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ArbitrLikvidat [17]
2 years ago
12

Albert Jr. takes his new skateboard out to the top of the loading ramp. The ramp is 30 m long and sloped at an angle of 15 degre

es relative to the horizontal. Assuming he starts from rest, and neglecting friction, about how fast is Albert Jr. going at the bottom of the ramp?
Physics
1 answer:
WINSTONCH [101]2 years ago
5 0

Answer:

vf = 12.343 m/s

Explanation:

Given:

- The Length of the ramp s = 30 m

- The angle of the ramp θ = 15 degrees

- Initial velocity at the top of ramp vi = 0

- Neglect Friction

Find:

How fast is Albert Jr. going at the bottom of the ramp?

Solution:

- We can use conservation of energy for the skateboard at top and bottom of the ramp.

                                ΔK.E = ΔP.E

Where, ΔK.E : Change in kinetic energy

            ΔP.E : Change in gravitational potential energy

                                0.5*m*(vf^2 - vi^2) = m*g*Δh

Where, m: The mass of the object

            Δh: Change in elevation from top to bottom

                                Δh = s*sin(θ)

- Substitute and simplify:

                                 0.5*(vf^2 - vi^2) = g*s*sin(θ)

                                 vf^2 = 2*g*s*sin(θ)

                                 vf = √(2*g*s*sin(θ)) = √(2*9.81*30*sin(15))

                                 vf = 12.343 m/s

- The velocity at the bottom of ramp would be 12.343 m/s

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Answer:

<em>mass</em>

<em></em>

Explanation:

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A magnetic field of 37.2 t has been achieved at the mit francis bitter national magnetic laboratory. Find the current needed to
jeka57 [31]

Answer:

Here is the complete question:

https://www.chegg.com/homework-help/questions-and-answers/magnetic-field-372-t-achieved-mit-francis-bitter-national-magnetic-laboratory-find-current-q900632

a) Current for long straight wire  =3.7\ MA

b) Current at the center of the circular coil =2.48\times 10^{5}\ A

c) Current near the center of a solenoid 236.8\ A

Explanation:

⇒ Magnetic Field due to long straight wire is given by (B),where B=\frac{\mu\times I}{2\pi(r) },so\ I=\frac{B\ 2\pi(r)}{\mu}

\mu=4\pi \times 10^{-7}\ \frac{henry}{m}

Plugging the values,

Conversion 1\times 10^6 A = 1\ MA,and 2cm=\frac{2}{100}=0.02\ m

I=\frac{37.2\times \ 2\pi(0.02)}{4\ \pi \times (10^{-7})}=3.7\ MA

⇒Magnetic Field at the center due to circular coil (at center) is given by,B=\frac{\mu\times I (N)}{2(a)}

So I= \frac{2B(a)}{\mu\ N} = \frac{2\times 37.2\times 0.42}{4\pi\times 10^{-7}\times 100}=2.48\time 10{^5}\ A

⇒Magnetic field due to the long solenoid,B=\mu\ nI=\mu (\frac{N}{l})I

Then I=\frac{B}{\mu(\frac{N}{L})} \approx 236.8\A  

So the value of current are  3.7 MA,2.48\times 10^{5} A and 236.8\ A respectively.

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