All categories

2 years ago

Suppose Earth's mass increased but Earth's diame-

Physics

1 answer:

navik [9.2K]2 years ago

7 0

Answer: It would increase.

Explanation:

The equation for determining the force of the gravitational pull between any two objects is:

$F = G \frac{m1m2}{r^2}$

Where G is the universal gravitational constant, m1 is the mass of one body, m2 is the mass of the other body, and r^2 is the distance between the two objects' centers squared.

Assuming the Earth's mass but not its diameter increased, in the equation above m1 (the term usually indicative of the object of larger mass) would increase, while the r^2 would not.

Thus, it goes without saying that, with some simple reasoning about fractions, an increasing numerator over a constant denominator would result in a larger number to multiply by G, thus also meaning a larger gravitational strength between Earth and whatever other object is of interest.

You might be interested in

A system of two paint buckets connected by a lightweight rope is released from rest with the 12.0-kg bucket 2.00 m above the flo

NISA [10]

Explanation:

The given data is as follows.

Mass of small bucket (m) = 4 kg

Mass of big bucket (M) = 12 kg

Initial velocity ( $v_{o}$ ) = 0 m/s

Final velocity ( $v_{f}$ ) = ?

Height $H_{o} = h_{f}$ = 2 m

and, $H_{f} = h_{o}$ = 0 m

Now, according to the law of conservation of energy

starting conditions = final conditions

$\frac{1}{2}MV^{2}_{o} + Mgh_{o} + \frac{1}{2}mv^{2}_{o} + mgh_{o} = \frac{1}{2}MV^{2}_{f} + Mgh_{f} + \frac{1}{2}mv^{2}_{f} + mgh_{f}$

$\frac{1}{2}(12)(0)^{2} + (12)(9.81)(2) + \frac{1}{2}(4)(0)^{2} + (4)(9.81)(0) = \frac{1}{2}(12)V^{2}_{f} + (12)(9.81)(0) + \frac{1}{2}(4)V^{2}_{f} + (4)(9.81)(2)$

235.44 = $8V^{2}_{f}$ + 78.48

$V_{f}$ = 4.43 m/s

Thus, we can conclude that the speed with which this bucket strikes the floor is 4.43 m/s.

3 0

1 year ago

The wavelength of light is 5000 angstrom. Express it in nm and m.

Ierofanga [76]

Answer:

1 angstrom = 0.1nm

5000 angstrom = 5000/1 × 0.1nm

$1 \: angstrom = 1 \times {10}^{ - 10} m$

5000 angstrom = 5000 × 1 × 10^-10

Hope this helps you

7 0

2 years ago

A spring driven dart gun propels a 10g dart. It is cocked by exerting a force of 20N over a distance of 5cm. With what speed wil

adelina 88 [10]

<span>14 m/s Assuming that all of the energy stored in the spring is transferred to dart, we have 2 equations to take into consideration. 1. How much energy is stored in the spring? 2. How fast will the dart travel with that amount of energy. As for the energy stored, that's a simple matter of multiplication. So: 20 N * 0.05 m = 1 Nm = 1 J For the second part, the energy of a moving object is expressed as KE = 0.5 mv^2 where KE = Kinetic energy m = mass v = velocity Since we now know the energy (in Joules) and mass of the dart, we can substitute the known values and solve for v. So KE = 0.5 mv^2 1 J = 0.5 0.010 kg * v^2 1 kg*m^2/s^2 = 0.005 kg * v^2 200 m^2/s^2 = v^2 14.14213562 m/s = v So the dart will have a velocity of 14 m/s after rounding to 2 significant figures.</span>

6 0

2 years ago

Read 2 more answers

If you have to apply 40n of force on a crowbar to lift a rock that weights 400n, what is the actual mechanical advantage of the

Mrrafil [7]

The mechanical advantage is defined as the ratio between the force produced by a machine and the force applied in input:
$MA= \frac{F_{out}}{F_{in}}$
For the crowbar of the problem, the force applied in input is 40 N, while the force produced in output is equal to the weight of the rock that is lifted, so 400 N. Therefore, the mechanical advantage is
$MA= \frac{400 N}{40 N}=10$

3 0

2 years ago

Determine the centripetal force upon a 40-kg child who makes 10 revolution around the cliffhanger in 29.3 seconds.the radius of

zysi [14]

Answer:

The centripetal force acting on the child is 39400.56 N.

Explanation:

Given:

Mass of the child is, $m=40\ kg$