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forsale [732]
3 years ago
8

A 3900 kg truck is moving at 6.0 m/s what is the kinetic energy

Physics
2 answers:
Stolb23 [73]3 years ago
3 0

Answer:

70200J

Explanation:

k.E = 1/2mv^2

K.E = 1/2(3900)(6)^2

finlep [7]3 years ago
3 0
Here you go, hope this helped

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What would happen if the earth's continents moved closer together
GarryVolchara [31]
Not much   The   Americas  are moving towards  Asia   due to continental drift.
The Atlantic is getting wider any 1/2" per year

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2 years ago
Find the acceleration of the system and the tension in the ropes for the system shown. The table mass is 30 kg and the hanging m
marusya05 [52]

The system's tension is 616 N and acceleration is 5.6 m / s^{2}

<u>Explanation:</u>

From newton’s second law of motion which state that net force acting on a body is product of mass of a body and acceleration of a body which is given as,

             F_{n e t}=m_{t o t} \times a

Where,

F_{n e t} is net force acting on body

m_{\mathrm{tot}} is mass of body

a is acceleration of body

Given values  

Table mass (m) = 30 kg

Hanging mass (m) = 40 kg

                a=\frac{F_{n e t}}{m_{\mathrm{tot}}}=\frac{m \times g}{m_{\mathrm{tot}}}

Put the value for m = hanging mass = 40 kg and g=9.8 \mathrm{m} / \mathrm{s}^{2}, we get

                  a=\frac{40 \times 9.8}{30+40}=\frac{392}{70}=5.6 \mathrm{m} / \mathrm{s}^{2}

The tension in the ropes,  T=(m \times g)+(m \times a)

Here, m as hanging mass

T = tension, N or  k g m / s^{2}

m = mass, kg  

g = gravitational force, 9.8 \mathrm{m} / \mathrm{s}^{2}

a = acceleration, m / s^{2}

          T = (40 \times 9.8)+(40 \times 5.6) = 392+224 = 616 N

3 0
3 years ago
under what conditions would a scale record a weight for an object which is equal to the magnitude of the gravitational force on
d1i1m1o1n [39]
In a non accelerated frame, also called inertial frame, when no additional forces are acting upon the body.

For instance, a scale in free fall will not record the weight, but zero, like the astronauts. But also, if someone is pulling the object or pushing it, the scale will record some other value.
4 0
3 years ago
Read 2 more answers
The 20-g bullet is travelling at 400 m/s when it becomes embedded in the 2-kg stationary block. The coefficient of kinetic frict
nikklg [1K]

Answer:

The distance the block will slide before it stops is 3.3343 m

Explanation:

Given;

mass of bullet, m₁ = 20-g = 0.02 kg

speed of the bullet, u₁ =  400 m/s

mass of block, m₂ = 2-kg

coefficient of kinetic friction,  μk = 0.24

Step 1:

Determine the speed of the bullet-block system:

From the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the speed of the bullet-block system after collision

(0.02 x 400) + (2 x 0) = v (0.02 + 2)

8 = v (2.02)

v = 8/2.02

v = 3.9604 m/s

Step 2:

Determine the time required for the bullet-block system to stop

Apply the principle of conservation momentum of the system

v(m_1+m_2) -F_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -N \mu_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -g(m_1 +m_2) \mu_kt = v_f(m_1 +m_2)\\\\3.9604(2.02)-9.8(2.02)0.24t = v_f(2.02)\\\\8 - 4.751t = 2.02v_f\\\\3.9604 - 2.352t = v_f

when the system stops, vf = 0

3.9604 -2.352t = 0

2.352t = 3.9604

t = 3.9604/2.352

t = 1.684 s

Thus, time required for the system to stop is 1.684 s

Finally, determine the distance the block will slide before it stops

From kinematic, distance is the product of speed and time

S = \int\limits {v} \, dt \\\\S = \int\limits^t_0 {(3.9604-2.352t)} \, dt\\\\ S = 3.9604t - 1.176t^2

Now, recall that t = 1.684 s

S = 3.9604(1.684) - 1.176(1.684)²

S = 6.6693 - 3.3350

S = 3.3343 m

Thus, the distance the block will slide before it stops is 3.3343 m

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3 years ago
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14) What is the velocity of a wave 5.35 x 10 cm long with a frequency of 16kHz
JulijaS [17]

Answer:

wavelength= 5.35×10cm=0.535m

f= 16000hz

v= wavelength × f= 0.535×16000=8560m/sec

6 0
3 years ago
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