Explain the function of power supply, readout, peripheral, microcomputer, transducer and processor
The function of the power supply is to provide electrical energy to the device or system that needs it. The power supply converts the incoming voltage from the power source into a form that is usable by the device, such as DC voltage.
The readout is a device or component that displays data or information to the user. The readout could be a simple LED display or a complex graphical display.
A peripheral is a device or component that connects to a computer or other electronic device to provide additional functionality. Examples of peripherals include printers, scanners, and external hard drives.
A microcomputer is a type of computer that is designed to fit on a single microchip. Microcomputers are found in a wide range of devices, including smart phones, tablets, and embedded systems.
A transducer is a device that converts one form of energy to another. In electronics, transducers are commonly used to convert electrical energy into mechanical energy, or vice versa.
The processor is the central component of a computer or electronic device. The processor is responsible for executing instructions and controlling the other components of the system. The performance and capabilities of a device are largely determined by the speed and power of the processor.
Which statement best describes the refraction of light as it moves from air to glass?
A. Light bends due to the difference in the speed of light in air and glass.
B. Although the light bends, its speed remains the same as before.
C. Although the light changes speed, it continues in the same direction as before.
D. Light undergoes diffraction due to the difference in the speed of light in air and glass.
One strategy in a snowball fight is to throw a snowball at a high angle over level ground. Then, while your opponent is watching that snowball, you throw a second one at a low angle timed to arrive before or at the same time as the first one. Assume both snowballs are thrown with a speed of 26.5 m/s. The first one is thrown at an angle of 58.0° with respect to the horizontal. Find a - At what angle should the second snowball be thrown to arrive at the same point as the first?, find b - How many seconds later should the second snowball be thrown after the first in order for both to arrive at the same time?
The second snowball should be thrown at an angle of approximately 48.196° with respect to the horizontal to arrive at the same point as the first snowball.
the second snowball should be thrown 4.582 seconds later in order for both to arrive at the same time.
To find the angle at which the second snowball should be thrown, we can use the fact that the horizontal displacement of both snowballs must be the same.
Let's first find the horizontal and vertical components of the velocity for the first snowball. The initial speed is 26.5 m/s, and the angle is 58.0° with respect to the horizontal.
The horizontal component of the velocity for the first snowball is given by:
V1x = V1 * cos(angle1)
= 26.5 m/s * cos(58.0°)
= 26.5 m/s * 0.530
= 14.045 m/s
Now, let's find the vertical component of the velocity for the first snowball:
V1y = V1 * sin(angle1)
= 26.5 m/s * sin(58.0°)
= 26.5 m/s * 0.848
= 22.472 m/s
Since the vertical acceleration is the same for both snowballs (gravity), the time it takes for both to arrive at the same point is the same. Therefore, we can use the time of flight of the first snowball to calculate the vertical displacement for the second snowball.
The time of flight for the first snowball can be calculated using the vertical component of velocity and the acceleration due to gravity:
t = (2 * V1y) / g
= (2 * 22.472 m/s) / 9.8 m/s²
≈ 4.582 s
Now, let's find the vertical displacement for the second snowball:
Δy = V1y * t - (0.5 * g * t²)
= 22.472 m/s * 4.582 s - (0.5 * 9.8 m/s² * (4.582 s)²)
≈ 103.049 m
To find the angle at which the second snowball should be thrown, we can use the horizontal displacement and the vertical displacement:
tan(angle2) = Δy / Δx
= 103.049 m / (2 * 14.045 m/s * t)
= 103.049 m / (2 * 14.045 m/s * 4.582 s)
≈ 1.085
Now, we can find the angle2 by taking the arctan of both sides:
angle2 ≈ arctan(1.085)
angle2 ≈ 48.196°
Therefore,
To find how many seconds later the second snowball should be thrown, we can simply use the time of flight of the first snowball, which is approximately 4.582 seconds.
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A ball is thrown vertically upward with a speed of 15.0 m/s. Find a - How high does it rise? in meters, find b - How long does it take to reach its highest point? in seconds, find c - How long does the ball take to hit the ground after it reaches its highest point? in seconds, find d - What is its velocity when it returns to the level from which it started? in m/s.
Given that the initial velocity at which the ball is thrown vertically upward is 15m/s. Let us also assume that the value of acceleration due to gravity (g) = 9.8m/s² and in this case, the value will be -9.8m/s² as the ball is moving against gravity.
a) To calculate how high the ball rises, we can use the kinematic equation:
v² = u² + 2gs......(i)
where v ⇒ final velocity
u ⇒ initial velocity
g ⇒ acceleration and,
s ⇒ displacement (the height)
The final velocity will be 0 when the ball reaches its maximum height.
Substituting the values in equation (i), we get
0² = 15² + (2*-9.8*s)
0 = 225 - 19.6s
Thus, s = 225/19.6 = 11.48 m.
Therefore, the ball rises approximately 11.48 meters.
b) To find the time taken to reach the highest point, we can use the kinematic equation,
v = u + gt......(ii)
where t = time
Substituting the values in equation (ii)
0 = 15 - 9.8*t
t = -15/ -9.8 = 1.53 seconds
Thus, the time taken to reach the highest point = 1.53 seconds.
c) To find the time taken for the ball to hit the ground after it reaches its highest point, we can use the equation,
s = ut +1/2gt².....(iii)
As the ball is moving downwards, the initial velocity, u will be 0m/s.
Thus, substituting the values in equation (iii), we get
11.48 = 0*t + 1/2*9.8*t²
11.48 = 4.9t²
t² = 2.34
Therefore t = 1.53 seconds
Thus, the time taken for the ball to hit the ground is 1.53 seconds.
d) To find the velocity at which the ball returns to the level from which it started, we can use the equation
v = u+ gt.....(iv)
v = 0 + 9.8*1.53
Thus, v = 14.99 ≅ 15 m/s
Therefore, the velocity when it returns to the level from which it started is 15m/s.
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