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exis [7]
4 years ago
11

If the distance traveled increases and the time of travel stays same, what happens to the speed

Physics
1 answer:
valkas [14]4 years ago
7 0
I think the answer is C. The speed increases. I hope this helps. I'm sorry if I'm wrong :)
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A 0.12 kg body undergoes simple harmonic motion of amplitude 8.5 cm and period 0.20 s. (a) What is themagnitude of the maximum f
Neporo4naja [7]

Answer:

a)F=698.83 N

b)K=8221.56 N/m

Explanation:

Given that

mass ,m = 0.12 kg

Amplitude ,A= 8.5 cm

time period ,T = 0.2 s

We know that

T=\dfrac{2\pi}{\omega}

{\omega}=\dfrac{2\pi }{0.2}\ rad/s

{\omega}=31.41\ rad/s

We know that

{\omega}^2=m\ K

K=Spring constant

K=\dfrac{\omega^2}{m}

K=\dfrac{31.41^2}{0.12}\ N/m

K=8221.56 N/m

The maximum force F

F= K A

F= 8221.56 x 0.085 N

F=698.83 N

a)F=698.83 N

b)K=8221.56 N/m

3 0
4 years ago
Why does wave height increase in shallow water??
Vitek1552 [10]
As the water russhes toward the shore, it rises because it is pushing against it.<span />
3 0
3 years ago
Alien A lifts a 500-newton child from the floor to a height of 0.40 meters in 2 seconds
vivado [14]
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4 0
4 years ago
An elevator provides 21 000 w of power during a 12 s ride. how much work does the elevator do?
Mice21 [21]

Answer:

252000 J

Explanation:

W = Pt

W = (21000 w)(12 s)

W =252000 J

4 0
3 years ago
A long, thin solenoid has 450 turns per meter and a radius of 1.06 . The current in the solenoid is increasing at a uniform rate
kirill115 [55]

Answer:9.34 A/s

Explanation:

Given

radius of solenoid R=1.06 m

Emf induced E=8.50\times 10^{-6} V/m

no of turns per meter n=450

we know Induced EMF is given by

\int Edl=-\frac{\mathrm{d} \phi}{\mathrm{d} t}=-\frac{\mathrm{d} B}{\mathrm{d} t}A

Magnetic Field is given by

B=\mu _0ni

thus \frac{\mathrm{d} B}{\mathrm{d} t}=-\mu _0n\frac{\mathrm{d} i}{\mathrm{d} t}

Area of cross-section

A=\pi R^2 where

solving integration we get

E.\cdot 2\pi r=\mu _0n\frac{\mathrm{d} i}{\mathrm{d} t}\pi R^2

where r=distance from axis

R=radius of Solenoid

\frac{\mathrm{d} i}{\mathrm{d} t}=\frac{Er}{\mu _0nR^2}

\frac{\mathrm{d} i}{\mathrm{d} t}=\frac{8.50\times 10^{-6}\times 3.49\times 10^{-2}}{4\pi \times 10^{-7}\times 450\times 1.06^2}

\frac{\mathrm{d} i}{\mathrm{d} t}=9.34 A/s

4 0
3 years ago
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