Answer:
(C) because the elevator is not accelerating
Note F = M a = M g (the resultant force on the elevator is due to gravity)
or Fup = Fc the force exerted on the elevator by the cable
and Fdown = Fe the force exerted on the elevator by gravity
F = M a = Fup - Fdown = zero resultant force on elevator
tank contains 335 kg of water at a uniform temperature of 60oC. The tank is insulated and not heated; it neither loses nor gains heat through the walls of the tank. A valve is opened and water exits the tank at a rate of 0.5 kg/sec and a temperature of 60oC. After 10 seconds the valve is closed again . Using the assumption that water at zero degrees centigrade contains zero energy and considering only internal, how much energy left the tank through the valve during this 10 second period; report as kJ.
Answer:
Explanation:
Thermal energy or internal energy gain or loss = mass x specific heat x temperature
specific heat of water = 4.2 kJ / kg degree Celsius
mass of water lost in 10 second = rate of loss x time = .5 x 10 = 5 kg .
heat energy associated with lost water = 5 x 4.2 x ( 60 - 0 ) = 1260 kJ .
Heat energy lost = 1260 kJ .
3. The car's mass is 400 kg. It moves at a velocity of 20 m/s. Calculate the car's momentum. *
(10 Points)
0.05 kg.m/s
8000 kg.m/s
80,000 kg.m/s
20 kg.m/s
Answer:
momentum=mass×velocity
momentum =400kg×20m/s=8000kg.m/s
Suppose you are standing at the earth's geographic north magnetic pole, the place on the earth's surface that compasses point toward. You have a compass that is free to swivel in any direction. Which way does your compass point? Suppose you are standing at the earth's geographic north magnetic pole, the place on the earth's surface that compasses point toward. You have a compass that is free to swivel in any direction. Which way does your compass point? It would point up. It would point east. It would point down. It would point west.
Answer:
It would point up.
Explanation:
Since I am at the earth's geographic north magnetic pole, the place on the earth's surface that compasses point toward, the north pole of the compass would also point towards the earth's geographic north magnetic pole, since all other compasses point toward there.
Since the compass is free to swivel in any direction, the compass would point up, since it is at the earth's geographic north magnetic pole, the place on the earth's surface that compasses point toward.
So, the compass would point up.
A plane takes off at St. Louis, flies straight to Denver, and then returns the same way. The plane flies at the same speed with respect to the ground during the entire flight, and there are no head winds or tail winds. Since the earth revolves around its axis once a day, you might expect that the times for the outbound trip and the return trip differ, depending on whether the plane flies against the earth's rotation or with it. Is this expectation true or false
Answer:
In the Both time
Explanation:
A plane takes off at St.Louis, flies straight to Denver, and then returns the same way. The plane flies at the same speed with respect to the ground during ...
Depending on whether the plane flies against the earth's rotation or with it. Is this expectation is true statement.
What is Plane?Physical quantities such as work, temperature, and distance can all be completely represented in daily life by their magnitude. The laws of arithmetic can, however, be used to explain how these physical values relate to one another.
Motion in two dimensions is another name for motion in a plane. For instance, a projectile moving in a circle. The origin, along with the two coordinate axes X and Y, will serve as the reference point for the investigation of this kind of motion.
Therefore, Depending on whether the plane flies against the earth's rotation or with it. Is this expectation is true statement.
To learn more about plane, refer to the link:
https://brainly.com/question/1962726
#SPJ2
12. An organ pipe that is 1.75 m long and open at both ends produces sound of
frequency 303 Hz when resonating in its second overtone. What is the speed of
sound in the room?
295 m/s
328 m/s
354 m/s
389 m/s
401 m/s
Answer:
354 m/s
Explanation:
For the second overtune (Third harmonic) of an open pipe,
λ = 2L/3................................ Equation 1
Where L = Length of the open pipe, λ = Wave length.
Given: L = 1.75 m.
Substitute into equation 1
λ = 2(1.75)/3
λ = 1.17 m.
From the question,
V = λf.......................... Equation 2
V = speed of sound in the room, f = frequency
Given: f = 303 Hz.
Substitute into equation 2
V = 1.17(303)
V = 353.5
V ≈ 354 m/s
Hence the right answer is 354 m/s
An old mining tunnel disappears into a hillside. You would like to know how long the tunnel is, but it's too dangerous to go inside. Recalling your recent physics class, you decide to try setting up standing-wave resonances inside the tunnel. Using your subsonic amplifier and loudspeaker, you find resonances at 5.0 Hz and 6.4 Hz , and at no frequencies between these. It's rather chilly inside the tunnel, so you estimate the sound speed to be 333 m/s .
