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77julia77 [94]
2 years ago
5

Which is a unique feature of Venus? A. Its surface is well below freezing. B. Its surface is hot enough to melt lead. C. It has

liquid water on the surface. D. It is heavily cratered.
Physics
2 answers:
Mkey [24]2 years ago
7 0
The correct answer is B. can you please help me answer my latest question too?
Sophie [7]2 years ago
6 0
It has a surface that is well below freezing.
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The waste product of photosynthesis is:
zalisa [80]

Answer:

The waste product of photosynthesis is oxygen

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2 years ago
A motorcycle accelerates uniformly from rest at 7.9\,\dfrac{\text{m}}{\text{s}^2}7.9 s 2 m ​ 7, point, 9, space, start fraction,
8090 [49]

Answer:

t = 3.516 s

Explanation:

The most useful kinematic formula would be the velocity of the motorcylce as a function of time, which is:

v(t) = v_0 +at

Where v_0 is the initial velocity and a is the acceleration. However the problem states that the motorcyle start at rest therefore v_0 = 0

If we want to know the time it takes to achieve that speed, we first need to convert units from km/h to m/s.

This can be done knowing that

1 km = 1000 m

1 h = 3600 s

Therefore

1 km/h = (1000/3600) m/s = 0.2777... m/s

100 km/h = 27.777... m/s

Now we are looking for the time t, for which v(t) = 27.77 m/s. That is:

27.777 m/s = 7.9 m/s^2 t

Solving for t

t = (27.7777 / 7.9) s = 3.516 s

6 0
3 years ago
a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it wi
Lelu [443]

a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 46.0 cm/s .The amplitude of the subsequent oscillations 48.13 cm/s

a 1.25 kilogram block is fastened to a spring with a 17.0 newtons per meter spring constant. Given that K is equal to 14 Newtons per meter and mass equals 10.5 kg. The block is then struck with a hammer by a student while it is at rest, giving it a speedo of 46.0 cm for a brief period of time. The required energy provided by the hammer, which is half mv squared, is transformed into potential energy as a result of the succeeding oscillations. This is because we know that energy is still available for consultation. So access the amplitude here from here. He will therefore be equal to and by. Consequently, the Newton's spring constant is 14 and the value is 10.5. The velocity multiplied by 0.49

Speed at X equals 0.35 into amplitude, or vice versa. At this point, the spirit will equal half of K X 1 squared plus half. Due to the fact that this is the overall energy, square is equivalent to half of a K square or an angry square. amplitude is 13 and half case 14 x one is 0.35. calculate that is equal to initial velocities of 49 squares and masses of 10.5. This will be divided in half and start at about 10.5 into the 49-square-minus-14. 13.42 into the entire square in 20.35. dividing by 10.5 and taking the square as a result. 231 6.9 Six centimeters per square second. 10.5 into 49 sq. 14. 2 into a 13.42 square entire. then subtract 10.5 from the result to get the square. So that is 48.13cm/s.

To learn more about oscillations Please click on the given link:

brainly.com/question/26146375

#SPJ4

This is incomplete question Complete Question is:

a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 46.0 cm/s . what are The amplitude of the subsequent oscillations?

4 0
10 months ago
Why do we need to put Airplane mode?
Serga [27]

Answer:

if we ever ride a airplane we dont mess up its signals and crash ,and its easier to ignore calls and texts

Explanation:

6 0
2 years ago
Read 2 more answers
The force of the added water produces a torque on the dam. In a simple model, if the torque due to the water were enough to caus
Fittoniya [83]

Answer:

The appropriate response is "\tau=\frac{1}{6} PgLh^3". A further explanation is described below.

Explanation:

The torque (\tau) produced by the force on the dam will be:

⇒  d \tau=XdF

On applying integration both sides, we get

⇒  \tau = \int_{0}^{a}x pgL(h-x)dx

⇒     = pgL\int_{0}^{h}(h-x)dx

⇒     =pgL[\frac{h^3}{2} -\frac{h^3}{3} ]

⇒     =\frac{1}{6} PgLh^3

8 0
2 years ago
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