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3 years ago

What is indicated by the slope of an acceleration vs. time graph?

Physics

1 answer:

Delicious77 [7]3 years ago

5 0

A slip is going up or down and a time graph is it keep take

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You jump off a truck and accelerate toward the surface of the Earth. Does the Earth accelerate toward you?

Triss [41]

When we jump from the truck and accelerate towards the earth surface, the earth also accelerates towards us but it's acceleration is very negligible.

To find the answer, we need to know about the acceleration of earth due to the gravitational attraction.

<h3>What's the gravitational force between the earth and a person?</h3>

Gravitational attraction force is GMm/r² between the earth and a person.
M= mass of the earth

m= mass of the person

r= separation between them.

<h3>What's the acceleration of the earth towards the person when he jumps from a truck?</h3>

According to Newton's second law, Force = M×acceleration
Acceleration= Force / M
Here, Force = GMm/r²,

so acceleration of earth= Gm/r²

As this acceleration is very small, so we can't notice it.

Thus, we can conclude that the earth also accelerates towards us.

Learn more about the gravitational force here:

brainly.com/question/72250

#SPJ4

7 0

2 years ago

A concave mirror has a radius of curvature of 0.60 meter.

love history [14]

Pls give me brainliest!!!

6 0

2 years ago

Why are there only two elements in the first period of the periodic table?(1 point)

Katen [24]

Answer:

because only two electrons can fit in the first orbit around the nucleus, and each period on the table is organized by number of orbits

6 0

2 years ago

How would you find the total energy stored in the

likoan [24]

Answer:

The energy of the capacitors connected in parallel is 0.27 J

Given:

C = $2.0\micro F = 2.0\times 10^{- 6} F$

C' = $4.0\micro F = 4.0\times 10^{- 6} F$

Potential difference, V = 300 V

Solution:

Now, we know that the equivalent capacitance of the two parallel connected capacitors is given by:

$C_{eq} = C + C' = 2.0 + 4.0 = 6.0\micro F = 6.0\times 10^{- 6} F$

The energy of the capacitor, E is given by;

$E = \frac{1}{2}C_{eq}V^{2}$

$E = \frac{1}{2}\times 6.0\times 10^{- 6}\times 300^{2} = 0.27 J$

6 0

2 years ago

A certain unfiltered full-wave rectifier with 120 V, 60 Hz input produces an output with a peak of 15 V. When a capacitor-input

KATRIN_1 [288]

Answer:

The peak-to-peak ripple voltage = 2V

Explanation:

120V and 60 Hz is the input of an unfiltered full-wave rectifier

Peak value of output voltage = 15V

load connected = 1.0kV

dc output voltage = 14V

dc value of the output voltage of capacitor-input filter

where

V(dc value of output voltage) represent V₀

V(peak value of output voltage) represent V₁

V₀ = 1 - ( $\frac{1}{2fRC}$ )V₁

make C the subject of formula

V₀/V₁ = 1 - (1 / 2fRC)

1 / 2fRC = 1 - (v₀/V₁)

C = 2fR ((1 - (v₀/V₁))⁻¹

Substitute for,

f = 240Hz , R = 1.0Ω, V₀ = 14V , V₁ = 15V

C = 2 * 240 * 1 (( 1 - (14/15))⁻¹

C = 62.2μf

The peak-to-peak ripple voltage

= (1 / fRC)V₁

= 1 / ( (120 * 1 * 62.2) )15V

= 2V

The peak-to-peak ripple voltage = 2V

3 0

3 years ago