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Hoochie [10]
3 years ago
15

A metal ball has a mass of 2.05 kg and a volume of 6.8 cm. What is its density? Remember

Physics
1 answer:
kogti [31]3 years ago
7 0

Answer:

the density is 0.301 :)

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Freeze wedging<span> is caused by the repeated freeze-thaw. </span>Frost wedging<span> occurs as the result of 9 % expansion of water when it is converted to ice. Cracks filled with water are forced further apart when it freezes. cycle.</span>
4 0
3 years ago
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A crane lifts an air conditioner to the top of a building. If the building is 12 m high, and the air conditioner has a mass of 2
andrey2020 [161]

Work needed = 23,520 J

<h3>Further explanation</h3>

Given

height = 12 m

mass = 200 kg

Required

work needed by the crane

Solution

Work is the transfer of energy caused by the force acting on a moving object  

Work is the product of force with the displacement of objects.  

Can be formulated  

W = F x d  

W = Work, J, Nm  

F = Force, N  

d = distance, m  

F = m x g

Input the value :

W = mgd

W = 200 kg x 9.8 m/s²x12 m

W = 23520 J

5 0
3 years ago
If a particle's position is given by x=4-12t+3t^2, where t is in seconds and x is in meters, what is its velocity at t=1 second?
andreyandreev [35.5K]

Answer:

v = -6m/s

Explanation:

x=4-12t+3t^2

\frac{dx}{dt}=-12+6t

For t = 1:

\frac{dx}{dt}=-6

3 0
3 years ago
How is momentum conserved in a Newton's cradle when one steel ball hits the other?
lora16 [44]

Answer:Both balls have momentum to start,and they share it after collision

Explanation:both balls have momentum to start and they share it after collision

7 0
3 years ago
Read 2 more answers
a ball rolls horizontally of the edge of the cliff at 4 m/s, if the ball lands at a distance of 30 m from the base of the vertic
algol13

Answer:

Approximately 281.25\; \rm m. (Assuming that the drag on this ball is negligible, and that g = 10\; \rm m \cdot s^{-2}.)

Explanation:

Assume that the drag (air friction) on this ball is negligible. Motion of this ball during the descent:

  • Horizontal: no acceleration, velocity is constant (at v(\text{horizontal}) is constant throughout the descent.)
  • Vertical: constant downward acceleration at g = 10\; \rm m \cdot s^{-2}, starting at 0\; \rm m \cdot s^{-1}.

The horizontal velocity of this ball is constant during the descent. The horizontal distance that the ball has travelled during the descent is also given: x(\text{horizontal}) = 30\; \rm m. Combine these two quantities to find the duration of this descent:

\begin{aligned}t &= \frac{x(\text{horizontal})}{v(\text{horizontal})} \\ &= \frac{30\; \rm m}{4\; \rm m \cdot s^{-1}} = 7.5\; \rm s\end{aligned}.

In other words, the ball in this question start at a vertical velocity of u = 0\; \rm m \cdot s^{-1}, accelerated downwards at g = 10\; \rm m \cdot s^{-2}, and reached the ground after t = 7.5\; \rm s.

Apply the SUVAT equation \displaystyle x(\text{vertical}) = -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t to find the vertical displacement of this ball.

\begin{aligned}& x(\text{vertical}) \\[0.5em] &= -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t\\[0.5em] &= - \frac{1}{2} \times 10\; \rm m \cdot s^{-2} \times (7.5\; \rm s)^{2} \\ & \quad \quad + 0\; \rm m \cdot s^{-1} \times 7.5\; s \\[0.5em] &= -281.25\; \rm m\end{aligned}.

In other words, the ball is 281.25\; \rm m below where it was before the descent (hence the negative sign in front of the number.) The height of this cliff would be 281.25\; \rm m\!.

5 0
3 years ago
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