900n pushes a wedge 0.10 m into a log, if the work done on the log is 50j what is the efficiency of the wedge

Two neutral metal spheres on wood stands are touching. A negatively charged rod is held directly above the top of the left spher

FrozenT [24]

Answer: Option (C) is the correct answer.

Explanation:

As we know that metals are able to conduct electricity so, when a negatively charges rod is kept closer to the left sphere then electrons will enter the sphere.

Since, like charges repel each other. Hence, some of the negative changes from the rod will repel the negative charges of left sphere.

As both left and right spheres are touching each other so, the electrons will move towards the right sphere. As a result, there will be too many electrons (negative charge) present on the right sphere and very less electrons present in the left sphere.

Thus, we can conclude that the statement right sphere is negatively charged, another is charged positively, is true.

7 0

3 years ago

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Which method can be used for separating a mixture of sand and salt?

Veronika [31]

Answers A and C can be automatically eliminated because evaporation has to do with the water cycle and magnetism has to do with electric current and such. Next we can eliminate B or distillation because it also has to do with water/liquids. Therefore, the answer is D or filtration because it is the only answer left.

I hope this helps!

6 0

3 years ago

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Astronomers discover an exoplanet, a planet obriting a star other than the Sun, that has an orbital period of 3.27 Earth years i

Naddik [55]

Answer:

r = 3.787 10¹¹ m

Explanation:

We can solve this exercise using Newton's second law, where force is the force of universal attraction and centripetal acceleration

F = ma

G m M / r² = m a

The centripetal acceleration is given by

a = v² / r

For the case of an orbit the speed circulates (velocity module is constant), let's use the relationship

v = d / t

The distance traveled Esla orbits, in a circle the distance is

d = 2 π r

Time in time to complete the orbit, called period

v = 2π r / T

Let's replace

G m M / r² = m a

G M / r² = (2π r / T)² / r

G M / r² = 4π² r / T²

G M T² = 4π² r3

r = ∛ (G M T² / 4π²)

Let's reduce the magnitudes to the SI system

T = 3.27 and (365 d / 1 y) (24 h / 1 day) (3600s / 1h)

T = 1.03 10⁸ s

Let's calculate

r = ∛[6.67 10⁻¹¹ 3.03 10³⁰ (1.03 10⁸) 2) / 4π²2]

r = ∛ (21.44 10³⁵ / 39.478)

r = ∛(0.0543087 10 36)

r = 0.3787 10¹² m

r = 3.787 10¹¹ m

7 0

3 years ago

You throw a ball from the balcony onto the court in the basketball arena. You release the ball at a height of 7.00 m above the c

Mariana [72]

Answer:

Your friend has to wait 0.26 s after you throw the ball to start running.

Explanation:

The equation that gives the position vector of the ball is as follows:

r = (x0 + v0 · t · cos α, y0 + v0 · t ·sin α + 1/2 · g · t²)

Where:

x0 = initial horizontal positon

v0 = initial velocity

t = time

α = throwing angle

y0 = initial vertical position

g = acceleration due to gravity

The equation of displacement of your friend is as follows:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of your friend at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

Please, see the attached figure for a description of the situation. Notice that the frame of reference is located at the throwing point.

Let´s find the time of flight of the ball. We know that at the final time, the y-component of the vector r has to be -6.00 m (1 m above the ground). Then:

y = y0 + v0 · t ·sin α + 1/2 · g · t²

-6.00 m = 0 m + 9.00 m/s · t · sin 33.0° - 1/2 · 9.8 m/s² · t²

0 = -4.9 m/s² · t² + 9.00 m/s · sin 33.0° · t + 6.00 m

Solving the quadratic equation:

t = 1.71 s

Now that we have the time of flight, we can calculate the x-component of the vector r (the horizontal distance traveled by the ball):

x= x0 + v0 · t · cos α

x = 0m + 9.00 m/s · 1.71 s · cos 33°

x = 12.9 m

Then, your friend will have to run (12.9 m - 11.0 m) 1.9 m to catch the ball 1 m above the ground.

Let´s see, how much time it takes your friend to run that distance:

x = x0 + v0 · t + 1/2 · a · t² (x0 = 0, v0 = 0)

x = 1/2 · a · t²

1.9 m = 1/2 · 1.80 m/s² · t²

Solving for t

t = 1.45 s

Then, since the time of flight of the ball is 1.71 s, your friend has to wait

1.71 s - 1.45 s = 0.26 s after you throw the ball to start running.

6 0

3 years ago

Tsunamis are fast-moving waves often generated by underwater earthquakes. In the deep ocean their amplitude is barely noticable,

Charra [1.4K]

To develop this problem it is necessary to apply the concepts related to the kinematic equations of motion. And from the speed found the relationships between wavelength, frequency and last of the period (which is inversely proportional to the frequency)

PART A) We know that the velocity of a body or a wave is equivalent to the distance traveled over a time interval. So,

$V = \frac{x}{t}$