Answer:
Products is the result. Reactants produce the result
Explanation:
,,
Describe how spatial awareness during physical activity impacts student performance
I uploaded the answer to a file hosting. Here's link:
tinyurl.com/wpazsebu
Spatial awareness may impact student performance (even physical performance) because spatial understanding is fundamental for visual processing.
Spatial awareness refers to the understanding of a person (in this case a student) and of the relationships between different objects within a specific space.It has been shown that spatial awareness can affect performance in mathematics, punctuation spelling, etc. An example of spatial awareness includes the visualization of different objects from different perspectives.In conclusion, spatial awareness may impact student performance (even physical performance) because spatial understanding is fundamental for visual processing.
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which causes magnets to stick to metal
Answer:
Steel
Explanation:
Steel is a metal that magnets stick to because iron can be found inside steel
Answer:Magnets stick to any metal that contains iron, cobalt or nickel.
Explanation:Iron is found in steel, so steel attracts a magnet and sticks to it. Stainless steel, however, does not attract a magnet.
A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surface of the water. It takes a time of 2.60 s for the boat to travel from its highest point to its lowest, a total distance of 0.700 m . The fisherman sees that the wave crests are spaced a horizontal distance of 6.00 m apart.
Required:
a. How fast are the waves traveling?
b. What is the amplitude of each wave?
c. If the total vertical distance traveled by the boat were 0.500 , but the other data remained the same, how fast are the waves traveling?
d. If the total vertical distance traveled by the boat were 0.500 , but the other data remained the same, what is the amplitude of each wave?
Answer:
a) 1.2 m/s
b) 0.350 m
c) 1.2 m/s
d) 0.250 m
Explanation:
a)
At any traveling wave, there exists a fixed relationship between the frequency, the speed and the wavelength, as follows:[tex]v = \lambda * f (1)[/tex]
We know that the frequency is equal to the number of cycles that the wave does in a second, so it's the inverse of the time needed to complete one cycle, that we call the period.In our case, since the wave completes one half cycle (from the highest point to the lowest one) in 2.60 s, this means that the time needed to complete a cycle, is just the double of it, i.e., 5.20 s .The frequency of the wave is just the inverse of this value:⇒ f = 1/T = 1/5.20 s = 0.2 1/s = 0.2 Hz (2)If the wave crests are spaced a horizontal distance of 6.0 m apart, this means that the wavelength λ is just 6.0 m.Replacing in (1) by (2) and the given λ, we can find the speed v, as follows:[tex]v = \lambda * f = 6.0 m * 0.2 1/s = 1.2 m/s (3)[/tex]
b)
By definition, the amplitude of the wave, is the absolute value of its highest value over its the zero reference level (in this case the surface of the water), so it is the half of the total vertical distance traveled by the boat:⇒ A = Δd/2 = 0.700m/2 = 0.350 m (4)
c)
Since the amplitude of the wave is independent of the frequency and the wavelength (that define the speed of the wave) the wave speed remains the same, i.e., 1.2 m/s.d)
If the total distance traveled by the boat were 0.500 m , the amplitude would be just half this value, as follows:⇒ A = Δd/2 = 0.500m/2 = 0.250 m (5)A 38.0 kg box initially at rest is pushed 4.00 m along a rough, horizontal floor with a constant applied horizontal force of 130 N. If the coefficient of friction between box and floor is 0.300, find the following. (a) the work done by the applied force J (b) the increase in internal energy in the box-floor system due to friction J (c) the work done by the normal force J (d) the work done by the gravitational force J (e) the change in kinetic energy of the box J (f) the final speed of the box m/s
Answer:
a) Wapp = 520 N
b) ΔUf = 447 N
c) Wn = 0
d) Wg = 0
e) ΔK = 73 J
f) v = 1.96 m/s
