# If the mass of the earth and all objects on it were suddenly doubled, but the size remained the same, the acceleration due to gravity at the surface would becomeA) 1/2 of what it now is.B) 2 times what it now is.C) 1/4 of what it now is.D) the same as it now is.E) 4 times what it now is.

B) 2 times what it now is

Explanation:

The acceleration due to gravity at the surface of the Earth is given by

where

G is the gravitational constant

M is the mass of the Earth

R is the Earth's radius

In this problem, the mass of the Earth is doubled:

M' = 2M

while the radius remains the same:

R' = R

so the new acceleration due to gravity would be

so, the acceleration due to gravity would become twice the current value.

Note also that the value of g does not depend on the mass of the objects involved.

## Related Questions

During the first stage of sleep deprivation, the subject may __________.

During the first stage of sleep deprivation the subject will become moody, grumpy or irritable.

Explanation:

There are two kinds of sleep deprivation which are classified as acute and chronic sleep deprivation. The acute one is the kind in which patient cannot sleep for one or two nights but in chronic sleep deprivation patient miss the seven hours of sleep per night in a week.

Dr. Ruiz further classified the stages of sleep deprivation as:

• Moodiness and irritability.
• Difficulty in concentrating.
• Difficulty in naming things.
• Lapse in memory.
• Hallucination.
• Paranoia
First stage of sleep Deprivation Subject

During the first stage of sleep deprivation, the subject is NREM which stands for non-rapid eye movement. In this condition, we are not sleeping in the depth. It can be said as dreamless sleep. On electro-encephalography recording, the brain waves are not fast so they have high voltage.

In this condition, the breathing, heart rate and blood pressure is low. The sleep is comparatively tranquil. NREM lasts for 90 minutes to 120 minutes. It accounts for about 75% of the normal sleep time. Rapid eye movements do not occur.

A small artery has a length of 1.10 × 10 − 3 m and a radius of 2.50 × 10 − 5 m. If the pressure drop across the artery is 1.45 kPa,what is the flow rate through the artery? Assume that the temperature is 37 °C and the viscosity of whole blood is 2.084 × 10 − 3 Pa·s.

The flow rate is

Solution:

As per the question:

Pressure drop,  = 1.45 kPa = 1450 Pa

Radius of the artery, R =

length of the artery, L =

Temperature, T =

Viscosity,

Now,

The flow rate is given by:

What is the net force on a 2 kg skateboard acceleration at a rate of 2 m/s2

Using the equasion F=ma F=2*2 F= 4N 4 Newtons or kg * m over s^2, the answer would be  12 m/s^2.

What change occurs to the mass of an object when an unbalanced force is applied to it?