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Oliga [24]
3 years ago
14

Learning Task 2: Prepare a basin with half-filled water and stone. Drop a stone

Physics
1 answer:
ELEN [110]3 years ago
6 0

Answer:

1 . What happens when you drop the stone?

Depending on the weight from which the stone was dropped, the glass might well break

2 depending on the size and weight and shape on the stone the glass might well break

3 depending on the density on the stone the stone might when float on the water

Explanition :

GIVE ME BRAINLESS PLEASE !!

You might be interested in
The Hubble Space Telescope is stabilized to within an angle of about 2-millionths of a degree by means of a series of gyroscopes
Likurg_2 [28]

Answer:

The answer to the question is;

The required torque that it would take to cause the gyroscopes to precess through an angle of 1.0×10−6 degree during a 5.0-hour exposure of a galaxy is 2.44 ×10⁻¹² N·m  

Explanation:

To solve the question we first resolve the units of the given quantities as follows

The gyroscopes spin at 19,200 rpm that is 19,200 revolutions per minute

1 revolution = 2π rad and

1 minute = 60 seconds

Therefore 19,200 revolutions per minute = 2π×19,200÷60 rad/s

= 2010.619 rad/s

The angle of precess is given as 1.0×10⁻⁶ °. We convert the angle to radians as follows

360 ° = 2π radians

1 ° = \frac{\pi }{180} radians and

1.0×10⁻⁶ ° =  \frac{\pi }{180} radians × 1.0×10⁻⁶ ° = 1.745×10⁻⁸ rad

To find the torque we note that the torque is given by

Precession angular speed × The moment of inertia × angular velocity

The precession angular speed is given by \frac{Precession. Angle}{time}

The precession angle was determined in rad as 1.745×10⁻⁸ rad

The precession time is 5 hours which is equal to 5×60×60 = 18000 s

Therefore the precession velocity = \frac{1.745*10^{-8} rad}{18000 s} =  9.696×10⁻¹³ rad/s

The moment of inertia is given by

Formula for the moment of inertia of a thin walled cylinder I = m·r²

Where:

r = Radius of the gyroscope = Diameter/2 = 5.0 cm/2 = 2.5 cm = ‪0.025‬ m

m = Mass of each gyroscopes = 2.0 kg

Therefore I = m·r² = 2.0 kg × (0.025‬ m)² = 0.00125 kg·m²

Torque, τ = Ω·I·ω

Where:

Ω = Precession velocity = 9.696×10⁻¹³ rad/s

I = Moment of inertia = 0.00125 kg·m²

ω = Angular speed = 2010.619 rad/s

τ = 9.696×10⁻¹³ rad/s × 0.00125 kg·m² × 2010.619 rad/s =

2.44 ×10⁻¹² kg·m²/ s² =   2.44 ×10⁻¹² N·m  

The required torque is 2.44 ×10⁻¹² N·m.

7 0
3 years ago
A toy car has an initial acceleration of 2 m/s" across a horizontal surface after it is released from rest. After the car travel
Ganezh [65]

Answer:

Closed system, because the speed of the car is as expected in the case where an object has uniform acceleration for a time t

Explanation:

Here in the question it is mentioned that a toy car has an initial acceleration of 2m/s²  across a horizontal surface so we can say that it is acted upon by an external force

Assuming that the acceleration is constant and the reason for this assumption is there at the last

The major difference between an open system and closed system is in case of open system there will be transfer of matter and in case of closed system there will be no change in matter of the system

If acceleration is constant in case of closed system we can expect the speed of the car after a time t by using the formula

  s = u×t + 0·5×a×t²

where s is the distance travelled

t is the time taken to travel that distance

u is the initial velocity

a is the acceleration of that system

But in case of open system as there will be a change of mass there will be a change in velocity of the system so in this case we cannot expect the speed of the car after a time t

And if the acceleration is not constant then we cannot say that the toy car is an open system or closed system, that is why we are assuming that the acceleration of the toy car is constant

3 0
4 years ago
Read 2 more answers
Saturn moves in an orbit around the Sun with radius 10 AU. How many degrees does it move on the Celestial in one year? (Hint: Ca
Lana71 [14]

Answer:

B. About 12 degrees

Explanation:

The orbital period is calculated using the following expression:

T = 2π*(\sqrt{\frac{r^3}{Gm}})

Where r is the distance of the planet to the sun, G is the gravitational constant and m is the mass of the sun.

Now, we don't actually need to solve the values of the constants, since we now that the distance from the sun to Saturn is 10 times the distance from the sun to the earth. We now this because 1 AU is the distance from the earth to the sun.

Now, we divide the expression used to calculate the orbital period of Saturn by the expression used to calculate the orbital period of the earth. Notice that the constants will cancel and we will get the rate of orbital periods in terms of the distances to the sun:

\frac{Tsaturn}{Tearth} = \sqrt{\frac{rSaturn^3}{rEarth^3} } = \sqrt{10^3}}

Knowing that the orbital period of the earth is 1 year, the orbital period of Saturn will be \sqrt{10^3}} years, or 31.62 years.

We find the amount of degrees it moves in 1 year:

1year * \frac{360degrees}{31.62years} = 11.38 degrees

or about 12 degrees.

6 0
3 years ago
How much heat energy (in megajoules) is needed to convert 7 kg of ice at -9°C C to water at 0°C?
Molodets [167]

Answer:

hence option A is correct

Explanation:

heat required from -9°C to 0°C ice = mass × specific heat of ice ×change in temperature

heat required from -9°C to 0°C ice = 7×2100×9 =132300 J =0.1323 MJ

( HERE SPECIFIC HEAT OF ICE IS A CONSTANT VALUE OF 2100

J/(kg °C )

heat required from  0°C ice to 0°C water = mass× specific heat of fusion of ice

                                                             = 7×3.36×10^5

                                                              = 2.352 × 10^6 J

                                                              = 2.352 MJ

TOTAL HEAT ENERGY REQUIRED = 0.1323 MJ +2.352 MJ

                                                          = 2.4843 MJ

hence option A is correct

5 0
3 years ago
If the pendulum took longer to complete one oscillation, how would the graph change?
pickupchik [31]

We don't know what kind of graph it is.

For example, it might be a graph of the pendulum's distance from center,

angle from center, speed, acceleration, total distance swung since it was

started, mass, weight, temperature, etc.


If the graph shows the pendulum's distance from center, angle from center,

speed, or acceleration, then the graph will look like a wave, with the period

of the wave being the period of the pendulum's oscillation. If the pendulum

took longer to complete one oscillation, that means its PERIOD increased,

and the distance between the peaks of the graph would be longer.


If it was a graph of total distance the pendulum swung since it was started,

the graph wouldn't look like a wave, just a steadily rising wiggle line. If the

pendulum took longer to complete one oscillation, the wiggles in the line

would be farther apart, and the average slope of any large section of the

line would be less.


If it was a graph of the pendulum's mass, weight, temperature, cost, etc.,

then the graph would be a horizontal line, and nothing that might change

the period of oscillation would have any effect on the graph.

7 0
3 years ago
Read 2 more answers
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