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OverLord2011 [107]
2 years ago
11

Science Help Please?? It is believed that solar nebular material came from which of the following:

Physics
2 answers:
kolbaska11 [484]2 years ago
7 0
B would be the answer to this
Ray Of Light [21]2 years ago
5 0
A is the answer, (I put B and got it wrong)
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Special ways that animals' bodies work to help them survive in whatev
alexira [117]

Answer:

Physiological – special ways that animals' bodies work to help them survive in whatever condition they're in, such as camels in the desert conserving water and being able to go days without drinking.

Explanation:

8 0
2 years ago
if you drop a stone from height of 2.5m. what is the speed of the stone right before it hits the ground?
KonstantinChe [14]
Since the stone was dropped from height, its initial velocity = 0 m/s

Using  v² = u² + 2gs.

Where g ≈ 10 m/s²,  u = initial velocity = 0 m/s, s = height from drop = 2.5 m

v² = u² + 2gs

v² = 0² + 2*10*2.5

v² = 0 + 50

v² = 50

v = √50

v ≈ 7.07 m/s

Hence velocity just before hitting the ground is ≈ 7.07 m/s 
6 0
2 years ago
A 4.4 nC charge exerts a repulsive force of 36 mN on a second charge which is located
zhenek [66]

The magnitude and sign of the second charge will be + 8.6241×10⁻¹⁹ C. The principal of the Columb's law is used in the given problem.

<h3>What is Columb's law?</h3>

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

Charges that are similar repel each other, whereas charges that are diametrically opposed attract each other.

They will repel, moving in opposite directions at the same speed. Because the magnitude and nature of the charge are the same.

The given data in the problem is;

q₁  is the charge 1 = 4.4 nC = 4.4 ×10⁻⁹ C

F is the repulsive force = 36 mN =36 ×10⁶ N

d is the distance = 0.70 m

The Coulomb force is found as;

\rm F = \frac{Kq_1q_2}{r^2}\\\\\ \rm 36\times 10^6 = \frac{9 \times 10^9 }{(0.7)^2} \times 4.4 \times 10^{-9} \times q_2\\\\\ q_2 = 8.6241  \times 10^{-19 } \ C

Hence, the magnitude and sign of the second charge will be + 8.6241×10⁻¹⁹ C.

To learn more about Coulomb's law, refer to the link;

brainly.com/question/1616890

#SPJ2

6 0
2 years ago
What constant acceleration (in ft/s2) is required to increase the speed of a car from 27 mi/h to 50 mi/h in 5 seconds
SIZIF [17.4K]

Answer:

6.746 ft/s^2

Explanation:

v(t)=50

v(0)=27

t=5/3600 = 1/720 hours

v(t)-v(0)= a(t-0)

50-27= a(1/720)

a= 23*720= 16560 mi/h^2

16560mi/h^2 * 5280/3600^2 (ft/s^2) =6.746 ft/s^2

4 0
3 years ago
A person is pulling a freight cart with a force of 58 pounds. how much work is done in moving the cart 70 feet if the cart's han
Kobotan [32]

<span>The person is dragging with a force of 58 lbs at an angle of 27 degrees relating to the ground. We want to use cosine function to look for the horizontal force component. And then we can compute for W = (Horizontal Force) x (Distance). We want the horizontal force component since that is the component that is parallel to the direction the cart is moving. </span><span>

(cos 27 degrees)(58 lbs) = 51.69 lbs (This is the horizontal force component.) 
W = (51.69 lbs) x (70 ft) = 3618.3 ft*lbs</span>

6 0
3 years ago
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