A lunar eclipse can occur during a full moon. Please select the best answer from the choices provided T F

Answers

Answer 1
Answer:

Answer:

True

Explanation:

Lunar eclipse occurs when the Sun, the Earth and the Moon align in a straight line. The Earth blocks the sunlight falling on moon. In this alignment, the moon is in full phase. During solar eclipse, the moon passes through the shadow of Earth. Lunar eclipse occurs always during full moon phase-when the Earth comes between sun and moon.

Hence, the given statement is true.

Answer 2
Answer:

True. Lunar eclipses only happen when there is a full moon.


Related Questions

A forest fire sends carbon monoxide and ash into the air. This is an example of the__1__ affecting the__2___ .1. A. HydrosphereB LithosphereC. Atmosphere2 A LithosphereB AtmosphereC Biosphere
What best describes the bromide ion that forms​
Captain Kiddo is trapped inside a submarine. In order to escape, she first needs to open a heavy revolving door. a. Would be easier for her to apply all her force close to the axle of the door or as far away as possible from it?b. To finally get the hatch open she needs to grab hold of the wheel with her hands on opposite sides and rotate the wheel by exerting a force toward the top of the door on one side while exerting the same force towards the bottom of the door on the opposite side. Is there a net force applied on the wheel? Is there a net torque applied? Explain your answers
OFFERING 60 POINTS IF YOU CAN SHOW THE WORK!!!!A 1000 kg roller coaster begins on a 10 m tall hill with an initial velocity of 6m/s and travels down before traveling up a second hill. As the coaster moves from its initial height to its lowest position, 1700J of energy is transformed to thermal energy by friction.
If I am given a total capacitence of two capacitors, their capacitence togather is 22 F. What capacitence would the individual capacitors have if they are connected in parallel or connected in series.

When you walk at an average speed (constant speed, no acceleration) of 24 m/s in 94.1 secyou will cover a distance of__?

Answers

Answer:

2258.4 m

Explanation:

Distance covered is a product of speed and time hence

s=vt where s is the displacement/distance covered, v is the speed and t is the time taken

s=24*94.1=2258.4 m

Therefore, the distance covered is 2258.4 m

Suppose that a solid ball, a solid disk, and a hoop all have the same mass and the same radius. Each object is set rolling without slipping up an incline with the same initial linear (translational) speed. Which object goes farthest up the incline?

Answers

Final answer:

Given the same initial linear speed, a solid ball, solid disk, and hoop will expend energy on both rotation and translation. The solid ball, having the lowest moment of inertia, uses the most energy for translation and, therefore, will travel the highest up an incline.

Explanation:

In the context of this problem related to physics, the deciding factor is the distribution of mass, which influences each object's moment of inertia. Objects set to roll tend to use energy in two ways: translation (moving along the incline) and rotation (spinning about their center). Moment of inertia essentially measures how much of the object's energy goes towards rotation.

For a solid ball, solid disk, and hoop with the same mass and radius, the hoop has the highest moment of inertia with all of its mass at the maximum distance from the center. Followed by the solid disk, with its mass spread evenly from the center to its edge. Lastly, the solid ball has the lowest moment of inertia as its mass is concentrated towards the center.

This means that, given the same initial linear speed, the hoop will expend most of its energy on rotation rather than moving up the incline (translation). The solid disk will have a more balanced distribution between translation and rotation, and finally, the solid ball will use the least amount of energy on rotation and the most on translation. As such, the solid ball will go the farthest up the incline.

Learn more about the Physics of Rotating Objects here:

brainly.com/question/11991361

#SPJ3

Two bodies, one hot and the other cold kept in vacuum.what will happen to the tempreture of bodies after some time.

Answers

Hot body will lose heat from it, and that heat will goes out from it through radiation, so it's temperature will decrease after some time.

