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

Physics

1 answer:

navik [9.2K]3 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.

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The length of the adult mosquito is typically between 3.0 mm and 6.0 mm. The smallest known mosquitoes are around 2.5 mm, and th

FrozenT [24]

0.74in

Explanation:

Given parameters:

length range of adult mosquito = 3.0 - 6.0mm

Smallest mosquito = 2.5mm

Largest mosquito = 19mm

Average mass range = 3.0 - 5.0mg

Unknown:

Length of largest known mosquito in inches = ?

Solution;

The length is the longest dimension. It is how long a body is.

The problem here is converting from mm to inches;

The length of the longest mosquito which is the largest is 19mm

19mm to inches;

1mm = 0.039inches

19mm = 19mm x $\frac{0.039in}{1mm}$ = 0.74in

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4 0

3 years ago

If he leaves the ramp with a speed of 31.0 m/s and has a speed of 29.5 m/s at the top of his trajectory, determine his maximum h

raketka [301]

Answer:

The maximum height reached is 4.63 m.

Explanation:

Given:

Initial speed of the man (u) = 31.0 m/s

Speed at the top of trajectory ( $u_x$ ) = 29.5 m/s

Acceleration due to gravity (g) = 9.8 m/s²

When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.

Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.

So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.

Now, initial velocity can be rewritten in terms of its components as:

$u^2=u_x^2+u_y^2$