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Lunna [17]
2 years ago
15

A dolphin swims due east for 1.90 km, then swims 7.20 km in the direction south of west. What are the magnitude and direction of

the vector that will take the dolphin back to its starting point? (Enter the magnitude in km and the direction in degrees north of east.)
Physics
1 answer:
kykrilka [37]2 years ago
3 0

Answer:

magnitude = 7.446 km, direction = 75.22° north of east

Explanation:

From the questions,

To get the the magnitude of the resultant vector we use Pythagoras theorem

a² = b²+c²

From the diagram,

y² = 1.9²+7.2²

y² = 55.45

y = √(55.45)

y = 7.446 km.

The direction of the dolphin is given as,

θ = tan⁻¹(7.2/1.9)

θ = tan⁻¹(3.7895)

θ = 75.22° north of east

Hence the magnitude of the resultant vector = 7.446 km, and it direction is 75.22° north of east

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Scilla [17]

Answer:

W=kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})

Explanation:

Position of charge q₁ is (0,0)

Position of charge q₂ is (x₁,0)

So, the electric potential energy between the charges is given by :

U_1=k\dfrac{q_1q_2}{x_1}

Now the position of charge q₂ has been changes from (x₁,0) to (x₂,y₂). Now, electric potential energy between the charges is :

U_2=k\dfrac{q_1q_2}{\sqrt{x_2^2+y_2^2}}

We know form the work energy theorem that, the change in potential energy is equal to the work done. Mathematically, it is given by :

W=-\Delta U

W=-(U_2-U_1)

W=(U_1-U_2)

W=(k\dfrac{q_1q_2}{x_1}-k\dfrac{q_1q_2}{\sqrt{x_2^2+y_2^2}})

W=kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})

Hence, the work done by the electrostatic force on the moving point charge is kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}}). Hence, this is the required solution.

3 0
2 years ago
If you have to apply 40n of force on a crowbar to lift a rock that weights 400n, what is the actual mechanical advantage of the
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The mechanical advantage is defined as the ratio between the force produced by a machine and the force applied in input:
MA= \frac{F_{out}}{F_{in}}
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2 years ago
An electron moving at right angles to a 0.1 T magnetic field experiences an acceleration of 6 × 1015 m.s-2. What is the speed of
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Explanation:

It is given that,

Magnetic field, B = 0.1 T

Acceleration, a=6\times 10^{15}\ m/s^2

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Mass of electron, m=9.1\times 10^{-31}\ kg    

(a) The force acting on the electron when it is accelerated is, F = ma

The force acting on the electron when it is in magnetic field, F=qvB\ sin\theta

Here, \theta=90

So, ma=qvB

Where

v is the velocity of the electron

B is the magnetic field

v=\dfrac{ma}{qB}

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or

v=3.41\times 10^5\ m/s

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José is pinned against the walls of the Rotor, a ride with a radius of 3.00 meters that spins so fast that the floor can be remo
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r = radius of the circle of the ride = 3.00 meters

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Angular momentum is given as

L = m v r

inserting the values

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m = 3570 kg m²/s/(51 m²/s)

m = 7 kg

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