All categories

3 years ago

A rock falls off a 15 meter cliff. How fast is the rock traveling vertically two seconds later?

1 answer:

8 0

The acceleration of gravity on Earth is 9.8 m/s². That means that
an object falling under the influence of gravity will move 9.8 m/s
faster than it was moving a second earlier.

Falling from rest, it will be moving 9.8 m/s after the first second,
and 19.6 m/s after the 2nd second.

The height from which it fell doesn't matter.

You might be interested in

Approximately how many Sun's are in the Milky way?

IgorC [24]

Answer:

there are uncountable sun are there but sun is star therefore there are uncountable stars are there

7 0

3 years ago

Read 2 more answers

It is observed that the number of asteroids (or meteoroids) of a given diameter is roughly inversely proportional to the square

MaRussiya [10]

Answer:

(a) M = =1.57 *10²¹kg

(b) M = =1.57 *10²⁰kg

(d) M = =1.57 *10¹⁸kg

Explanation:

(a) For 1000km body, diameter D = 1000km = 1,000,000 m

and N = (1000/D)² = (1000/1000km)² = 1

substituting into the formula, we have

M = 1 * 4/3 π*3000*[1000000/2]³

=1.57 *10²¹kg

(b) For 100km body, diameter D = 100km = 100,000 m

and N = (1000/D)² = (1000/100km)² = 100

substituting into the formula, we have

M = 1 * 4/3 π*3000*[100000/2]³

=1.57 *10²⁰kg

and N = (1000/D)² = (1000/10km)² = 100

substituting into the formula, we have

M = 1 * 4/3 π*3000*[10000/2]³

=1.57 *10¹⁹kg

and N = (1000/D)² = (1000/1km)² = 100

substituting into the formula, we have

M = 1 * 4/3 π*3000*[1000/2]³

=1.57 *10¹⁸kg

4 0

3 years ago

Pepe and Alfredo are resting on an offshore raft after a swim. They estimate that 3.0 m separates a trough and an adjacent crest

Agata [3.3K]

Answer:

The velocity ( $v$ ) of the wave is 3.08 $ms^{-1}$ .

Explanation:

According to the figure, the distance ( $\large{L}$ ) between a trough and its adjacent crest is $\large{L = 3 m}$ . Also the wavelength ( $\large{\lambda}$ ) of the wave is $\large{\lambda = 2L}$ . Pepe and Alfredo count 11 crests to pass the raft in $\large{t}$ = 21.5 s.

So, the time period ( $\large{T}$ ) of oscillation of the wave is

$\large{T} = \dfrac{t}{11} = \dfrac{21.5}{11} = 1.95s$

So, the velocity ( $\large{V}$ ) of the wave is

$\large{V = \dfrac{\lambda}{T} = \dfrac{2 \times L}{T} = \dfrac{2 \times 3}{1.95}= 3.08 ms^{-1}}$

4 0

3 years ago

A 0.25 kg ball is suspended from a light 0.65 m string as shown. The string makes an angle of 31° with the vertical. Let U = 0 w

steposvetlana [31]

Explanation:

a) The height of the ball h with respect to the reference line is

$h = L - L\cos{31°} = L(1 - \cos{31°})$

so its initial gravitational potential energy $U_0$ is

$U = mgh = mgL(1 - \cos{31°})$

$\:\:\:\:\:=(0.25\:\text{kg})(9.8\:\text{m/s}^2)(0.65\:\text{m})(1 - \cos{31})$

$\:\:\:\:\:=0.23\:\text{J}$

b) To find the speed of the ball at the reference point, let's use the conservation law of energy:

$\Delta{K} + \Delta{U} = 0 \Rightarrow K_0 + U_0 = K + U$

We know that the initial kinetic energy $K_0,$ as well as its final gravitational potential energy $U$ are zero so we can write the conservation law as