All categories

3 years ago

Why does the value of g vary from place to place?

Physics

1 answer:

NemiM [27]3 years ago

8 0

Answer:

The variation in apparent gravitational acceleration (g) at different locations on Earth is caused by two things (as you implied). ... The distance between the centers of mass of two objects affects the gravitational force between them, so the force of gravity on an object is smaller at the equator compared to the poles.

You might be interested in

According to the graph, how many students received an 90 for their first semester grade?

SCORPION-xisa [38]

Answer:

Explanation:

4 0

3 years ago

Estimate the water volume in the graduated cylinder to the nearest 0.1 mL.

earnstyle [38]

Answer:

With a 50-mL graduated cylinder, read and record the volume to the nearest 0.1 mL. The 10-mL graduated cylinder scale is read to the nearest 0.01 mL and the 500-mL graduated cylinder scale is read to the nearest milliliter (1 mL). A buret is a scaled cylindrical tube attached to a stopcock, or valve.

5 0

4 years ago

a ball is projected upward at time t = 0.00 s from a point on a roof 70 m above the ground. The ball rises, then falls and strik

grin007 [14]

Answer: 17.68 s

Explanation:

This problem is a good example of Vertical motion, where the main equation for this situation is:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

Where:

$y=0$ is the height of the ball when it hits the ground

$y_{o}=70 m$ is the initial height of the ball

$V_{o}=82m/s$ is the initial velocity of the ball

$t$ is the time when the ball strikes the ground

$g=9.8m/s^{2}$ is the acceleration due to gravity

Having this clear, let's find $t$ from (1):

$0=70m+(82m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2}$ (2)

Rewritting (2):

$-\frac{1}{2}(9.8m/s^{2})t^{2}+(82m/s)t+70m=0$ (3)

This is a quadratic equation (also called equation of the second degree) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ (4)

Where:

$a=-\frac{1}{2}(9.8m/s^{2}$

$b=82m/s$

$c=70m$

Substituting the known values:

$t=\frac{-82 \pm \sqrt{82^{2}-4(-\frac{1}{2}(9.8)(70)}}{2a}$ (5)

Solving (5) we find the positive result is:

$t=17.68 s$

7 0

3 years ago

If you have a 1.0 m aqueous solution of NaCl, by how much will it increase the water’s boiling point, if KB = 0.512 °C/m? In oth

fgiga [73]

Answer: The elevation in boiling point is 1.024°C.

Explanation:

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=ik_b\times m$

where,

i = Van't Hoff factor = 2 (for NaCl)

$\Delta T_b$ = change in boiling point = ?

$k_b$ = boiling point constant = $0.512^oC/m$

m = molality = 1.0 m

Putting values in above equation, we get:

$\Delta Tb=2\times 0.512^oC/m\times 1.0m\\\\\Delta Tb=1.024^oC$

Hence, the elevation in boiling point is 1.024°C.

5 0

4 years ago

Hitungkan pecutan bagi blok di bawah: / Cal (a) m= 2 kg F= 8.0 N

ioda

Answer:

Acceleration = 4 m/s²

Explanation:

Given the following data;

Force = 8 N

Mass = 2 kg

To find the acceleration of the block;

Newton's Second Law of Motion states that the acceleration of a physical object is directly proportional to the net force acting on the physical object and inversely proportional to its mass.

Mathematically, it is given by the formula;

$Acceleration = \frac {Net \; force}{mass}$

Substituting into the formula, we have;

$Acceleration = \frac {8}{2}$

Acceleration = 4 m/s²

4 0

3 years ago