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guapka [62]
2 years ago
5

The radius of a Circular racetrack is 50 m. if the race car can complete one lap in 20 seconds what is the speed of the race car


Physics
1 answer:
jonny [76]2 years ago
8 0

Answer:

15.71 m/s

Explanation:

circumference of the racetrack = 2πr

= 2πx50

= 314.16m

the circumference becomes the distance

speed= distance/time

= 314.14/20

= 15.71 m/s

kindly mark Brainiest

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The Neutrons does not have a charge.

The Protons are positively charge.

Hence the charge on the Nucleus, would be the charge of the proton, which is positive.

Hence Nucleus is Positively Charged.
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If F1 = 104 and F2 = 104, what will be the net force?
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Read 2 more answers
Estimate the electric field at a point 2.40 cm perpendicular to the midpoint of a uniformly charged 2.00-m-long thin wire carryi
nadya68 [22]

Answer:

E = 1.85*10^{12}\frac{N}{C}

Explanation:

Hi!

The perpendicular distance 2.4cm, is much less than the distance to both endpoints of the wire, which is aprox 1m. Then the edge effect is negligible at this field point, and we can aproximate the wire as infinitely long.

The electric filed of an infinitely long wire is easy to calculate. Let's call z the axis along the wire. Because of its simmetry (translational and rotational), the electric field E must point in the radial direction,  and it cannot depende on coordinate z. To calculate the field Gauss law is used, as seen in the image, with a cylindrical gaussian surface. The result is:

E = \frac{\lambda}{2\pi \epsilon_0 r}\\\lambda=\text{charge per unit length}=\frac{4.95 \mu C}{2 m} = 2.475 \frac{C}{m}\\r=\text{perpendicular distance to wire}\\\epsilon_0=8.85*10^{-12}\frac{C^2}{Nm^2}

Then the electric field at the point of interest is estimated as:

E = \frac{\22.475}{2\pi*( 8.85*10^{-12})*(2.4*10^{-2})}\frac{N}{C}=1.85*10^{12}\frac{N}{C}

6 0
4 years ago
You are a member of an alpine rescue team and must get a box of supplies, with mass 3.00 kg , up an incline of constant slope an
Sever21 [200]

Answer:

v = 8.45 m/s

Explanation:

given,

mass  = 3 kg

angle = 30.0°

vertical distance = 3.3 m

μ = 0.06

according to conservation of energy

KE(loss) = PE(gain) + Work done (against\ friction)..............(1)

frictional Force

F_f = \mu N

F_f = \mu m g cos \theta

work against friction

W = F d

W = \mu m g cos \theta \times h l sin\theta

W = \dfrac{\mu m g \times h}{tan\theta}

Potential energy

PE = mgh

\dfrac{1}{2}mv^2= \dfrac{\mu mgh}{tan \theta}+ mgh

\dfrac{1}{2}v^2= \dfrac{0.06 \times 9.81 \times 3.3}{tan 30^0}+ 9.8\times 3.3

v = 8.45 m/s

the minimum speed is equal to 8.45 m/s

6 0
3 years ago
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