = 6.67 * 10⁻¹¹ ; M = 7.5 * 10²³ ; a = 12).

$\displaystyle r =\sqrt{\frac{(6.67\times 10^{-11})(7.5 \times 10^{23}}{12} }$
$r=2.04 \times 10^6$

The radius of the new planet is ~2.04 * 10⁶ m.

3 0

2 years ago

Find the length of a pendulum that oscillates with a frequency of 0.16 hz. the acceleration due to gravity is 9.81 m/s 2 . answe

Vsevolod [243]

The period of the pendulum is the reciprocal of the frequency:
$T= \frac{1}{f}= \frac{1}{0.16 Hz}=6.25 s$

The period of the pendulum is given by
$T=2 \pi \frac{L}{g}$
where L is the length of the pendulum, and g the acceleration of gravity. By re-arranging the formula and using the value of T we found before, we can calculate the length of the pendulum L:
$L=g \frac{T^2}{(2 \pi)^2}=(9.81 m/s^2) \frac{(6.25 s)^2}{(2 \pi)^2}=9.71 m$

7 0

2 years ago