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Sunny_sXe [5.5K]
3 years ago
8

What is the kinetic energy of an object with a mass of 50kg and is and it is traveling at a rate of 60m/s.

Physics
1 answer:
liq [111]3 years ago
6 0

Answer:

90,000 J

Explanation:

Kinetic energy can be found using the following formula.

KE=\frac{1}{2}mv^2

where <em>m </em>is the mass in kilograms and <em>v</em> is the velocity in m/s.

We know the object has a mass of 50 kilograms. We also know it is a traveling at a rate of 60 m/s. Velocity is the speed of something, so the velocity of the object is 60 m/s.

<em>m</em>=50

<em>v</em>=60

Substitute these values into the formula.

KE=\frac{1}{2}*50*60^2

First, evaluate the exponent: 60^2. 60^2 is the same as multiplying 60, 2 times.

60^2=60*60=3,600

KE=\frac{1}{2}*50*3,600

Multiply 50 and 3,600

KE=\frac{1}{2}*180,000

Multiply 1/2 and 3,600, or divide 3,600 by 2.

KE=90,000

Add appropriate units. Kinetic energy uses Joules, or J.

KE=90,000 Joules

The kinetic energy of the object is 90,000 Joules

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A train reaches a speed of 35.0 m/s after accelerating at a rate of 5.00 m/s2 over a distance of 40.0 m. What was the train’s in
MatroZZZ [7]

Answer:

Initial velocity, U = 28.73m/s

Explanation:

Given the following data;

Final velocity, V = 35m/s

Acceleration, a = 5m/s²

Distance, S = 40m

To find the initial velocity (U), we would use the third equation of motion.

V² = U² + 2aS

Where;

V represents the final velocity measured in meter per seconds.

U represents the initial velocity measured in meter per seconds.

a represents acceleration measured in meters per seconds square.

S represents the displacement measured in meters.

Substituting into the equation, we have;

35² = U + 2*5*40

1225 = U² + 400

U² = 1225 - 400

U² = 825

Taking the square root of both sides, we have;

Initial velocity, U = 28.73m/s

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3 years ago
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A certain type of laser emits light that has a frequency of 4.2 × 1014 Hz. The light, however, occurs as a series of short pulse
bogdanovich [222]

Explanation:

It is given that,

Frequency of the laser light, f=4.2\times 10^{14}\ Hz

Time, t=3.2\times 10^{-11}\ s

(a) Let \lambda is the wavelength of this light. It can be calculated as :

\lambda=\dfrac{c}{f}

\lambda=\dfrac{3\times 10^8}{4.2\times 10^{14}}

\lambda=7.14\times 10^{-7}\ m

or

\lambda=714\ nm

(b) Let n is the number of the wavelengths in one pulse. It can be calculated as :

n=f\times t

n=4.2\times 10^{14}\times 3.2\times 10^{-11}

n = 13440

Hence, this is the required solution.

8 0
2 years ago
Now let’s apply the work–energy theorem to a more complex, multistep problem. In a pile driver, a steel hammerhead with mass 200
andrew11 [14]

Answer:

a) v = 7.67

b) n = 81562 N

Explanation:

Given:-

- The mass of hammer-head, m = 200 kg

- The height at from which hammer head drops, s12 = 3.00 m

- The amount of distance the I-beam is hammered, s23 = 7.40 cm

- The resistive force by contact of hammer-head and I-beam, F = 60.0 N

Find:-

(a) the speed of the hammerhead just as it hits the I-beam and

(b) the average force the hammerhead exerts on the I-beam.

Solution:-

- We will consider the hammer head as our system and apply the conservation of energy principle because during the journey of hammer-head up till just before it hits the I-beam there are no external forces acting on the system:

                                   ΔK.E = ΔP.E

                                  K_2 - K_1 = P_1- P_2

Where,  K_2: Kinetic energy of hammer head as it hits the I-beam

             K_1: Initial kinetic energy of hammer head ( = 0 ) ... rest

             P_2: Gravitational potential energy of hammer head as it hits the I-beam. (Datum = 0)

             P_1: Initial gravitational potential energy of hammer head      

- The expression simplifies to:

                                K_2 = P_1

Where,                     0.5*m*v2^2 = m*g*s12

                                v2 = √(2*g*s12) = √(2*9.81*3)

                                v2 = 7.67 m/s

- For the complete journey we see that there are fictitious force due to contact between hammer-head and I-beam the system is no longer conserved. All the kinetic energy is used to drive the I-beam down by distance s23. We will apply work energy principle on the system:

                               Wnet = ( P_3 - P_1 ) + W_friction

                               Wnet = m*g*s13 + F*s23

                               n*s23 = m*g*s13 + F*s23

Where,    n: average force the hammerhead exerts on the I-beam.

               s13 = s12 + s23

Hence,

                             n = m*g*( s12/s23 + 1) + F

                             n = 200*9.81*(3/0.074 + 1) + 60

                             n = 81562 N

                               

                                                   

6 0
3 years ago
How does metamorphic rock change into another type of metamorphic rock?
Marrrta [24]
Metamorphic rocks are formed by tremendous heat, great pressure, and chemical reactions. To change it into another type of metamorphic rock you have to reheat it and bury it deeper again beneath the Earth's surface.

Hope this helped! :)
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