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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?

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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Determine the voltage ratings of the high-and-low voltage windings for this connection and the MVA rating of the autotransformer

SIZIF [17.4K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The High Voltage Rating for Auto - Transformer is 86kV

The Low Voltage Rating for Auto - Transformer is 78kV

The MVA rating is 268.75 $MVA$

b

The efficiency is 99.4%

Explanation:

From the question we are given are given that

The transformer has Mega Volt Amp rating of 25MVA

The frequency is 60-Hz

Voltage rating 8.0kV : 78kV

The short circuit test gives : 453kV,321A,77.5kW

The open circuit test gives : 8.0kV, 39.6A, 86.2kW

This can be represented on a diagram shown on the second uploaded image

From this diagram we can deduce that the The High Voltage Rating for Auto - Transformer is 86kV and the Low Voltage Rating for Auto - Transformer is 78kV

Now to obtain the current flowing through the 8kV coil in the Auto-transformer we have

$\frac{25 \ Mega \ Volt\ Ampere }{8\ Kilo Volt}$

The volt will cancel each other

$\frac{25*10^6}{8*10^3} = 3125\ A$

Now to obtain the required MVA rating we would multiply the value of Power obtained during the open circuit test by the value of the current calculated.we are making use of the power obtain during open circuit testing because the transformer at this point is not under any load.

$MVA \ rating = (86*10^3)(3125) =268.75$

We need to understand that Iron losses is due to open circuit test which has power = 86.2kW

While copper loss is due to short circuit test which has power = 77.5kW

The the current flowing through the secondary coil $I_2$ as shown in the circuit diagram can be obtained as

$I_2 = \frac{25*10^6}{78*10^3} =320.52 A \approx 321$

Now the efficiency can be obtained as thus

$\frac{(operational \ MVA )*(Power factor \pf))}{(operational\ MVA (power factor pf) + copper loss + Iron loss)}*\frac{100}{1}$

=99.941%

8 0

2 years ago

A. The potential energy stored in the compressed spring of a dart gun, with a spring constant of 62.00 N/m, is 0.940 J. Find by

valina [46]

a. The spring is compressed by 0.174 m.

b. The vertical distance the dart travels from its position when the spring is compressed to its highest position is 9.6 m.

c. The horizontal velocity of the dart at that time is 13.74 m/s.

d. The horizontal distance from the equilibrium position at which the dart hits the ground is 19.236 m.

<h3>What is potential energy?</h3>

The energy by virtue of its position is called the potential energy.

a. Given is the potential energy stored in the compressed spring of a dart gun, with a spring constant of 62.00 N/m, is 0.940 J.

PE of spring = 1/2 kx²

Put the values, we get

P.E = 0.940 = 1/2 x 62 x²

x = 0.174 m

Thus, the spring is compressed by 0.174 m.

b. Given is a 0.010 kg dart is fired straight up.

The vertical height is find out by

0.940 J = (0.010 kg) (9.8 m/s²) h

h = 9.6 m

Thus, the vertical distance the dart travels from its position is 9.6 m

c. From the conservation of energy principle, total mechanical energy is conserved.

1/2 mv² =mgh

v = √2gh

Plug the values, we get

v = √2 x 9.8x 9.6

v = 13.74 m/s

Thus, the horizontal velocity is 13.74 m/s.

d. Time that dart spends in air, t = √2h/g

t = √(2x9.6)/9.81

t = 1.4 s

The horizontal distance from the equilibrium position at which the dart hits the ground.

Horizontal distance = (Velocity on x direction) x time

Horizontal distance = 13.74 m/s x 1.4s

Horizontal distance = 19.236 m

Thus, the horizontal distance is 19.236 m.

Learn more about potential energy.

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7 0

2 years ago

The frequency of violet light is 7.5x10 to the 14th power hertz. What is the wavelength?

Verizon [17]

Wavelength times frequency = speed of light
7.5E14 x wavelength = 300,000 m/s

Wavelength in meters = 300,000 divided by 7.5E14

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2 years ago

The separation in time between the arrival of primary and secondary waves is

Gre4nikov [31]

The separation in time between the arrival of primary and secondary wave is called LAG TIME.
The time difference between the arrival of primary wave and secondary wave in a seismogram is called lag time. The primary wave always travels faster than the secondary wave, thus the difference between the two can be obtained by estimating the difference between the arrival time of the two waves/.

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3 years ago

g You shine orange laser light that has a wavelength of 600 nm through a narrow slit. The slit forms a diffraction pattern on a

zimovet [89]

Answer:

λ = 3 10⁻⁷ m, UV laser

Explanation:

The diffraction phenomenon is described by the expression

a sin θ = m λ

let's use trigonometry

tan θ = y / L

as in this phenomenon the angles are small

tan θ = $\frac{sin \ \theta}{cos \ \theta}$ = sin θ

sin θ = y / L

we substitute

a y / L = m λ

let's apply this equation to the initial data

a 0.04 / L = 1 600 10⁻⁹

a / L = 1.5 10⁻⁵

now they tell us that we change the laser and we have y = 0.04 m for m = 2

a 0.04 / L = 2 λ

a / L = 50 λ

we solve the two expression is

1.5 10⁻⁵ = 50 λ

λ = 1.5 10⁻⁵ / 50

λ = 3 10⁻⁷ m

UV laser

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2 years ago