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3 years ago

What is normal distribution in statistics?

1 answer:

8 0

In statistics there are different types of distributions, like rectangular, or Poisson, A normal distribution is bellshaped

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A rock has a mass of 25kg. The acceleration due to gravity is 10m/s and the height of the rock is 42 m. Find the potential energ

alisha [4.7K]

Answer:

The potential energy of the rock = 10.5 kN

Explanation:

Mass of rock = 25 kg

Acceleration due to gravity = 10 m/s²

Height = 42 m

Potential energy, PE = mgh, where m is the mass, g is acceleration due to gravity and h is the height.

PE = 25 x 10 x 42 = 10500 N = 10.5 kN

The potential energy of the rock = 10.5 kN

7 0

3 years ago

Read 2 more answers

Select the correct answer?

yKpoI14uk [10]

Answer:

The answer is d i think so yea

Explanation:

hope u hae a really good day

8 0

3 years ago

An electron passes location < 0.02, 0.04, -0.06 > m and 5 us later is detected at location < 0.02, 1.62,-0.79 > m (1

hram777 [196]

Answer:

(a)

Average velocity = < 0, 316000, 146000> m/s

(b) < 0, 2.844, 1.314 > m

Explanation:

r1 = < 0.02, 0.04, - 0.06 > m

r2 = < 0.02, 1.62, - 0.79 > m

time, t = 5 micro second = 5 x 10^-6 s

(a) Average velocity is defined as the ratio of total displacement to the total time taken.

Displacement = r = r2 - r1

r = < 0.02 - 0.02, 1.62 - 0.04, - 0.79 + 0.06 > m

r = < 0, 1.58, - 0.73 > m

So. Average velocity = $\frac{< 0, 1.58, - 0.73 >}{5 \times 10^{-6}}$

Average velocity = $< 0, 316000, 146000> m/s$

Average velocity = < 0, 316000, 146000> m/s

(b)

Distance = velocity x time

Here time, t = 9 micro second

d = < 0, 316000, 146000> x 9 x 10^-6 m

d = < 0, 2.844, 1.314 > m

5 0

3 years ago

A 1300 kg steel beam is supported by two ropes. (Figure

Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

the net horizontal force acting on the beam is

$R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0$

where $R_1,R_2$ are the magnitudes of the tensions in ropes 1 and 2, respectively;

the net vertical force acting on the beam is

$R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0$

where $m=1300\,\rm kg$ and $g=9.8\frac{\rm m}{\mathrm s^2}$ .

Eliminating $R_2$ , we have

$\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)$

$R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2$

$R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2$

$-R_1 \sin(50^\circ) = -\dfrac{mg}2$

$R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}$

Solve for $R_2$ .

$\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0$

$\dfrac{R_2}2 = -mg\cot(110^\circ)$

$R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}$

8 0

2 years ago

How do surface currents affect climate

Bumek [7]

Ocean currents act much like a conveyer belt, transporting warm water and precipitation from the equator toward the poles and cold water from the poles back to the tropics. Therefore, currents regulate global climate, helping to counteract the uneven distribution of solar radiation reaching Earth's surface.

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4 years ago