Explanation:
a)
Applying the definition of work, as the dot product between the applied force and the displacement, since both are parallel each other, the work done on the box by the applied force can be written as follows:[tex]W_{app} = F_{app} * \Delta X = 130 N * 4.0 m = 520 N (1)[/tex]
b)
The change in the internal energy due to the friction, is numerically equal to the work done by the force of friction.This work is just the product of the kinetic force of friction, times the displacement, times the cosine of the angle between them.As the friction force opposes to the displacement, we can find the work done by this force as follows:[tex]W_{ffr} = F_{fr} * \Delta X * cos (180) (2)[/tex]
The kinetic force of friction, can be expressed as the product of the kinetic coefficient of friction times the normal force.If the surface is level, this normal force is equal to the weight of the object, so we can write (2), as follows:[tex]W_{ffr} = F_{fr} * \Delta X * cos (180) = -\mu_{k} * m* g* \Delta X = \\ -0.300*38.0kg9.8 m/s2*4.0m = -447 J (3)[/tex]
So, the increase in the internal energy in the box-floor system due to the friction, is -Wffr = 447 Jc)
Since the normal force (by definition) is normal to the surface, and the displacement is parallel to the surface, no work is done by the normal force.d)
Since the surface is level, the displacement is parallel to it, and the gravitational force is always downward, we conclude that no work is done by the gravitational force either.e)
The work-energy theorem says that the net work done on the object, must be equal to the change in kinetic energy.We have two forces causing net work, the applied force, and the friction force.So the change in kinetic energy must be equal to the sum of the work done by both forces, that we found in a) and b).So, we can write the following expression:[tex]\Delta K = W_{app} + W_{ffr} = 520 J - 447 J = 73 J (4)[/tex]
f)
Since the object starts at rest, the change in kinetic energy that we got in e) is just the value of the final kinetic energy.So, replacing in (4), we finally get:[tex]\Delta K = 73 J = \frac{1}{2}*m*v^{2} (5)[/tex]
Solving for v:[tex]v_{f} = \sqrt{\frac{2*\Delta K}{m} } = \sqrt{\frac{2*73J}{38.0kg}} = 1.96 m/s (6)[/tex]a) The work done by the applied force [tex]W_{AP}=520\ J[/tex]
b) The change in the internal energy [tex]\Delta U=447\ J[/tex]
c) Work done by normal force [tex]W_n=0[/tex]
d) Work done by gravitation [tex]W_g=0[/tex]
e) The change in KE will be [tex]\Delta KE=73\ J[/tex]
f) The final speed v = 1.96 m/s
What will be the work done?The work done on any object can be defined as the force applied on the object and its displacement due the effect of the force.
If the object achieve movement due to the work then the energy in the object will be kinetic energy.
If the object attains some height against the gravity then the energy in the object will be potential energy.
Now it is given in the question that
The horizontal force [tex]F_h=130\N[/tex]
mass of the object m= 38 kg
Coefficient of friction [tex]\mu=0.3[/tex]
Displacement of the object [tex]\delta x=4\ m[/tex]
(a) The work done will be
[tex]W=F_h\tines \Delta x[/tex]
[tex]W=130\times 4=520\ J[/tex]
(b) The increase in the internal energy
The increase in the internal energy of the box is due to the energy generated by the force of friction so
[tex]W_f=F_f\times \Delta x\times Cos(180)[/tex]
here [tex]F_f[/tex] is the frictional force and is given as
[tex]\mu=\dfrac{F_f}{R}[/tex]
Here R is the normal reaction and its value will be weight of the box in opposite direction.
[tex]\mu=\dfrac{F_f}{-mg}[/tex]
[tex]F_f=-mg\times \mu[/tex]
[tex]W_f=F_f\times \Delta x\times Cos180=-mg\times\mu \times cos180[/tex]
[tex]W_f=-38\times 9.81\times 0.3\times4=-447\J\ J[/tex]
(c) The work done by the normal force will be zero because the displacement is horizontal against the normal work so the work done will be zero.
(d) The work done by the gravitational force will also be zero. Because the displacement is horizontal and the gravitational force acts downward.
(e) The change in the KE of the box.
