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Tcecarenko [31]
2 years ago
11

the gravitational force between the sun and earth is larger than sun and the moon. what do you think the reason for this is?

Physics
2 answers:
Tems11 [23]2 years ago
7 0

Answer: The greater an object's mass, the more gravitational force it exerts.

Explanation: So, to begin answering your question, Earth has a greater gravitational pull than the moon simply because the Earth is more massive. Sorry if I get this wrong. I am in 5th grade! ♥

ivann1987 [24]2 years ago
4 0

hi i need help with one of my question but i don’t know how else to contact you :/. anyway i posted the question and i added the text so i was wondering if u could help me !

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How polar bear can survive in the polar region<br>​
ladessa [460]

Answer:

Hey mate.....

Explanation:

This is ur answer...

<em>Polar bears are well adapted for survival in the Arctic. Their adaptations include: a white appearance - as camouflage from prey on the snow and ice. thick layers of fat and fur - for insulation against the cold.</em>

Hope it helps!

Brainliest pls!

Follow me! ◇

7 0
2 years ago
Two identical tiny spheres of mass m =2g and charge q hang from a non-conducting strings, each of length L = 10cm. At equilibriu
Citrus2011 [14]

Answer:

0.247 μC

Explanation:

As both sphere will be at the same level at wquilibrium, the direction of the electric force will be on the x axis. As you can see in the picture below, the x component of the tension of the string of any of the spheres should be equal to the electric force of repulsion. And its y component will be equal to the weight of one sphere. We can use trigonometry to find the components of the tensions:

F_y:  T_y - W = 0\\T_y = m*g = 0.002 kg *9.81m/s^2 = 0.01962 N

T_y = T_*cos(50)\\T = \frac{T_y}{cos(50)} = 0.0305 N

T_x = T*sin(50) = 0.0234 N

The electric force is given by the expression:

F = k*\frac{q_1*q_2}{r^2}

In equilibrium, the distance between the spheres will be equal to 2 times the length of the string times sin(50):

r = 2*L*sin(50) = 2 * 0.1m * sin(50) 0.1532 m

And k is the coulomb constan equal to 9 *10^9 N*m^2/C^2. q1 y q2 is the charge of each particle, in this case, they are equal.

F_x = T_x - F_e = 0\\T_x = F_e = k*\frac{q^2}{r^2}

q = \sqrt{T_x *\frac{r^2}{k}} = \sqrt{0.0234 N * \frac{(0.1532m)^2}{9*10^9 N*m^2/C^2} } = 2.4704 * 10^-7 C

O 0.247 μC

8 0
2 years ago
Please help. I don’t understand this
skad [1K]

The short answer is that the displacement is equal tothe area under the curve in the velocity-time graph. The region under the curve in the first 4.0 s is a triangle with height 10.0 m/s and length 4.0 s, so its area - and hence the displacement - is

1/2 • (10.0 m/s) • (4.0 s) = 20.00 m

Another way to derive this: since velocity is linear over the first 4.0 s, that means acceleration is constant. Recall that average velocity is defined as

<em>v</em> (ave) = ∆<em>x</em> / ∆<em>t</em>

and under constant acceleration,

<em>v</em> (ave) = (<em>v</em> (final) + <em>v</em> (initial)) / 2

According to the plot, with ∆<em>t</em> = 4.0 s, we have <em>v</em> (initial) = 0 and <em>v</em> (final) = 10.0 m/s, so

∆<em>x</em> / (4.0 s) = (10.0 m/s) / 2

∆<em>x</em> = ((4.0 s) • (10.0 m/s)) / 2

∆<em>x</em> = 20.00 m

5 0
2 years ago
Find equation tangent to a circle at given point
vlabodo [156]

the equation of the tangent line must be passed on a point A (a,b) and perpendicular to the radius of the circle. <span>
I will take an example for a clear explanation:
let x² + y² = 4 is the equation of the circle, its center is C(0,0). And we assume that the tangent line passes to the point A(2.3).

</span>since the tangent passes to the A(2,3), the line must be perpendicular to the radius of the circle. 

<span>Let's find the equation of the line parallel to the radius.</span>

<span>The line passes to the A(2,3) and C (0,0). y= ax+b is the standard form of the equation. AC(-2, -3) is a vector parallel to CM(x, y).</span>

det(AC, CM)= -2y +3x =0, is the equation of the line // to the radius.

let's find the equation of the line perpendicular to this previous line.

let M a point which lies on the line. so MA.AC=0 (scalar product), 

it is (2-x, 3-y) . (-2, -3)= -4+4x + -9+3y=4x +3y -13=0 is the equation of tangent



7 0
3 years ago
A gun is fired on a day when the speed of sound is 335 m/s and an echo is heard 0.75 seconds later. How far away is the object t
cricket20 [7]

Answer:

v= 335 m/s

2∆t= 0.75 s

∆x= v.∆t → ∆x= 335×½×0.75 = 125.625 m

8 0
2 years ago
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