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Tanya [424]
3 years ago
8

You drive a car 1600 ft to the east, then 2500 ft to the north. The trip took 2.5 minutes. What was the magnitude of your averag

e velocity?
Physics
1 answer:
elena-14-01-66 [18.8K]3 years ago
7 0

Answer:

Velocity=6.03m/s

Explanation:

Given data

Time t=2.5 minutes=150 seconds

Distance A=1600 ft=487.68 m........east

Distance B=2500 ft=762m ........north

To find

Average velocity

Solution

First we need to find the resultant distance magnitude.To find that we apply Pythagorean theorem to find hypotenuse

So

A^{2}+B^{2}=C^{2}\\  C=\sqrt{A^{2}+B^{2}}\\ C=\sqrt{(487.68m)^{2}+(762m)^{2}}\\ C=904.7m

Velocity=\frac{Distance}{Time}\\Velocity=\frac{904.7m}{150s}\\Velocity=6.03m/s

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A loop of wire is perpendicular to a magnetic field such that the field is coming straight at you through the loop. The field be
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Answer:

Inducted Magnetic field will be toward from you

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Explanation:

Lenz's law states that the direction of the current induced in a wire by a changing magnetic field is such that the magnetic field created by the induced current opposes the initial changing magnetic field.

So if the field begins to decrease, the induced magnetic field would try to stop this, so  its direction will be the same as the magnetic field, toward from you.

This induced magnetic field is produced by the current in the wire. If the inducted magnetic field will be toward you, the right hand rule says that the direction from the inducted current will be counter clockwise.

5 0
2 years ago
A ball was dropped from a height of 10 feet. Each time it hits the ground, it bounces 4/5 of its previous height. Find the total
Shtirlitz [24]

Answer:

d = 90 ft

Explanation:

As we know that after each bounce it reaches to 4/5 times of initial height

so we can say

h_2 = \frac{4}{5}h

so the distance covered is given as

d = h + 2(\frac{4}{5}h) + 2(\frac{4}{5})^2h + 2(\frac{4}{5})^3h........

here we know that

h = 10 feet

d = h + 2(\frac{4}{5}h)(1 + \frac{4}{5} + (\frac{4}{5})^2 + ...........)

d = 10 + 2(\frac{4}{5}(10))(\frac{1}{1 - \frac{4}{5}})

d = 90 ft

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3 years ago
A thin spherical shell of radius R has a total charge +Q uniformly distributed over its surface. Of the following distance r fro
grigory [225]

Answer:

The correct answer is B

Explanation:

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         Φ._{E} = ∫ E. dA = q_{int} / ε₀

For this case we create a Gaussian surface that is a sphere.  We can see that the two of the sphere and the field lines from the spherical shell grant in the direction whereby the scalar product is reduced to the ordinary product

        ∫ E dA = q_{int} / ε₀

The area of ​​a sphere is

     A = 4π r²

   

    E 4π r² =q_{int} / ε₀

    E = (1 /4πε₀ )  q / r²

Having the solution of the problem let's analyze the points:

A   ) r = 3R / 4  = 0.75 R.

  In this case there is no charge inside the Gaussian surface therefore the electric field is zero

        E = 0

B) r = 5R / 4 = 1.25R

In this case the entire charge is inside the Gaussian surface, the field is

    E = (1 /4πε₀ )  Q / (1.25R)²

    E = (1 /4πε₀ )  Q / R2 1 / 1.56²

    E₀ = (1 /4π ε₀ )  Q / R²

   E_{B} =  Eo /1.56 ²

  E_{B}  = 0.41 Eo

C) r = 2R

All charge inside is inside the Gaussian surface

    E_{B} =(1 /4π ε₀ ) Q    1/(2R)²

    E_{B} = (1 /4π ε₀ ) q/R²   1/4

    E_{B} = Eo  1/4

    E_{B} = 0.25 Eo

D) False the field changes with distance

The correct answer is B

4 0
3 years ago
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