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2 years ago

On a distance-time graph, time is shown on the y-axis. A) True B) False

Physics

1 answer:

Artist 52 [7]2 years ago

7 0

Answer:

false : In distance time graph,time is shown on the x -axis

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Brlan is repairing an old alarm clock. He needs to replace a device that converts the electric energy into sound energy. Which o

Brums [2.3K]

The answer is buzzer

3 0

2 years ago

Assume that the average mass of each of the approximately 1 billion people in China is 55 kg.Assume that they all gather in one

Harman [31]

Answer: $5.9(10)^{-8} m$

Explanation:

The equation to calculate the center of mass $C_{M}$ of a particle system is:

$C_{M}=\frac{m_{1}r_{1}+m_{1}r_{1}+...+m_{n}r_{n}}{m_{1}+m_{2}+...+m_{n}}$

In this case we can arrange for one dimension, assuming the geometric center of the Earth and the ladder are on a line, and assuming original center of mass located at the Earth's geometric center:

$C_{M}=\frac{m_{E}(0 m) + m_{p} r_{E-p}}{m_{E}+m_{p}}$

Where:

$m_{E}=5.9(10)^{24} kg$ is the mass of the Earth

$m_{p}=55(10)^{9} kg$ is the mass of 1 billion people

$r_{E}=6371000 m$ is the radius of the Earth

$r_{E-p}=6371000 m- 2m=6370998 m$ is the distance between the center of the Earth and the position of the people (2 m above the Earth's surface)

$C_{M}=\frac{m_{p}55(10)^{9} kg (6370998 m)}{5.9(10)^{24} kg+55(10)^{9} kg}$

$C_{M}=5.9(10)^{-8} m$ This is the displacement of Earth's center of mass from the original center.

8 0

3 years ago

How large a force is necessary to stretch a 4.0-mm-diameter steel wire from its original length by 1.0%?

jekas [21]

The force needed to stretch the steel wire by 1% is 25,140 N.

The given parameters include;

diameter of the steel, d = 4 mm

the radius of the wire, r = 2mm = 0.002 m

original length of the wire, L₁

final length of the wire, L₂ = 1.01 x L₁ (increase of 1% = 101%)

extension of the wire e = L₂ - L₁ = 1.01L₁ - L₁ = 0.01L₁

the Youngs modulus of steel, E = 200 Gpa

The area of the steel wire is calculated as follows;

$A = \pi r^2\\\\ A= 3.142 \times (0.002)^2\\\\ A= 1.257 \times 10^{-5} \ m^2$

The force needed to stretch the wire is calculated from Youngs modulus of elasticity given as;

$E = \frac{stress}{strain} = \frac{F/A}{e/L} = \frac{FL}{Ae} \\\\F = \frac{EAe}{L}$

$F = \frac{200 \times 10^9\ \times\ 1.257\times 10^{-5}\ \times \ 0.01l_1}{l_1} \\\\F = 25,140\ N$

Thus, the force needed to stretch the steel wire by 1% is 25,140 N.

Learn more here: brainly.com/question/21413915

4 0

2 years ago

Explain how the data collected and the calculations for the first and second resonance points in today's experiment would change

grandymaker [24]

Answer:

tssths

Explanation:

hgst

8 0

3 years ago

How much power does a hair dryer require to transform 198 000 J of energy in 15 mins?

Marina86 [1]

(198,000 joule / 15 minute) x (minute / 60 sec) = 220 joule/sec = <em>220 watts

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7 0

3 years ago