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Hoochie [10]
2 years ago
9

Earth's gravity is pulling a ball downward toward Earth. What is the reaction force?

Physics
1 answer:
weeeeeb [17]2 years ago
4 0

Answer: Air resistance is "pushing" the ball upwards.

Explanation:

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What is another unit for momentum besides kg-m/s?<br> a. N<br> b. N-s<br> c. N-s2<br> d. N/s
lara [203]
'Newton-second' is dimensionally equivalent to 'kilogram-meter/second'.
8 0
2 years ago
If an object in motion, experiences a net zero unbalanced force, the the body
RSB [31]

Answer:

4) moves with constant velocity

Explanation:

If an object is in motion, three things can happen:

1. The object slows down in speed.

This means a net force is needed to be able to slow the object down in speed.

2. The object increases in speed.

This means a net force is needed to be able to increase the speed of the object.

3. The object remains in motion at the same speed.

No net zero unbalanced force is needed.

This is the given situation. So this is the right answer.

6 0
2 years ago
The frequency of which type of electromagnetic wave is just higher than that of visible light?
Reptile [31]

When you sort the ranges of EM waves from lowest to highest frequency, the visible light range is just higher than infrared (think "below red") and just lower than ultraviolet (think "above violet").

Choice E.

3 0
3 years ago
Spitting cobras can defend themselves by squeezing muscles around their venom glands to squirt venom at an attacker. Suppose a s
Alenkasestr [34]

Answer: 1.289 m

Explanation:

The path the cobra's venom follows since it is spitted until it hits the ground, is described by a parabola. Hence, the equations for parabolic motion (which has two components) can be applied to solve this problem:

<u>x-component: </u>

x=V_{o}cos\theta t  (1)

Where:

x is the horizontal distance traveled by the venom

V_{o}=3.10 m/s is the venom's initial speed

\theta=47\° is the angle

t is the time since the venom is spitted until it hits the ground

<u>y-component: </u>

y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}   (2)

Where:

y_{o}=0.44 m  is the initial height of the venom

y=0  is the final height of the venom (when it finally hits the ground)

g=-9.8m/s^{2}  is the acceleration due gravity

Let's begin with (2) to find the time it takes the complete path:

0=0.44 m+3.10 m/s sin\theta(47\°)+\frac{-9.8m/s^{2} t^{2}}{2}   (3)

Rewritting (3):

-4.9 m/s^{2} t^{2} + 2.267 m/s t + 0.44 m=0   (4)

This is a quadratic equation (also called equation of the second degree) of the form at^{2}+bt+c=0, which can be solved with the following formula:

t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a} (5)

Where:

a=-4.9 m/s^{2

b=2.267 m/s

c=0.44 m

Substituting the known values:

t=\frac{-2.267 \pm \sqrt{2.267^{2}-4(-4.9)(0.44)}}{2(-4.9)} (6)

Solving (6) we find the positive result is:

t=0.609 s (7)

Substituting (7) in (1):

x=(3.10 m/s)cos(47\°)(0.609 s)  (8)

We finally find the horizontal distance traveled by the venom:

x=1.289 m  

7 0
3 years ago
Calculate the force of gravity between planet X and planet y if both planets are 3.75 X 10^11 m apart, planet X has a mass of 1.
GenaCL600 [577]

So, the force of gravity that the asteroid and the planet have on each other approximately \boxed{\sf{2.9 \times 10^{17} \: N}}

<h3>Introduction</h3>

Hi ! Now, I will help to discuss about the gravitational force between two objects. The force of gravity is not affected by the radius of an object, but radius between two object. Moreover, if the object is a planet, the radius of the planet is only to calculate the "gravitational acceleration" on the planet itself,does not determine the gravitational force between the two planets. For the gravitational force between two objects, it can be calculated using the following formula :

\boxed{\sf{\bold{F = G \times \frac{m_1 \times m_2}{r^2}}}}

With the following condition :

  • F = gravitational force (N)
  • G = gravity constant ≈ \sf{6.67 \times 10^{-11}} N.m²/kg²
  • \sf{m_1} = mass of the first object (kg)
  • \sf{m_2} = mass of the second object (kg)
  • r = distance between two objects (m)

<h3>Problem Solving</h3>

We know that :

  • G = gravity constant ≈ \sf{6.67 \times 10^{-11}} N.m²/kg²
  • \sf{m_X} = mass of the planet X = \sf{1.55 \times 10^{22}} kg.
  • \sf{m_Y} = mass of the planet Y = \sf{3.95 \times 10^{28}} kg.
  • r = distance between two objects = \sf{3.75 \times 10^{11}} m.

What was asked :

  • F = gravitational force = ... N

Step by step :

\sf{F = G \times \frac{m_X \times m_Y}{r^2}}

\sf{F = 6.67 \cdot 10^{-11} \times \frac{1.55 \cdot 10^{22} \cdot 3.95 \times 10^{28}}{(3.75 \times 10^{11})^2}}

\sf{F \approx \frac{40.84 \times 10^{-11 + 22 + 28}}{14.0625 \times 10^{22}}}

\sf{F \approx 2.9 \times 10^{39 - 22}}

\sf{F \approx 2.9 \times 10^{17} \: N}

<h3>Conclusion</h3>

So, the force of gravity that the asteroid and the planet have on each other approximately

\boxed{\sf{2.9 \times 10^{17} \: N}}

<h3>See More</h3>
  • Gravity is a thing has depends on ... brainly.com/question/26485200
8 0
1 year ago
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