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3 years ago

The average gorilla can travel 40.23 meters in 4 seconds, calculate the average speed in meters per second (m/s)

Physics

2 answers:

Masja [62]3 years ago

8 0

Answer:

10.0575 m/s

Explanation:

40.23 divide by 4

Murrr4er [49]3 years ago

8 0

Answer:

it is 10.0575 m/s

Explanation:

The average speed is calculated by total distance/total time taken. Here, the distance is 40.23 meters and time is 4 seconds.

So, average speed = 40.23/4 m/s

= 10.0575 m/s

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In his explanation of the threshold frequency in the photoelectric field, Einstein reasoned that the absorbed photon must have t

Doss [256]

Answer:

4.6×10^-7 m or 0.46nm

Explanation:

From

Wo= hc/λ

Where:

Wo= work function of the metal

h= planks constant

c= speed of light

λ= wavelength

λ= hc/Wo

λ= 6.6×10^-34 × 3×10^8/4.30×10^-19

λ= 4.6×10^-7 m

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3 years ago

A hunter stood at about 60 m away from a tree. He used the bow to release the arrow in order to shoot a coconut held by a monkey

Sergeeva-Olga [200]

Answer:

$v = u + at (upward)\\ 0 = 45 \sin(20) - 9.81t \\ t = 1.56s$

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3 years ago

A typical ceiling fan running at high speed has an airflow of about 1.85 ✕ 103 ft3/min, meaning that about 1.85 ✕ 103 cubic feet

densk [106]

Answer:

0.8726 $m^3/s$

Explanation:

We are to convert 1.85 x $10^3$ $ft^3/min$ to $m^3/s$

First, let us convert the numerator from ft3 to m3

1 ft3 = 0.0283 m3

Hence,

1.85 x $10^3$ ft3 = 1.85 x $10^3$ x 0.0283 m3

= 52.355 m3

Now, let us convert the denominator from minutes to seconds

1 min = 60 sec

Therefore;

1.85 x $10^3$ $ft^3/min$ = 52.355/60 $m^3/s$

= 0.8726 $m^3/s$

7 0

3 years ago

In midair an M = 145 kg bomb explodes into two pieces of m1 = 115 kg and another, respectively. Before the explosion, the bomb w

Daniel [21]

Answer:

$v_2=-133.17m/s$ , the minus meaning west.

Explanation:

We know that linear momentum must be conserved, so it will be the same before ( $p_i$ ) and after ( $p_f$ ) the explosion. We will take the east direction as positive.

Before the explosion we have $p_i=m_iv_i=Mv_i$ .

After the explosion we have pieces 1 and 2, so $p_f=m_1v_1+m_2v_2$ .

These equations must be vectorial but since we look at the instants before and after the explosions and the bomb fragments in only 2 pieces the problem can be simplified in one dimension with direction east-west.

Since we know momentum must be conserved we have:

$Mv_i=m_1v_1+m_2v_2$