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3 years ago

5

What human activity uses the most water worldwide

2 answers:

Ronch [10]3 years ago

7 0

Answer:

agriculture

Explanation:

Sergeu [11.5K]3 years ago

7 0

The human activity that uses the most water worldwide is for agriculture. In other words, watering crops.

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The key difference between the binomial and hypergeometric distribution is that

Ivenika [448]

Explanation:

Both distributions describe the number of times an event occurs in a givn number of trials. In the binomial distribution, the probability is the same for each trial. While in the hypergeometric distribution, each trial changes the probability of each subsequent trial, since there is no replacement.

5 0

3 years ago

Two basketballs of equal mass are rolling toward each other at constant velocities. The first basketball (B1) has a velocity of

slamgirl [31]

$v'_2 = \frac{2m_1}{m_1+m_2} (4.3) - \frac{m_1-m_2}{m_1+m_2} (4.3)\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} (4.3) + \frac{2m_2}{m_1+m_2} (4.3)$

<u>Explanation:</u>

Velocity of B₁ = 4.3m/s

Velocity of B₂ = -4.3m/s

For perfectly elastic collision:, momentum is conserved

$m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2$

where,

m₁ = mass of Ball 1

m₂ = mass of Ball 2

v₁ = initial velocity of Ball 1

v₂ = initial velocity of ball 2

v'₁ = final velocity of ball 1

v'₂ = final velocity of ball 2

The final velocity of the balls after head on elastic collision would be

$v'_2 = \frac{2m_1}{m_1+m_2} v_1 - \frac{m_1-m_2}{m_1+m_2} v_2\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} v_1 + \frac{2m_2}{m_1+m_2} v_2$

Substituting the velocities in the equation

$v'_2 = \frac{2m_1}{m_1+m_2} (4.3) - \frac{m_1-m_2}{m_1+m_2} (4.3)\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} (4.3) + \frac{2m_2}{m_1+m_2} (4.3)$

If the masses of the ball is known then substitute the value in the above equation to get the final velocity of the ball.

5 0

3 years ago

If a force is 100 N and is pointing 37 degrees north of east. (a) Draw a diagram of this force. (b) Draw the force's x and y com

sasho [114]

Answer

given,

force = 100 N

Point 37 degrees north of east

a) and b) part is shown in the diagram attached below.

c) to find the x and y component of the force

x- component of the force

$F_x = F cos \theta$

$F_x = 100\times cos 37^0$

$F_x = 79.86 N$

y- component of the force

$F_y = F sin \theta$

$F_y = 100\times sin 37^0$

$F_y = 60.18 N$

3 0

3 years ago

A sinusoidal transverse wave travels along a long, stretched string. The amplitude of this wave is 0.08190.0819 m, its frequency

Anvisha [2.4K]

Answer:

The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.

Explanation:

Given that,

Amplitude = 0.08190 m

Frequency = 2.29 Hz

Wavelength = 1.87 m

(a). We need to calculate the shortest transverse distance between a maximum and a minimum of the wave

Using formula of distance

$d=2A$

Where, d = distance

A = amplitude

Put the value into the formula

$d=2\times0.08190$

$d=0.1638\ m$

Hence, The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.

5 0

3 years ago

NEED HELP FAST ILL MARK BRAINLIEST AND RATE 5 STARS AND SAY THANK YOU

sertanlavr [38]

3 is the answer to your question

4 0

3 years ago

Read 2 more answers