The change in the KE of the box will be the net energy of the box so from the work energy theorem the net energy will be
[tex]\Delta KE =W_{AP}-W_f=520-447=73\ J[/tex]
(f) The speed of the box
The KE of the box will be
[tex]KE=\dfrac{1}{2} mv^2[/tex]
[tex]v=\sqrt{ \dfrac{2\times KE}{m}[/tex]
[tex]v=\sqrt{\dfrac{2\times73}{38} }=1.96\ \dfrac{m}{s}[/tex]
Thus
a) The work done by the applied force [tex]W_{AP}=520\ J[/tex]
b) The change in the internal energy [tex]\Delta U=447\ J[/tex]
c) Work done by normal force [tex]W_n=0[/tex]
d) Work done by gravitation [tex]W_g=0[/tex]
e) The change in KE will be [tex]\Delta KE=73\ J[/tex]
f) The final speed v = 1.96 m/s
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Mechanical energy is the most concentrated form of energy.
a. true
b. false
how can we minimise the use of plastic
Answer:
recyclable toothbrushes, cardboard or metal straws, cardboard gallons instead of plastic, use glass and reusable cups
Explanation:
Adding resistors in series changes the total resistance of a circuit by
(5 points)
O increasing the resistance
O decreasing the resistance
o it does not affect the resistance
o decreasing the resistance if the value of the resistor added is less than the
greatest resistor in the circuit
Answer:
Increasing the resistance
Explanation:
Answer: A
Explanation:
increasing the resistance
When6-2 He He-6 undergoes beta decay, the daughter is?
Answer: The daughter is named Susie.
Explanation: LIL SUSIE!!!
HUH? DIDN'T UNDERSTAND THE QUESTION!
HAVE A GREAT DAY!!!!!
Answer:6/3 Li
Explanation:
I’m not sure what the person under me is talking about but yeah
A 416 kg merry-go-round in the shape of a horizontal disk with a radius of 1.7 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. How large a torque would have to be exerted to bring the merry-go-round from rest to an angular speed of 3.7 rad/s in 2.9 s
Answer:
The torque exerted on the merry-go-round is 766.95 Nm
Explanation:
Given;
mass of the merry-go-round, m = 416 kg
radius of the disk, r = 1.7 m
angular speed of the merry-go-round, ω = 3.7 rad/s
time of motion, t = 2.9 s
The torque exerted on the merry-go-round is calculated as;
[tex]\tau = Fr= I\alpha\\\\\tau = (\frac{1}{2} m r^2)(\frac{\omega }{t} )\\\\\tau = (\frac{1}{2} \times 416 \times 1.7^2)( \frac{3.7}{2.9} )\\\\\tau = 766.95 \ Nm[/tex]
Therefore, the torque exerted on the merry-go-round is 766.95 Nm
Which of the following have frequencies greater than orange light Your answer:
radio waves
purple light
ultraviolet rays
red light
green light
gamma rays
microwaves
infrared rays
Answer:
Gamma Rays have the highest frequencies
Explanation:
This is because Gamma rays have the highest energies, the shortest wavelengths, and the highest frequencies compared to the other light frequency. Radio waves, on the other hand, have the lowest energies, longest wavelengths, and lowest frequencies of any type of EM radiation which means the answer has to be gamma rays. Brainly Please!!!! Here are screenshots that may help
Calculate the kinetic energy of an 80,000 kg airplane that is flying with a velocity of 167 m/s.
Answer:
1115560000 J
Explanation:
1/2 * 80,000 * 167^2 m/s = 1115560000 J
Car B is rounding the curve with a constant speed of 54 km/h, and car A is approaching car B in the intersection with a constant speed of 72 km/h. The x-y axes are attached to car B. The distance separating the two cars at the instant depicted is 40 m. Determine:
This question is incomplete, the complete question is;
Car B is rounding the curve with a constant speed of 54 km/h, and car A is approaching car B in the intersection with a constant speed of 72 km/h. The x-y axes are attached to car B. The distance separating the two cars at the instant depicted is 40 m. Determine: the angular velocity of Bxy rotating frame (ω).
Answer:
the angular velocity of Bxy rotating frame (ω) is 0.15 rad/s
Explanation:
Given the data in the question and image below and as illustrated in the second image;
distance S = 40 m
V[tex]_B[/tex] = 54 km/hr
V[tex]_A[/tex] = 72 km/hr
α = 100 m
now, angular velocity of Bxy will be;
ω[tex]_B[/tex] = V[tex]_B[/tex] / α
so, we substitute
ω[tex]_B[/tex] = ( 54 × 1000/3600) / 100
ω[tex]_B[/tex] = 15 / 100
ω[tex]_B[/tex] = 0.15 rad/s
Therefore, the angular velocity of Bxy rotating frame (ω) is 0.15 rad/s
Please help me!!!!!!
