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3 years ago

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The position of an object in simple harmonic motion is described by the following equation: y=(0.50 m)cos(pi/2)t. What is the am

plitude of vibration?
A. 0.75 m
B. 0.25 m
C. 0.50 m
D. 1.0 m

2 answers:

Kazeer [188]3 years ago

5 0

Answer:

C. 0.50 m

Hope this helps,

Please mark me as the brainliest

Thank you

andre [41]3 years ago

3 0

Answer:

C. 0.50 m

Explanation:

mark me as brainliest

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Jazz is a 172 lb athlete who exercised at 7.6 METs. At this workload, what is his energy expenditure in kcals/min.? Round to the

dezoksy [38]

Answer:

Energy expenditure in K cals/min = 10 K cals /min (approximately)

Explanation:

As we know

Energy expenditure in Kcal/min= METs x 3.5 x Body weight (kg) / 200

Given is METs=7.6

Weight of Jazz= 172lb=78.02kg

putting the values in formula,

Energy expenditure in K cals/min= 7.6 x 3.5 x 78.02 / 200

=10.38 K cals /min

=10 K cals /min (approximately)

Therefore, Energy expenditure in K cals/min by Jazz will be approximately 10 K cals /min

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3 years ago

A proton, which is the nucleus of a hydrogen atom, can be modeled as a sphere with a diameter of 2.4 fm and a mass of 1.67 10-27

Step2247 [10]

For the answer to the question above asking to d<span>etermine the density of the proton.
</span>Density is mass over volume.

The volume of a sphere is 4πr³/3. r is half the diameter.

So the density would be 2.3×10¹⁷ kg/m³.
I hope my answer helped you. Feel free to ask more questions. Have a nice day!

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4 years ago

Read 2 more answers

A solenoidal coil with 26 turns of wire is wound tightly around another coil with 350 turns. The inner solenoid is 20.0 cm long

noname [10]

Answer:

Part a)

$\phi = 2.76 \times 10^{-7} T m^2$

Part B)

$M = 5.52 \times 10^{-5} H$

Part C)

$EMF = 0.1 V/s$

Explanation:

Part a)

Magnetic field due to a long ideal solenoid is given by

$B = \mu_0 n i$

n = number of turns per unit length

$n = \frac{N}{L}$

$n = \frac{350}{0.20}$

$n = 1750 turn/m$

now we know that magnetic field due to solenoid is

$B = (4\pi \times 10^{-7})(1750)(0.100)$

$B = 2.2 \times 10^{-4} T$

Now magnetic flux due to this magnetic field is given by

$\phi = B.A$

$\phi = (2.2 \times 10^{-4})(\pi r^2)$

$\phi = (2.2 \times 10^{-4})(\pi(0.02)^2)$

$\phi = 2.76 \times 10^{-7} T m^2$

Part B)

Now for mutual inductance we know that

$\phi_{total} = M i$

$\phi_{total} = N\phi$

$\phi_{total} = 20(2.76 \times 10^{-4})$

$\phi_{total} = 5.52 \times 10^{-6}$

now we have

$M = \frac{5.52 \times 10^{-6}}{0.100}$

$M = 5.52 \times 10^{-5} H$

Part C)

As we know that induced EMF is given as

$EMF = M \frac{di}{dt}$

$EMF = 5.52 \times 10^{-5} (1800)$

$EMF = 0.1 V/s$

3 0

3 years ago

If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?

vovikov84 [41]

Complete Question

In an action movie, the villain is rescued from the ocean by grabbing onto the ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his briefcase of counterfeit money when he is 130 m above the water. If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?

Answer:

The speed of the helicopter is $u = 7.73 \ m/s$

Explanation:

From the question we are told that

The height at which he let go of the brief case is h = 130 m

The time taken before the the brief case hits the water is t = 6 s

Generally the initial speed of the briefcase (Which also the speed of the helicopter )before the man let go of it is mathematically evaluated using kinematic equation as

$s = h+ u t + 0.5 gt^2$

Here s is the distance covered by the bag at sea level which is zero

$0 = 130+ u * (6) + 0.5 * (-9.8) * (6)^2$

=> $0 = 130+ u * (6) + 0.5 * (-9.8) * (6)^2$

=> $u = \frac{-130 + (0.5 * 9.8 * 6^2) }{6}$

=> $u = 7.73 \ m/s$

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