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Alchen [17]
3 years ago
6

A squirrel, named jumping Jacky rides on a remote-controlled toy car in a circle in the horizontal x-y plane of radius of 45.0 m

. If the car is travelling with a constant tangential velocity of 36.0 m/s, what is the magnitude of the centripetal acceleration vector and its direction (does it point toward or away from the center of the circle or along the path that Jackie takes)?
Physics
1 answer:
viktelen [127]3 years ago
8 0

Answer:  28.8 m/sec², towards the center of the circle.

Explanation:

By definition, acceleration is defined as the quotient between a change in velocity (in value or direction) over time.

In this case, even the velocity value is constant, is changing direction all the time, so there is an acceleration present.

This acceleration is called the centripetal acceleration, and always aims to the center of the circle, because otherwise the toy car will move in a straight line at constant speed, like is dictated by Newton's First law.

It can be showed that it can be calculated as follows:

a = V² /  R  =  (36.0)² (m/s)² /  45.0 m = 28.8 m/s, pointing towards the center of the circle.

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Write two units for electric field intensity and show thier equivalence?​
nataly862011 [7]

Answer:

Electric field intensity is the force experienced by a test charge q in a electric field E.

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4 years ago
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A flat, circular loop has 18 turns. The radius of the loop is 15.0 cm and the current through the wire is 0.51 A. Determine the
Ostrovityanka [42]

Answer:

The magnitude of the magnetic field at the center of the loop is 3.846 x 10⁻⁵ T.

Explanation:

Given;

number of turns of the flat circular loop, N = 18 turns

radius of the loop, R = 15.0 cm = 0.15 m

current through the wire, I = 0.51 A

The magnetic field through the center of the loop is given by;

B = \frac{N\mu_o I}{2R}

Where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

B = \frac{N\mu_o I}{2R} \\\\B = \frac{18*4\pi*10^{-7} *0.51}{2*0.15} \\\\B = 3.846 *10^{-5} \ T

Therefore, the magnitude of the magnetic field at the center of the loop is 3.846 x 10⁻⁵ T.

6 0
4 years ago
An electric dipole is formed from two charges, ±q, spaced 0.800 cm apart. The dipole is at the origin, oriented along the y-axis
Simora [160]

Answer: q = 2.781e-9C = 2.781nC

E=200C

Explanation:

E = Qd/(2πEor^3)

Where

E=Electric field intensity

Q=Charge

d=distance between the dipole=0.008m

Eo=permitivitty

400 N/C = Q(0.80e-2 m)/(2πε*(10e-2 m)^3)

Q= (400* 2* 3.142 * 8.85 x 10-12 * 0.1^3)/0.008

q = 2.781e-9C = 2.781nC

b)

Though the dipole are two separate charges. And since the point is on the x-axis, the electric field strengths are equivalent. The magnitude of the vector sum is:

E = kq*2sin θ/r^2

= 2(8.99e9 N*m^2/C^2)(2.781e-9 C)*sin(arctan(.4/10))/(10e-2 m)^2

= 2(8.99e9) * (2.781e-9) * sin(2.290)/(10e-2 m)^2

=200 C

3 0
3 years ago
What do electricians call disconnects or disconnecting?
muminat
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4 0
2 years ago
please help !!!!!!!!!!!!!!!!!! give the answer to the question i. which lighthouse will be warmer during the day time and why ?
Lostsunrise [7]

Answer:

I. light house 1 will be warmer during the day ii. light house 2 will be warmer at night.

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