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kvv77 [185]
2 years ago
13

Which of the following is NOT a product of science?

Physics
2 answers:
emmasim [6.3K]2 years ago
5 0
Answer: D. None of these
Marina86 [1]2 years ago
3 0

Answer:

D, None of these.

Explanation:

They have science in them but its not a product.

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1. What do you think determines the type of traits you inherit?
Ainat [17]

Answer: How dominant one persons genes are.

Explanation:

The more dominant a trait is, the more likely someone will inherit them.

6 0
3 years ago
Please help do not guess
Georgia [21]

Answer:

358.9 (+/- 0.4) million years ago

Holocene Epoch, of the Quaternary Period

Devonian period

66 million years ago (prox)

521 million years ago

110,000 years ago

NW

North america became more cold when it moved NW

Explanation:

8 0
3 years ago
What is the restoring force of a spring with a spring constant of 4a and a stretched displacement of 3b? A. –7 ab B. `-7 a/b ` C
algol13

Answer:

C. -12 ab

Explanation:

The restoring force on a spring is given by Hooke's law:

F=-kx

where

k is the spring constant

x is the stretched (or compressed) displacement of the spring

In this problem we have:

k = 4a

x = 3b

Substituting into the equation, we find:

F=-(4a)(3b) = -12 ab

And the negative sign means that the direction of the force (negative) is opposite to the direction of the displacement (positive).

3 0
3 years ago
A satellite of mass 1000kg is in a circular orbit around a planet. The centripetal acceleration of the satellite in its orbit is
ludmilkaskok [199]

Answer: 5000N

Explanation:

The basic principle of a circular orbit is that Fg = m × ac, so as we have the mass and the centripetal acceleration (also called normal acceleration) we just have to operate. Fg = 1000kg × 5m/s² = 5000N

8 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
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