In same manner, cold body will take the heat, and it's temperature will increase

Hope this helps!

g During a collision with a wall, the velocity of a 0.4 KgKg ball changes from 25 m/sm/s towards the vall to 12 m/sm/s away from the wall. If the time the ball was in contact with the call was 0.5 secsec , what was the magnitude of the avarage force applied to the ball

Answers

Answer:

F = 10.8N

Explanation:

Given the mass m = 0.4kg, v1 = 25m/s, v2 = 12m/s and t =0.5s

From Newtown's second law of motion the average force can be found. This law states that the product of the force experienced by a body and the time t of the force acting on the body is equal to the change in momentum of the body. Mathematically it can be stated as follows

F×t = m(v2 – v1)

F = m(v2 – v1)/t = 0.4(25 – 12)/0.5 = 10.8N

A 0.47 kg block of wood hangs from the ceiling by a string, and a 0.070kg wad of putty is thrown straight upward, striking the bottom of the block with a speed of 5.60 m/s. The wad of putty sticks to the block. (Answer on previous exams) How high does the putty-block system rise above the original position of the block Is the kinetic energy of the system conserved during the collision Is the mechanical energy of the system conserved during the collision Is the mechanical energy conserved after the collision

Answers

Answer:

The height is  h =0.0269 \ m

The kinetic energy during collision is not conserved

The Mechanical energy during the collision is not conserved

The  mechanical energy after the collision is not conserved

Explanation:

From the question we are told that

    The mass of the block is  m_b = 0.47\ kg

      The mass of the wad of putty is  m_p =   0.070 \ kg

      The speed o the wad of putty is  v_p = 5.60 \ m/s

 

The law of momentum conservation can be mathematically represented as

          p_i = p_f

Where p_i is the initial momentum which is mathematically represented as

            p_i =m_p * v_p

While  p_f is the initial momentum which is mathematically represented as

            p_f = (m_b + m_p)v_f

Where  v_f s the final velocity

       So

             m_p v_p = (m_p + m_b) * v_f

Making  v_f the subject

               v_f = (m_p v_p)/(m_b +m_p)

substituting values

               v_f = ((0.070)*(5.60))/(0.47  + 0.070)

                v_f = 0.726 \ m/s

According to the law of energy conservation

       KE = PE

Where KE is the kinetic energy of the system which is mathematically represented as

           KE =  (1)/(2)  (m_p + m_b)v_f^2

And PE is the potential energy of the system which is mathematically represented as

             PE = (m_p +m_b) gh

So

         (1)/(2)  (m_p + m_b)v_f^2 = (m_p +m_b) gh

Making h the subject of the formula

         h = (v_f^2)/(2g)

substituting values

         h = ((0.726 )^2 )/(2 * 9.8)

         h =0.0269 \ m

Now the kinetic energy is conserved during collision because the system change it height during which implies some of the kinetic energy was converted to potential energy during collision

The the mechanical energy of the system during the collision  is conserved because this energy consists of the kinetic and the potential energy.

Now after the collision the mechanical energy is not conserved because the external force like air resistance has reduced the mechanical energy of that system

Hot air enters a rectangular duct (20cm wide, 25cm high, and 5m long) at 100 kPa and 60 degrees C at an average velocity of 5 m/s. While air flows the duct, it gets cool down (loses energy) so that air leave the duct at 54 degrees C. Determine the rate of heat loss from the air under steady condition

Answers

Answer:

1.57 kW

Explanation:

The rate of heat loss is given by:

q = Gm * Cp * (tfin - ti)

Where

q: rate of heat loss

Gm: mass flow

Cp: specific heat at constant pressure

The Cp of air is:

Cp = 1 kJ/(kg*K)

The mass flow is the volumetric flow divided by the specific volume

Gm = Gv / v

The volumetric flow is the air speed multiplied by the cruss section of the duct.

Gv = s * h * w (I name speed s because I have already used v)

The specific volume is obtained from the gas state equation:

p * v = R * T

60 C is 333 K

The gas constant for air is 287 J/(kg*K)

Then:

v = (R * T)/p

v = (287 * 333) / 100000 = 0.955 m^3/kg

Then, the mass flow is

Gm = s * h * w / v

And rthe heat loss is of:

q = s * h * w * Cp * (tfin - ti) / v

q = 5 * 0.25 * 0.2 * 1 * (54 - 60) / 0.955 = -1.57 kW (negative because it is a loss)