The mass of Brachiosaurus moves with a momentum of 134,054 kgm/s and the velocity is 3.9 m/s, which is 34.37 kg.
The momentum is the product of mass and velocity. The momentum is the vector quantity and the unit of momentum is Kgm/s.
Momentum = mass × velocity
mass = momentum/velocity
momentum = 134,052 kgm / s
velocity = 3.9 m/s
mass = 134052 / 3.9
= 34.37 kg
Thus, the mass of Brachiosaurus is 34.37 kg.
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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 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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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.
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.
How can you determine the number of neutrons in an atom?
A. Mass number plus number of electrons
B. Atomic number minus mass number
C. Mass number minus atomic number
D. Atomic number plus mass number
Answer:
B. Atomic number minus mass number
Explanation:
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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can you please tell me where does 1-14 i really need help thanks :) god bless you all
The above has to do with the study of the earth's lithospheric plates. See the attached image and the explanation below.
What are the processes of the movement of lithospheric plates?The movement of lithospheric plates is a geological process that occurs due to the motion of hot, molten material in the Earth's mantle. The lithosphere, which is the rigid outer layer of the Earth's surface, is divided into several large plates that move relative to each other.
These movements are caused by the convection of material in the mantle and the forces that arise at the boundaries between the plates.
There are three main types of plate boundaries: divergent, convergent, and transform. Divergent boundaries occur where plates move apart from each other, creating new oceanic crust. Convergent boundaries arise where plates collide, leading to subduction, volcanic activity, and the formation of mountains. Transform boundaries occur where plates slide past each other.
The movement of lithospheric plates gives rise to various geological phenomena, such as earthquakes, volcanic activity, and the formation of mountain ranges and ocean basins.
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As a 5.00-kg sample of liquid mercury is cooled into a solid, it liberates 157 kJ of energy. What is the original temperature of the mercury? For mercury, the melting point is 234 K, the heat of fusion is 11.3 kJ/kg,
and the specific heat is 140 J/kg . K.
378 K
690 K
157 K
410 K
The original temperature of the mercury is 260.6K
Here is how to arrive at temperature of the mercuryTo solve this problem, we can use the formula for the heat released during the solidification of a substance:
Q = m * Lf
where Q is the heat released, m is the mass of the substance, and Lf is the heat of fusion of the substance.
In this case, Q = 157 kJ, m = 5.00 kg, and Lf = 11.3 kJ/kg.
We also need to use the formula for the heat absorbed or released during a temperature change:
Q = m * c * ΔT
where Q is the heat absorbed or released, m is the mass of the substance, c is the specific heat of the substance, and ΔT is the change in temperature.
We can use this formula to calculate the heat released as the mercury cools from its original temperature to its melting point, and then use the formula for solidification to calculate the heat released as the mercury solidifies.
Let T be the original temperature of the mercury.
The heat released as the mercury cools from its original temperature to its melting point is:
Q1 = m * c * (T - 234)
The heat released as the mercury solidifies is:
Q2 = m * Lf
The total heat released is:
Q = Q1 + Q2 = m * c * (T - 234) + m * Lf
Substituting the values given in the problem, we get:
157 kJ = 5.00 kg * 140 J/kg . K * (T - 234) + 5.00 kg * 11.3 kJ/kg
Simplifying and solving for T, we get:
T = 260.6 K
Therefore, the original temperature of the mercury was 260.6 K.
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