The force needed to lift the 1800 kg car with the small piston is 2,649 N or approximately 270 kg (since 1 kg is equal to 9.81 N).
The hydraulic lift works based on Pascal's principle, which states that the pressure applied to a confined fluid is transmitted equally in all directions throughout the fluid.
Assuming there is no loss of energy due to friction or other factors, the force exerted on the small piston will be equal to the force exerted on the large piston. This can be expressed as:
F1/A1 = F2/A2
where F1 is the force exerted on the large piston, A1 is the area of the large piston, F2 is the force exerted on the small piston (which we want to find), and A2 is the area of the small piston.
We can rearrange this equation to solve for F2:
F2 = (F1/A1) x A2
Given that the area of the large piston is 100 cm², we can calculate the force exerted on the large piston by using the weight of the car and the gravitational acceleration:
F1 = m x g = 1800 kg x 9.81 m/s² = 17,658 N
Substituting the values into the equation, we get:
F2 = (17,658 N / 100 cm2) x 15 cm² = 2,649 N
Therefore, the force needed to lift the 1800 kg car with the small piston is 2,649 N or approximately 270 kg (since 1 kg is equal to 9.81 N).
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How loud in Decibels would a sound be with an intensity of 7.8x10^-4 W/m2? (write your answer to one decimal space)
A sound that is 7.8x10-4 W/m2 in intensity is equal to (10 dB)log3.2106 W/m21012 W/m2=185 dB.
How can you determine the relative volume of a sound?The decibel, often known as the db or 0.1 bel, is the standard measurement unit. Hence, b = 10 log10 (I/I0) can be used to express the relationship between relative intensities, or b, in decibels. This equation can be used to determine that one decibel equals a 26 percent intensity variations.
What does physics mean by relative intensity?The "decibel level" of a sound is a less formal term for relative intensity level. It is not the same as energy; relative intensity level reflects loudness more faithfully by using a logarithmic scale.
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A 0.80kg block of carbon (solid) is dropped into 1.4kg of water. If the carbon starts at -20C, the water starts at 92C, and they have equal final temperatures, what is the final temperature of the system?
The system's final temperature is roughly 16.7°C.
What is a system's final temperature?You may determine your substance's final heat by multiplying the temperature change by the initial temperature. Your water's final temperature would be 24 + 6, or 30 degrees Celsius, for instance, if it started off at 24 degrees Celsius.
The following is the formula for energy conservation:
Q1 + Q2 = 0
Q = mcΔT
Q1 + Q2 = 0
568.8
Simplifying and solving for
6394.4 - 106768 = 0
= 16.7°C
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If the sun were more massive, what would happen to Earth’s gravity with the sun?
A. decrease
B. would be infinite
C. would be 0
D. increase
Answer: d. increase
Explanation:
If the sun were more massive, the gravitational force between the sun and Earth would increase. This means that Earth's gravity with the sun would also increase. Therefore, the correct answer is (D) increase.
The gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. So, if the mass of one of the objects increases, the gravitational force between them will also increase. In this case, if the mass of the sun were to increase, the gravitational force between the sun and Earth would become stronger, and hence, Earth's gravity with the sun would also increase.
A rock climber stands on top of a 59 m -high cliff overhanging a pool of water. He throws two stones vertically downward 1.0 s apart and observes that they cause a single splash. The initial speed of the first stone was 1.7 m/s . Include value and units.
a) How long after the release of the first stone does the second stone hit the water?
b) What was the initial speed of the second stone?
c) What is the speed of the first stone as it hits the water?
d) What is the speed of the second stone as it hits the water?
a) The time after the release of the first stone that the second stone hits the water is 2.0 s.
b) 15.7 m/s is the initial speed of the second stone.
c) The speed of the first stone as it hits the water is 15.7 m/s.
d) The speed of the second stone as it hits the water is 28.2 m/s.
What is velocity?Velocity is a vector quantity that measures both the speed and direction of an object's motion. It is equal to the rate of change of an object's position with respect to time. Velocity is usually represented by the symbol v and is measured in meters per second (m/s).
a) The time between first and second stone's release is 1.0 s. Since the time of release of first stone and the time of splash of both stones are same, the time between the release of second stone and the splash of both stones is 1.0 s.