Answer:
Gamma ray will get to the Earth first due to his frequency of transmission
A reaction occurs when a compound breaks down. This reaction has one reactant and two or more products. Energy, as from a battery, is usually needed to break the compound apart.
Answer:
decomposition
Explanation:
A 41.0-kg crate, starting from rest, is pulled across level floor with a constant horizontal force of 135 N. For the first 15.0 m the floor is essentially frictionless, whereas for the next 12.0 m the coefficient of kinetic friction is 0.320. (a) Calculate the work done by all the forces acting on the crate, during the entire 27.0 m path. (b) Calculate the total work done by all the forces. (c) Calculate the final speed of the crate after being pulled these 27.0 m.
Answer:
Explanation:
From the information given;
mass of the crate m = 41 kg
constant horizontal force = 135 N
where;
[tex]s_1 = 15.0 \ m \\ \\ s_2 = 12.0 \ m[/tex]
coefficient of kinetic friction [tex]u_k[/tex] = 0.28
a)
To start with the work done by the applied force [tex](W_f)[/tex]
[tex]W_F = F\times (s_1 +s_2) \times cos(0) \ J[/tex]
[tex]W_F = 135 \times (12 +15) \times cos(0) \ J \\ \\ W_F = (135 \times 37 )J \\ \\ W_F =4995 \ J[/tex]
Work done by friction:
[tex]W_{ff} = -\mu\_k\times m \times g \times s_2 \\ \\ W_{ff} = -0.320 \times 41 \times 9.81 \times 12 \ J \\ \\ W_{ff} = -1544.49 \ J[/tex]
Work done by gravity:
[tex]W_g = mg \times (s_1+s_2) \times cos (90)} \ J \\ \\ W_g = 0 \ j[/tex]
Work done by normal force;
[tex]W_n = N \times (s_1 + s_2) \times cos (90) \ J[/tex]
[tex]W_n = 0 \ J[/tex]
b)
total work by all forces:
[tex]W = F \times (s_1 + s_2) + \mu_k \times m \times g \times s_2 \times 180 \\ \\ W = 135 \times (15+12) \ J - 0.320 \times 41 \times 9.81 \times 12[/tex]
W = 2100.5 J
c) By applying the work-energy theorem;
total work done = ΔK.E
[tex]W = \dfrac{1}{2}\times m \times (v^2 - u^2)[/tex]
[tex]2100.5 = 0.5 \times 41 \times v^2[/tex]
[tex]v^2 = \dfrac{2100.5}{ 0.5 \times 41 }[/tex]
[tex]v^2 = 102.46 \\ \\ v = \sqrt{102.46} \\ \\ \mathbf{v = 10.1 \ m/s}[/tex]
Find the momentum of a 15 kg object traveling at 7 m/s
What is the momentum
What is the velocity
What is the mass
What equation did you use to solve?
Find the momentum of a 15 kg object traveling at 7 m/s.
The momentum of an object is found by using the following formula:
[tex]\displaystyle p=mv[/tex]
P is the momentum and is measured in kg · m/sm is the mass and is measured in kgv is the velocity and is measured in m/sIn this question, the object is 15 kg and is travelling at 7 m/s. That means the mass is 15 kg and the velocity is 7 m/s.
Since all the needed variables are found, substitute it into the equation:
[tex]\displaystyle p=mv \rightarrow p=15 \times 7[/tex]
Multiply:
[tex]\displaystyle p=105\ kg \times m/s[/tex]
__________________________________________________________
What is the momentum? 105 kg · m/s
What is the velocity? 7 m/s
What is the mass? 15 kg
What equation did you use to solve? p = mv
__________________________________________________________
A motorist is driving at 15 m/s when she sees that a traffic light 315m ahead has just turned red. She knows that this light stay red for 25 s, and she wants to be 20 m from the ligt when it turns green again. Taht way, she will still be able to stop if the light stays red longer than expected. She applies the brake gradually such that her acceleration is ax(t)= c + bt, where c and b are constant. Assume she starts with a constant speed at the origin.
Find the values of c a b and any other unknown constants in order to answer the following questions.