Answer:
L = 116.6 m
Explanation:
For this exercise we approximate the tunnel as a tube with one end open and the other closed, at the open end there is a belly and at the closed end a node, therefore the resonances occur at
λ = 4L 1st harmonic
λ = 4L / 3 third harmonic
λ = 4L / 5 fifth harmonic
General term
λ = 4L / n n = 1, 3, 5,... odd
n = (2n + 1) n are all integers
They indicate that two consecutive resonant frequencies were found, the speed of the wave is related to the wavelength and its frequency
v = λ f
λ = v / f
we substitute
[tex]\frac{v}{f} = \frac{4L}{n}[/tex]
L = [tex]n \frac{ v}{4f}[/tex]
for the first resonance n = n
L = (2n + 1) [tex]\frac{v}{4f_1}[/tex]
for the second resonance n = n + 1
L = (2n + 3) [tex]\frac{v}{4f_2}[/tex]
we have two equations with two unknowns, let's solve by equating
(2n + 1) \frac{v}{4f_1}= (2n + 3) \frac{v}{4f_2}
(2n + 1) f₂ = (2n +3) f₁
2n + 1 = (2n + 3) [tex]\frac{f_1}{f_2}[/tex]
2n (1 - \frac{f_1}{f_2}) = 3 \frac{f_1}{f_2} -1
we substitute the values
2n (1- [tex]\frac{5}{6.4}[/tex]) = 3 [tex]\frac{5}{6.4}[/tex] -1
2n 0.21875 = 1.34375
n = 1.34375 / 2 0.21875
n = 3
remember that n must be an integer.
We use one of the equations to find the length of the Tunal
L = (2n + 1) \frac{v}{4f_1}
L = (2 3 + 1) [tex]\frac{333}{4 \ 5.0}[/tex]
L = 116.55 m
Which type of biological molecule would contain fats?
A) Amino Acids.
C) Nucleic Acids.
B) Lipids.
D) Carbohydrates.
B
Explanation:
lipids contains fat
hope it helps
Which molecules are in put in photosynthesis
Answer:
During the process of photosynthesis, cells use carbon dioxide and energy from the Sun to make sugar molecules and oxygen. These sugar molecules are the basis for more complex molecules made by the photosynthetic cell, such as glucose.
Explanation:
yes.
Answer:
Sunlight, Carbon Dioxide, and Water
Explanation:
Technically minerals are in there too but when I learned this it was just Sunlight, Carbon Dioxide, and Water
Fifty grams of ice at 0◦ C is placed in a thermos bottle containing one hundred grams of water
at 6◦ C. How many grams of ice will melt? The heat of fusion of water is 333 kJ/kg and the
specific heat is 4190 J/kg · K.Immersive Reader
Answer:
7.55 g
Explanation:
Given that:
Heat of fusion = 333kj/kg
Heat capacity, c = 4190 j/kg /k
The Number of grams of ice that will melt can be represented as y:
Number of grams of ice that will melt * heat of fusion = specific heat capacity * temperature change
y * 333 * 10^3 J = (4190) * (6 - 0)
333000y = 25140
y = 25140 / 333000
y = 0.0754954 kg
y = 0.0754954 * 100
y = 7.549 g
Hence, Number of grams of ice that will melt = 7.55 g
d. Two point charges, q1 = +25 nC and q2 = -75 nC, are separated by a distance of 3.0 cm. Find the magnitude and direction of; i. the electric force q1 exerts on q2 [5] ii. the force that q2 exerts on q1 [4] (take k = 9.0 x 109 N.m2 /C2 )
Answer:
a) F₂₁ = 0.02 N, attracting.
b) F₁₂ = 0.02 N, attracting.
Explanation:
a)
The magnitude of the force that q₁ exerts on q₂ (F₂₁) is given by Coulomb's Law, as follows:[tex]F_{21} = k * \frac{q_{1} *q_{2}}{r_{12}^{2} } = 9e9 N.m2/C2 * \frac{(25e-9C)*(75e-9C)}{(0.03m)^{2}} = 0.02 N (1)[/tex]
Since q₁ and q₂ have opposite signs, the force between them will be always attractive, i.e., from q₂ towards q₁, along the line that joins both charges.b)
The magnitude of the force on q₁ due to q₂ can be obtained applying Newton's 3rd Law, or using (1), because all parameters are the same, so F₁₂ (in magnitude) = F₂₁ = 0.02 NAs we have already said, it must be opposite to the one found in a) so it must go from q₁ towards q₂, it is an attracting force also.