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2 years ago

How does size of an object affect its gravity

Physics

1 answer:

irakobra [83]2 years ago

6 0

Answer:

the more particles packed together the faster it falls

Explanation:

the mass + the 1 constant g-force = the speed without adding air resistance

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Pls help i dont know this

tatuchka [14]

Answer:

can you make it bigger

Explanation:

i can't see it

5 0

2 years ago

If a car has a mass of 1,000 kilograms, and a velocity of 35 m/s. What is its momentum?

fenix001 [56]

$\vec{p}=m\vec{v}\\p = 1000 * 35=35000[\frac{kg*m}{s}]$

7 0

3 years ago

At which point will the riders experience centripetal acceleration?

Korvikt [17]

the answer would be (X)

8 0

2 years ago

By what factor will the Electrostatic Force between two charged objects change when the amount of charge on both objects doubles

Mademuasel [1]

Answer:

F' = (4/9)F

Explanation:

The electrostatic force between two charged objects is given by Coulomb's Law:

F = kq₁q₂/r² -------------------- equation (1)

where,

F = Electrostatic Force

k = Coulomb's Constant

q₁ = magnitude of first charge

q₂ = magnitude of second charge

r = distance between charges

Now, when the charges and distance altered as follows:

q₁' = 2q₁

q₂' = 2q₂

r' = 3r

Then,

F' = kq₁'q₂'/r'²

F' = k(2q₁)(2q₂)/(3r)²

F' = (4/9)kq₁q₂/r²

using equation (1):

7 0

2 years ago

A dart is thrown at a dartboard 3.66 m away. When the dart is released at the same height as the center of the dartboard, it hit

lapo4ka [179]

Answer:

The angle is $\theta = 15.48^o$

Explanation:

From the question we are told that

The distance of the dartboard from the dart is $d = 3.66 \ m$

The time taken is $t = 0.455 \ s$

The horizontal component of the speed of the dart is mathematically represented as

$u_x = ucos \theta$

where u is the the velocity at dart is lunched

$distance = velocity \ in \ the\ x-direction * time$

substituting values

$3.66 = ucos \theta * (0.455)$

=> $ucos \theta = 8.04 \ m/s$

From projectile kinematics the time taken by the dart can be mathematically represented as

$t = \frac{2usin \theta }{g}$

=> $usin \theta = \frac{g * t}{2 }$

$usin \theta = \frac{9.8 * 0.455}{2 }$

$usin \theta = 2.23$

=> $tan \theta = \frac{usin\theta }{ucos \theta } = \frac{2.23}{8.04}$

$\theta = tan^{-1} [0.277]$

$\theta = 15.48^o$

4 0

3 years ago