1. Given the motorist's acceleration as a function of time, what are her position and velocity fucntions? - Do not use numbers for any constant here. Only derive the position and velocity functions.
2. What is her speed as she reaches the light?
Answer:
1) x = x₀ + vot - ½ c t² - 1/6 bt³, v = v₀ - ct - ½ b t²
2) v₁ = 5.25 m/s, v₂ = -8 m/s
Explanation:
1) For this exercise, the relationship of the body is not constant, so you must use the definition of speed and position to find them.
acceleration is
a = c + bt
a) the relationship between velocity and acceleration
a = [tex]\frac{dv}{dt}[/tex]
dv = -a dt
The negative sign is because the acceleration is contrary to the speed to stop the vehicle.
we integrate
∫ dv = - ∫ a dt
∫ dv = -∫ (c + bt) dt
v = -c t - ½ b t²
This must be valued from the lower limit, the velocity is vo, up to the upper limit, the velocity is v for time t
v - v₀ = -c (t-0) - ½ b (t²-0)
v = v₀ - ct - ½ b t²
b) the velocity of the body is
v = [tex]\frac{dx}{dt}[/tex]
dx = v dt
we replace and integrate
∫ dx = ∫ (v₀ - c t - ½ bt²) dt
x-x₀ = v₀ t - ½ c t² - ½ b ⅓ t³
Evaluations from the lower limit the body is at x₀ for t = 0 and the upper limit the body is x = x for t = t
x - x₀ = v₀ (t-0) - ½ c (t²-0) + [tex]\frac{1}{6}[/tex] (t³ -0)
x = x₀ + vot - ½ c t² - 1/6 bt³
2) The speed when you reach the traffic light
Let's write the data that indicates, the initial velocity is vo = 15 m / s, the initial position is xo = 315m, let's use the initial values to find the constants.
t = 25 s x = 20
we substitute
20 = 315 + 15 25 - ½ c 25² - 1/6 b 25³
0 = 295 + 375 - 312.5 c - 2604.16 b
670 = 312.5 c + 2604.16 b
we simplify
2.144 = c + 8.33 b
Now let's use the equation for velocity,
v = v₀ - ct - ½ b t²
v = 15 - c 25 - ½ b 25²
v = 15 - 25 c - 312.5 b
let's write our two equations
2.144 = c + 8.33 b
v = 15 - 25 c - 312.5 b
Let's examine our equations, we have two equations and three unknowns (b, c, v) for which the system cannot be solved without another equation, in the statement it is not clear, but the most common condition is that if the semaphore does not change, it follows with this acceleration (constant) to a stop
a = c + b 25
from the first equation
c = 8.33 / 2.144 b
C = 3.885 b
we substitute in the other two
v = 15 - 25 (3.885 b) - 312.5 b
v = 15 - 409.6 b
final acelearation
a = 28.885 b
let's use the cinematic equation
[tex]v_{f}^2[/tex]= v² - 2 a x
0 = v² - 2a 20
0 = v² - (28.885b) 40
v² = 1155.4 b
we write the system of equations
v = 15 - 409.6 b
v² = 1155.4 b
resolve
v²= 1155.4 ( [tex]\frac{15 -v }{409.6}[/tex] )
v² = 2.8 ( 15 -v)
v² + 2.8 v - 42.3 = 0
v= [ -2.8 ±[tex]\sqrt {2.8^2 + 4 \ 42.3) }[/tex] ]/2 = [-2.8 ± 13.3]/2
v₁ = 5.25 m/s
v₂ = -8 m/s
Steel beams are used for load bearing supports in a building. Each beam is 4.0 m long with a cross-sectional area of 7.5 10-3 m2 and supports a load of 4.5 104 N. Young's modulus for steel is 210 ✕ 109 N/m2.
