The ball reaches the top of its trajectory after 1.25 s
Explanation:
The motion of the ball is a projectile motion, which consists of two independent motions:
- A uniform motion (at constant velocity) along the horizontal direction
- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction
To solve the problem we are only interested in the vertical motion of the ball. The vertical velocity of the ball at time t is given by
where
is the initial vertical velocity
is the acceleration of gravity
t is the time
The ball reaches its maximum heigth when its vertical velocity is zero:
Also, the initial vertical velocity is given by
where
u = 18 m/s is the initial speed
is the angle of projection
Solving the equation for t, we find the time at which the ball reaches the maximum height:
Learn more about projectile motion:
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Answer:</h3>
B. (PE)beginning = (KE)end
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Explanation:</h3>
<u>We are given;</u>
- Mass of an object is 72.0 kg
- Velocity of the body before it hits the ground is 79.0 m/s
We are required to determine the relationship between the potential energy and kinetic energy before the fall.
- When an object is at the highest point, it has maximum potential energy and minimum kinetic energy.
- This is because potential energy is directly proportional to the height of an object above the earth's surface.
- On the other hand, when an object attains the highest speed it has maximum kinetic energy and minimum kinetic energy.
In this case;
- The velocity of the object when hitting the ground is maximum and thus the object will have maxim,um kinetic energy.
- As the object falls towards the ground the potential energy is being converted to kinetic energy.
- Therefore, the potential energy at the beginning will be equal to the kinetic energy at the end when the object is on the ground.
- We can therefore, conclude that, (PE)beginning = (KE)end
Answer:
To identify the problem
Explanation:
The first step in the problem-solving process is to identify the problem. It is not as simple as it sounds as different people may have different ideas of what the "problem" is. Clearly stating the problem and getting everyone involved to agree is an important first step.
Answer:
See explanation
Explanation:
Notice that the condenser section includes both the hot water and space heater and station (3) is specified as being in the Quality region. Assume that 50°C is a reasonable maximum hot water temperature for home usage, thus at a high pressure of 1.6 MPa, the maximum power available for hot water heating will occur when the refrigerant at station (3) reaches the saturated liquid state. (Quick Quiz: justify this statement). Assume also that the refrigerant at station (4) reaches a subcooled liquid temperature of 20°C while heating the air.
Using the conditions shown on the diagram and assuming that station (3) is at the saturated liquid state
a) On the P-h diagram provided below carefully plot the five processes of the heat pump together with the following constant temperature lines: 50°C (hot water), 13°C (ground loop), and -10°C (outside air temperature)
b) Using the R134a property tables determine the enthalpies at all five stations and verify and indicate their values on the P-h diagram.
c) Determine the mass flow rate of the refrigerant R134a. [0.0127 kg/s]
d) Determine the power absorbed by the hot water heater [2.0 kW] and that absorbed by the space heater [0.72 kW].
e) Determine the time taken for 100 liters of water at an initial temperature of 20°C to reach the required hot water temperature of 50°C [105 minutes].
f) Determine the Coefficient of Performance of the hot water heater [COPHW = 4.0] (defined as the heat absorbed by the hot water divided by the work done on the compressor)
g) Determine the Coefficient of Performance of the heat pump [COPHP = 5.4] (defined as the total heat rejected by the refrigerant in the hot water and space heaters divided by the work done on the compressor)
h) What changes would be required of the system parameters if no geothermal water loop was used, and the evaporator was required to absorb its heat from the outside air at -10°C. Discuss the advantages of the geothermal heat pump system over other means of space and water heating
Answer:
T = 0.01 s
Explanation:
Given that,
The frequency of the beats of a hummingbird, f = 100 Hz
We need to find the period of the hummingbirds flaps. Let the time is t. We know that the relation between frequency and time period is given by :
T = 1/f
Put all the values,
T = 1/100 = 0.01 s
So, the time period of the humming bird is 0.01 s.