Thus, the time after the release of the first stone that the second stone hits the water is 2.0 s.
b) The initial speed of the second stone can be calculated using the equation of motion,
v² = u² + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity (9.8 m/s²), and s is the displacement.
Substituting the values,
v² = (1.7)² + 2(9.8) * 59
v = 15.7 m/s
c) The speed of the first stone as it hits the water can be calculated using the equation of motion,
v² = u² + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity (9.8 m/s²), and s is the displacement.
Substituting the values,
v² = (1.7)² + 2(9.8) * 59
v = 15.7 m/s
d) The speed of the second stone as it hits the water can be calculated using the equation of motion,
v² = u² + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity (9.8 m/s²), and s is the displacement.
Substituting the values,
v² = (15.7)² + 2(9.8) * 59
v = 28.2 m/s
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How long does it take for radiation from a cesuim-133 atom to complete 1.5 million cycles
A cesium-133 atom's radiation goes through 1.5 million cycles in around 0.1633 microseconds (or 163.3 nanoseconds).
What frequency does one kind of radiation that cesium-133 emits have?9,192,631,770 hertz (cycles per second) is the frequency of the microwave spectral line that the isotope cesium-133 emits. The basic unit of time is provided by this. Cesium clocks have an accuracy and stability of 1 second in 1.4 million years.
The radiation emitted by cesium-133 has a frequency of 9,192,631,770 cycles per second, or 9.192631770 109 Hz.
The following formula may be used to determine how long 1.5 million radiation cycles take to complete:
Time is equal to the frequency of cycles.
Plugging in the numbers, we get:
time = 1.5 million / 9.192631770 × 10^9 Hz
time = 1.632995101 × 10^-7 seconds
So it takes approximately 0.1633 microseconds (or 163.3 nanoseconds) for radiation from a cesium-133 atom to complete 1.5 million cycles.
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According to this graph, the acceleration
is approximately:
A. 12 m/s²
C. 4 m/s²
Velocity (m/s)
14
12
10
12 2 3 4
Time t (s)
B. 1.5 m/s2
D. 3 m/s2
Help please
Answer:
Explanation:
Because you have velocity along the y axis and time along the x axis, this is a velocity v time graph which is an acceleration graph. The slope of the line in this graph IS the acceleration. We can use 2 points and the slope formula to solve for the acceleration:
(0, 0) and (1, 3):
[tex]m=\frac{3-0}{1-0}=3[/tex] m/s squared, choice D.
A light ray passing through air strikes the surface of a glass block (n=1.5) and makes 30° angle of incidence. How many degrees will the light ray deviate from its original path after refraction?
The light ray will deviate from its original path with 19.5° after refraction.
How do we calculate?Applying Snell's law to calculate the angle of refraction:
n1 sin θ1 = n2 sin θ2
where n1 and θ1 = the refractive index and the angle of incidence in the first medium (air),
n2 and θ2 = the refractive index and the angle of refraction in the second medium (glass).
In this example,
n1 = 1.00 (refractive index of air), θ1 = 30°, and
n2 = 1.5 (refractive index of glass).
We then calculate for θ2:
n1 sin θ1 = n2 sin θ2
1.00 * sin 30° = 1.5 * sin θ2
0.5 = 1.5 * sin θ2
sin θ2 = 0.5 / 1.5 = 1/3
θ2 = sin^-1(1/3)
θ2 = 19.5°
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The attractive electric force between the point charges q and −2q has a magnitude of 2.2 N when the separation between the charges is 1.4 m . k=8.99×109N⋅m2/C2
What is the magnitude of charge q?
The electric force between two point charges is given by the equation
[tex]F=k*q_1*q_2/r^2[/tex]
What is force?The interaction between two things is measured by the physical quantity known as force. It is a vector quantity, and the sign F is frequently used to denote it. When an object interacts with another object, it feels a push or a pull.
where r is the distance between the charges, q1 and q2 are their magnitudes, and k is the Coulomb constant.
When we enter the problem's specified values, we obtain
[tex]2.2N=8.99*10^9\ N*m^2/C^2*q*-2q/(1.4 m)^2[/tex]
which simplifies to
q = -0.500 N/C.
Thus, the magnitude of charge q is 0.500 N/C.
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