A physicist drives through a stop light. When he is pulled over, he tells the police officer that the Doppler shift made the red light of wavelength 635 nm appear green to him, with a wavelength of 550 nm. The police officer writes out a traffic citation for speeding. How fast was the physicist traveling, according to his own testimony
Answer:
Explanation:
[tex]\lambda[/tex] = Observed wavelength = 550 nm
[tex]\lambda'[/tex] = Actual wavelength = 635 nm
c = Speed of light = [tex]3\times 10^8\ \text{m/s}[/tex]
v = Velocity of the physicist
Doppler shift is given by
[tex]f=\sqrt{\dfrac{c+v}{c-v}}f'\\\Rightarrow \dfrac{c}{\lambda}=\sqrt{\dfrac{c+v}{c-v}}\dfrac{c}{\lambda'}\\\Rightarrow \dfrac{\lambda'^2}{\lambda^2}=\dfrac{c+v}{c-v}\\\Rightarrow \dfrac{\lambda'^2}{\lambda^2}=\dfrac{1+\dfrac{v}{c}}{1-\dfrac{v}{c}}\\\Rightarrow \dfrac{\lambda'^2}{\lambda^2}(1-\dfrac{v}{c})=1+\dfrac{v}{c}\\\Rightarrow \dfrac{\lambda'^2}{\lambda^2}(1-\dfrac{v}{c})=1+\dfrac{v}{c}\\\Rightarrow \dfrac{\lambda'^2}{\lambda^2}-1=\dfrac{v}{c}(1+\dfrac{\lambda'^2}{\lambda^2})\\\Rightarrow v=\dfrac{c(\dfrac{\lambda'^2}{\lambda^2}-1)}{1+\dfrac{\lambda'^2}{\lambda^2}}[/tex]
[tex]\Rightarrow v=\dfrac{3\times 10^8\times (\dfrac{635^2}{550^2}-1)}{1+\dfrac{635^2}{550^2}}\\\Rightarrow v=42817669.77\ \text{m/s}[/tex]
The physicist was traveling at a velocity of [tex]42817669.77\ \text{m/s}[/tex].
(5 Points)
a) At ground level, the pressure of the helium in a balloon is 1x105
Pa. The volume occupied by the helium is 9.6m The balloon is
released and it rises quickly through the atmosphere. Calculate
the pressure of the helium when it occupies a volume of 12m3.
(3 Marks)
b) A box is 15m below the surface of the sea. The density of sea-
water is 1020 kg/m.
Calculate the pressure on the box due to the sea-water.
(2 Marks)
Answer:
1. [tex]P_{2}[/tex] = 8 x [tex]10^{4}[/tex] Pa
2. P = 1.5 x [tex]10^{5}[/tex] N/[tex]m^{2}[/tex]
Explanation:
1. From Boyles' law;
[tex]P_{1}[/tex][tex]V_{1}[/tex] = [tex]P_{2}[/tex][tex]V_{2}[/tex]
[tex]P_{1}[/tex] = 1 x [tex]10^{5}[/tex] Pa
[tex]V_{1}[/tex] = 9.6 [tex]m^{3}[/tex]
[tex]V_{2}[/tex] = 12 [tex]m^{3}[/tex]
Thus,
1 x [tex]10^{5}[/tex] x 9.6 = [tex]P_{2}[/tex] x 12
[tex]P_{2}[/tex] = [tex]\frac{100000 x 9.6}{12}[/tex]
= 80000
[tex]P_{2}[/tex] = 8 x [tex]10^{4}[/tex] Pa
2. Pressure, P = ρhg
where: ρ is the density of the fluid, h is the height/ depth and g is the acceleration due to gravity (9.8 m/[tex]s^{2}[/tex]).
Thus,
P = 1020 x 15 x 9.8
= 149940
P = 1.5 x [tex]10^{5}[/tex] N/[tex]m^{2}[/tex]
What type of weather would a continental Polar air mass bring
Answer:
Continental polar ( cp):
Explanation:
Cold and dry, originating from high latitudes, typically as air flowing out of the polar highs. This air mass often brings the rattleing cold, dry and clear weather on a perfect winter day and also dry and warm weather on a pleasant day in summer.
A cryogenic vacuum pump works by condensing vapors onto some absorbent medium. This is an efficient and clean way to pump a system in a research environment. The term cryo means cold, which indicates that these types of vacuum pumps contain a refrigerant cycle to cool the internal parts. The temperature difference between the inside and outside of a typical cryogenic pump is Δ=303 ∘C . Derive an expression to convert this difference into Fahrenheit and express the answer.
