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olga2289 [7]
3 years ago
10

Susan, driving north at 53 mphmph , and Shawn, driving east at 63 mphmph , are approaching an intersection. Part A What is Shawn

's speed relative to Susan's reference frame?
Physics
1 answer:
mafiozo [28]3 years ago
5 0

Answer:

Shawn's speed relative to Susan's speed = 10 mph

Resultant velocity = 82.32 mph

Explanation:

The given data :-

i) Susan driving in north and speed of Susan is ( v₁ ) = 53 mph.

ii) Shawn driving in east and speed of Shawn is ( v₂ ) = 63 mph.

iii) The speed of both Susan and Shawn is relative to earth.

iv) The angle between Susan in north and Shawn in east is 90°.

We have to find Shawn's speed relative to Susan's speed.

v₂₁ = v₂ - v₁   = 63 - 53 = 10 mph

Resultant velocity,

v = \sqrt{v_{2} ^{2}+ v_{1} ^{2}  }  =\sqrt{63^{2} +53^{2} }

v = 82.32 mph

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A magnetic field is entering into a coil of wire with radius of 2(mm) and 200 turns. The direction of magnetic field makes an an
frozen [14]

Answer:

a) <em>2.278 x 10^-5 volts</em>

b) <em>1.139 x 10^-6 Ampere</em>

c) <em>2.59 x 10^-11 W</em>

Explanation:

The radius of the wire r = 2 mm = 0.002 m

the number of turns N = 200 turns

direction of the magnetic field ∅ = 25°

magnetic field strength B = 0.02 T

varying time = 2 sec

The cross sectional area of the wire = \pi r^{2}

==> A = 3.142 x 0.002^{2} = 1.257 x 10^-5 m^2

Field flux Φ = BA cos ∅ = 0.02 x 1.257 x 10^-5 x cos 25°

==> Φ = 2.278 x 10^-7 Wb

The induced EMF is given as

E = NdΦ/dt

where dΦ/dt = (2.278 x 10^-7)/2 = 1.139 x 10^-7

E = 200 x 1.139 x 10^-7 = <em>2.278 x 10^-5 volts</em>

<em></em>

<em></em>

b) If the two ends are connected to a resistor of 20 Ω, the current through the resistor is given as

I = E/R

where R is the resistor

I = (2.278 x 10^-5)/20 = <em>1.139 x 10^-6 Ampere</em>

<em></em>

<em></em>

<em> </em>c) power delivered to the resistor is given as

P = IE

P = (1.139 x 10^-6) x (2.278 x 10^-5) = <em>2.59 x 10^-11 W</em>

4 0
3 years ago
The magnitude of the electrostatic force between two identical ions that are separated by a distance of 5.0A is 3.7×10^-9N.a) wh
jeyben [28]

Explanation:

Given that,

Electrostatic force, F=3.7\times 10^{-9}\ N

Distance, r=5\ A=5\times 10^{-10}\ m

(a) F=\dfrac{kq^2}{r^2}, q is the charge on the ion              

q=\sqrt{\dfrac{Fr^2}{k}}

q=\sqrt{\dfrac{3.7\times 10^{-9}\times (5\times 10^{-10})^2}{9\times 10^9}}      

q=3.2\times 10^{-19}\ C

(b) Let n is the number of electrons are missing from each ion. It can be calculated as :

n=\dfrac{q}{e}

n=\dfrac{3.2\times 10^{-19}}{1.6\times 10^{-19}}

n = 2

Hence, this is the required solution.                        

8 0
3 years ago
A cat dozes on a stationary merry-go-round, at a radius of 4.4 m from the center of the ride. The operator turns on the ride and
monitta

Answer:

The coefficient of static friction is 0.29

Explanation:

Given that,

Radius of the merry-go-round, r = 4.4 m

The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.

We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

\mu mg=\dfrac{mv^2}{r}

v is the speed of cat, v=\dfrac{2\pi r}{t}

\mu=\dfrac{4\pi^2r}{gt^2}\\\\\mu=\dfrac{4\pi^2\times 4.4}{9.8\times (7.7)^2}\\\\\mu=0.29

So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.

4 0
3 years ago
How can the frequency with which the direction of a current changes be regulated?
sesenic [268]

Answer:

In an inductive circuit, when frequency increases, the circuit current decreases and vice versa.

Explanation:

5 0
3 years ago
What did Charles Darwin’s conclude on the Galapagos Island? Plz answer fast
kvasek [131]
He discovered several species of finches that varied from island to island and it helped him make his theory of natural selection.

hope this helps ! 
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