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2 years ago

The velocity of a car is 65 m/s and it’s mass is 2515 kg. What is it’s KE?

2 answers:

4 0

Example Problems. Kinetic Energy (KE = ½ m v2). 1) The velocity of a car is 65 m/s and its mass is 2515 kg. What is its KE? 2) If a 30 kg child were running at a rate of 9.9 m/s, what is his KE? Practice Problems. IN THIS ORDER…. Page 2: #s 6, 7, 8, 5. Potential Energy. An object can store energy as the result of its position.

spin [16.1K]2 years ago

3 0

Answer:

The KE will be 5312.9375 KJ

Explanation:

Mass of car = m = 2515 kg

Velocity of car = v = 65 m/s

Kinetic Energy = KE = ?

We know that,

KE = 1/2 mv^2 ...... (i)

By putting values in equation (i)

KE = 1/2 (2515)(65)^2

KE = 5312937.5 J

KE = 5312.9375 KJ

Hence, KE of the car will be 5312.9375 KJ.

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The force of Earths gravity keeps earth in orbit true or false

faust18 [17]

Answer:

True

Explanation:

The force of gravity keeps all of the planets in orbit around the sun

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2 years ago

You are designing a generator with a maximum emf 8.0 V. If the generator coil has 200 turns and a cross-sectional area of 0.030

shutvik [7]

Answer:

7.1 Hz

Explanation:

In a generator, the maximum induced emf is given by

$\epsilon= 2\pi NAB f$

where

N is the number of turns in the coil

A is the area of the coil

B is the magnetic field strength

f is the frequency

In this problem, we have

N = 200

$A=0.030 m^2$

$\epsilon=8.0 V$

B = 0.030 T

So we can re-arrange the equation to find the frequency of the generator:

$f=\frac{\epsilon}{2\pi NAB}=\frac{8.0 V}{2\pi (200)(0.030 m^2)(0.030 T)}=7.1 Hz$

4 0

2 years ago

Read 2 more answers

As a result of fringing effect :

PSYCHO15rus [73]

i thinks answers is gap relutance increases linearly with magnetic density

5 0

3 years ago

active and passive secruity measures are employed to identify, detect, classify and analyze possible threats inside of which zon

Juli2301 [7.4K]

Answer:

Assessment zone

Explanation:

It is the assessment zone in various security zones where active and passive security measures are employed to identify, detect, classify and analyze possible threats inside the assessment zones.

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2 years ago

Sometimes a person cannot clearly see objects close up or far away. To correct this type of vision, bifocals are often used. The

Rudik [331]

Answer:

1) P₁ = -2 D, 2) P₂ = 6 D

Explanation:

for this exercise in geometric optics let's use the equation of the constructor

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where f is the focal length, p and q are the distance to the object and the image, respectively

1) to see a distant object it must be at infinity (p = ∞)

$\frac{1}{f_1} = \frac{1}{q}$

q = f₁

2) for an object located at p = 25 cm

$\frac{1}{f_2} = \frac{1}{25} + \frac{1}{q}$

We can that in the two expressions we have the distance to the image, this is the distance where it can be seen clearly in general for a normal person is q = 50 cm

we substitute in the equations

1) f₁ = -50 cm

$\frac{1}{f_2} = \frac{1}{25} + \frac{1}{50}$

$\frac{1}{f_2}$ = 0.06

f₂ = 16.67 cm

the expression for the power of the lenses is

P = $\frac{1}{f}$

where the focal length is in meters

1) P₁ = 1/0.50

P₁ = -2 D

2) P₂ = 1 /0.16667

P₂ = 6 D

4 0

2 years ago