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Bond [772]
3 years ago
11

A car is initially traveling at 12 m/s when the driver sees a yellow light ahead. He accelerates at a constant 7 m/s^2 for 6 s i

n order to make it through the intersection. What was the velocity at the end of this time?
Physics
1 answer:
Vitek1552 [10]3 years ago
3 0

Answer:

v = 54m/s

Explanation:

a =  \frac{v - u}{t}

a = 7m/s²

u = 12m/s

t = 6s

7 = (v-12)/6

v - 12 = 42

v = 54m/s

(Correct me if i am wrong)

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Neporo4naja [7]
Cathy Rigby had a eating disorder.
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3 years ago
An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.50 mm. A 25
torisob [31]

Answer:

The electric field is 16666.66 V/m.

The surface charge density is 1.474\times10^{-7}\ C/m^2

The capacitance is 4.484\times10^{-12}\ F

The charge on each plate is 112.1\times10^{-12}\ C

Explanation:

Given that,

Area =7.70 cm²

Distance = 1.50 mm

Potential difference = 25.0 V

Suppose  we find the electric field between the plates, the surface charge density, the capacitance and the charge on each plates.

We need to calculate the electric field

Using formula of electric field

E=\dfrac{V}{d}

Put the value into the formula

E=\dfrac{25.0}{1.50\times10^{-3}}

E=16666.66\ V/m

We need to calculate the charge density

Using formula of charge density

\sigma=E\times\epsilon_{0}

Put the value into the formula

\sigma=16666.66\times8.85\times10^{-12}

\sigma=1.474\times10^{-7}\ C/m^2

We need to calculate the capacitance

Using formula of capacitance

C=\dfrac{\epsilon_{0}A}{d}

Put the value into the formula

C=\dfrac{8.85\times10^{-12}\times7.60\times10^{-4}}{1.50\times10^{-3}}

C=4.484\times10^{-12}\ F

We need to calculate the charge

Using formula of charge

q=CV

Put the value into the formula

q=4.484\times10^{-12}\times25.0

q=112.1\times10^{-12}\ C

Hence, The electric field is 16666.66 V/m.

The surface charge density is 1.474\times10^{-7}\ C/m^2

The capacitance is 4.484\times10^{-12}\ F

The charge on each plate is 112.1\times10^{-12}\ C

8 0
3 years ago
Will the positioning of your Slinky along the east-west, north-south, or some other orientation affect your readings? Why?
Oxana [17]

Positioning your Slinky along any direction different from its initial position will affect your reading, because there will be change in the magnetic field.

<h3>Effect of magnet on Slinky</h3>

If the Slinky is made of an iron alloy, it can be magnetized by itself. Moving the Slinky around can cause a change in the magnetic field, even if no current is flowing.

When there is a change in the magnetic field, the reading changes.

At any point, you change the orientation of the Slinky, you will need to zero the reading or adjust the Slinky back to its initial position, even if the sensor does not move.

Thus, Positioning your Slinky along any direction that is different to its initial position will affect your reading because there will be change in the magnetic field.

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5 0
2 years ago
A mass is attached to a spring with an unknown spring constant. The spring gains 10 J of elastic potential energy if stretched b
Firlakuza [10]

From the given information in the question, the correct option is Option 1: 14 cm.

A non-stretched elastic spring has a conserved potential energy which gives it the ability to perform work. The elastic potential energy can be expressed as:

PE = \frac{1}{2} k x^{2}

Where PE is the energy, k is the spring constant and x is extension.

i. Given that: PE = 10 J and x = 10 cm, then;

PE = \frac{1}{2} k x^{2}

10 = \frac{1}{2} k 10^{2}

20 = 100k

k = 0.2 J/cm

ii. To determine how far the spring is needed to be stretched, given that PE = 20 J.

PE = \frac{1}{2} k x^{2}

20 = \frac{1}{2} (0.2) x^{2}

40 = 0.2 x^{2}

x^{2} = 200

x = \sqrt{200}

  = 14.1421

x = 14.14 cm

So that;

x is approximately 14.00 cm.

Thus, the spring need to be stretched to 14.00 cm to give the spring 20 J of elastic potential energy.

For more information, check at: brainly.com/question/1352053.

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Answer:

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