All categories

3 years ago

Determine the acceleration due to gravity?

Physics

1 answer:

MissTica3 years ago

8 0

That's a weird graph, but judging from the units the acceleration is the slope of the graph.

a = (0.8 - 0.3)/(0.16 - 0.055) = 4.76 m/s²

You might be interested in

One way to force air into an unconscious person's lungs is to squeeze on a balloon appropriately connected to the subject. What

frozen [14]

Answer:

The force that you must exert on the balloon is 1.96 N

Explanation:

Given;

height of water, h = 4.00 cm = 4 x 10⁻² m

effective area, A = 50.0 cm² = 50 x 10⁻⁴ m²

density of water, ρ = 1 x 10³ kg/m³

Gauge pressure of the balloon is calculated as;

P = ρgh

where;

ρ is density of water

g is acceleration due to gravity

h is height of water

P = 1 x 10³ x 9.8 x 4 x 10⁻²

P = 392 N/m²

The force exerted on the balloon is calculated as;

F = PA

where;

P is pressure of the balloon

A is the effective area

F = 392 x 50 x 10⁻⁴

F = 1.96 N

Therefore, the force that you must exert on the balloon is 1.96 N

3 0

3 years ago

Please help! Thanks so much! :)

notsponge [240]

Answer: B

Explanation: I'm not 100% sure tho sorry if i'm wrong

7 0

3 years ago

Fluid pressure changes with depth are assumed to be linear. Which statement best explains why this does not hold true for atmosp

ankoles [38]

Answer:

Explanation:

Pressure due to fluid is directly proportional to the depth of fluid, density of the fluid and the value of acceleration due to gravity.

P = h d g

Where, h is the depth, d be the density and g be the acceleration due to gravity.

If we talk about teh atmospheric pressure, the density of air goes on decreasing as we go up and up. o we cannot say that it is directly depends only on the depth of air, it also depends on the changing density of air.

4 0

3 years ago

Think about the electricity sent to your home from a power plant. How does the voltage of the electricity that leaves the plant

tatiyna

C. High voltage to low voltage

6 0

3 years ago

An electric dipole is formed from ± 5.0 nC point charges spaced 3.0 mm apart. The dipole is centered at the origin, oriented alo

Ymorist [56]

Answer:

The electric field strength at point (x,y) = ( 20 mm ,0cm) is =16321.0769 N/C

The electric field strength at point (x,y) = (0cm, 20 mm) is =35321.58999 N/C

Explanation:

Question: What is the electric field strength at point (x,y) = ( 20 mm ,0cm)?

Answer:

The electric field at any given point of the dipole is given as:

E= (KP) ÷ (r^2 + a^2)^3/2

Where:

K = 9x10^9 Nm^2/c^2 (coloumb constant)

P = (0.003) (5x10^-9c) which is the movement of the dipole

(0.003) is arrived at when mm is converted to m. 3.0 mm space apart was converted to a meter.

r= the point, in the question above is 20mm = 0.02m

Now, the electric field, E can be calculated by putting the values in the formula above:

E = (KP) ÷ (r^2 + a^2)^3/2

= (9x10^9 Nm^2/c^2) (0.003 m) (5x10^-9c) ÷ [ (0.02m)^2 + (0.003)^2]^3/2

= 0.135 ÷ (8.271513x10^-6)

=16321.0769 N/C

Question: What is the electric field strength at point (x,y) = (0cm, 20 mm )?

Answer:

Here, the electric field, E= 2krp ÷ (r^2 - a^2)^2

E= 2 (9x10^9 Nm^2/c^2) (0.02m) (0.003 m) (5x10^-9c) ÷ [(0.02m)^2 - (0.003)^2]^2

= 0.0054 ÷ 0.000000152881

=35321.58999 N/C

8 0

2 years ago