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ad-work [718]
3 years ago
15

Pls help ASAP

Physics
1 answer:
Zinaida [17]3 years ago
7 0

Answer:

The rocket will appear larger than it actually is

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Cloud of dust and gas!
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Which of the following statements must be true?
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your answer is B. The velocity could be in any direction, but the acceleration is in the direction of the resultant force

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A neutral boron atom has 6 neutrons and 5 electrons. How many protons does it have? Select one: a. 7 b. 6 c. 5 d. 4
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Just about everyone at one time or another has been burned by hot water or steam. This problem compares the heat input to your s
kkurt [141]

Answer:

B. Steam burns the skin worse than hot water because the latent heat of vaporization is released as well.

Explanation:

It is given that both steam and the boiling water when in contact with the skin cools down from 100 to 34 degrees Celsius.

For any substance of mass m, the heat required to change the temperature by \Delta T is mC\Delta T (S.I. unit = Joules).

where C, the specific heat capacity is the same and a constant for both the condensed steam and the boiling water.  

But, there is a "hidden" energy (heat) released by the steam called latent heat

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3 0
3 years ago
child slides down a snow‑covered slope on a sled. At the top of the slope, her mother gives her a push to start her off with a s
Strike441 [17]

Answer:

θ = 13.7º

Explanation:

  • According to the work-energy theorem, the change in the kinetic energy of the combined mass of the child and the sled, is equal to the total work done on the object by external forces.
  • The external forces capable to do work on the combination of child +sled, are the friction force (opposing to the displacement), and the component of the weight parallel to the slide.
  • As this last work is just equal to the change in the gravitational potential energy (with opposite sign) , we can write the following equation:

       \Delta K + \Delta U = W_{nc} (1)

  • ΔK, is the change in kinetic energy, as follows:

       \Delta K = \frac{1}{2}* m* (v_{f} ^{2}  - v_{0} ^{2}) (2)

  • ΔU, is the change in the gravitational potential energy.
  • If we choose as our zero reference level, the bottom of the slope, the change in gravitational potential energy will be as follows:

        \Delta U = 0 - m*g*h = -m*g*d* sin\theta (3)

  • Finally, the work done for non-conservative forces, is the work done by the friction force, along the slope, as follows:

        W_{nc} = F_{f} * d * cos 180\º \\\\  = 0.2*m*g*d* cos 180\º = -0.2*m*g*d (4)

  • Replacing (2), (3), and (4) in (1), simplifying common terms, and rearranging, we have:

      \frac{1}{2}* (v_{f} ^{2}  - v_{0} ^{2}) = g*d* sin\theta -0.2*g*d

  • Replacing by the givens and the knowns, we can solve for sin θ, as follows:              \frac{1}{2}*( (4.30 m/s) ^{2}  - (0.75 m/s)^{2}) = 9.8 m/s2*25.5m* sin\theta -0.2*9.8m/s2*25.5m\\ \\ 8.56 (m/s)2 = 250(m/s)2* sin \theta -50 (m/s)2\\ \\ sin \theta = \frac{58.6 (m/s)2}{250 (m/s)2}  = 0.236⇒ θ = sin⁻¹ (0.236) = 13.7º
8 0
3 years ago
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