Answer: Temperature in Fahrenheit is 577.4
Explanation:
The conversion factor for converting celcius to Fahrenheit is:
[tex]F=\frac{9}{5}\times C+32[/tex]
where F = temperature in Fahrenheit
C = Temperature in Celcius
Given : Temperature difference in Celcius = [tex]303^0C[/tex]
Putting in the values we get:
[tex]F=\frac{9}{5}\times 303+32[/tex]
[tex]F=577.4[/tex]
Thus the answer in Fahrenheit is 577.4
The required conversion of temperature from Celsius to Fahrenheit is 577.4 Degree Fahrenheit.
Given data:
The temperature difference between the inside and outside of a typical cryogenic pump is, [tex]\Delta T=303^{\circ}\rm C[/tex].
The degree measure of the flow of heat from one point to another is known as the temperature at a specific point. The temperature measurements are done in Celsius, Fahrenheit and Kelvin scales. The expression to convert the Celsius scale to Fahrenheit scale is,
[tex]T=(C \times \dfrac{9}{5} + 32)\;\rm ^{\circ}F[/tex]
Solving as,
[tex]T=(303 \times \dfrac{9}{5}+32 )\;\rm ^{\circ}F\\\\T=(303 \times \dfrac{9}{5}+32 )\;\rm ^{\circ}F\\\\T=577.4\;\rm ^{\circ}F[/tex]
Thus, we can conclude that the required conversion of temperature from Celsius to Fahrenheit is 577.4 Degree Fahrenheit.
Learn more about the Temperature conversion here:
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A potter's wheel is rotating around a vertical axis through its center at a frequency of 2.0 rev/srev/s . The wheel can be considered a uniform disk of mass 4.5 kgkg and diameter 0.30 mm . The potter then throws a 2.8-kgkg chunk of clay, approximately shaped as a flat disk of radius 8.0 cmcm , onto the center of the rotating wheel. Part A What is the frequency of the wheel after the clay sticks to it
Answer:1.7 rev/s
Explanation:
Given
Frequency of wheel [tex]N_1=2\ rev/s[/tex]
angular speed [tex]\omega_1=2\pi N_1=4\pi\ rad/s[/tex]
mass of wheel [tex]m_1=4.5\ kg[/tex]
diameter of wheel [tex]d_1=0.30\ m=30\ cm[/tex]
radius of wheel [tex]r_1=\frac{d_1}{2}=\frac{30}{2}=15\ cm[/tex]
mass of clay [tex]m_2=2.8\ kg[/tex]
the radius of the chunk of clay [tex]r_2=8\ cm[/tex]
Moment of inertia of Wheel
[tex]I_1=\dfrac{m_1r_1^2}{2}=\dfrac{4.5\times 15^2}{2}\ kg-cm^2[/tex]
Combined moment of inertia of wheel and clay chunk
[tex]I_2=\dfrac{m_1r_1^2}{2}+\dfrac{m_2r_2^2}{2}=\dfrac{4.5\times 15^2}{2}+\dfrac{2.8\times 8^2}{2}\ kg-cm^2[/tex]
Conserving angular momentum
[tex]\Rightarrow I_1\omega_1=I_2\omega_2\\\Rightarrow \dfrac{4.5\times 15^2}{2}\cdot 4\pi=(\dfrac{4.5\times 15^2}{2}+\dfrac{2.8\times 8^2}{2})\omega_2\\\\\Rightarrow \omega _2=\dfrac{4\pi }{1+\dfrac{2.8}{4.5}\times (\dfrac{8}{15})^2}=\dfrac{4\pi}{1+0.1769}=0.849\times 4\pi[/tex]
Common frequency of wheel and chunk of clay is
[tex]\Rightarrow N_2=\dfrac{4\pi \times 0.849}{2\pi}=1.698\approx 1.7\ rev/s[/tex]
Which of the following best describes gravitational potential energy,
O The energy stored in an object when you lift it,
O The energy stored in an object due to its chemical structure,
The energy of an object in motion.
The energy stored in an object due to its electric charges.
a boy standing by a lake sees a fish in the pond and tries to thrust a spear into it he will success or not explain with reason
He will not probably success to thrust a spear into the fish in the pond because when light travels from water to air , it bends due to refractive property of light.
What is refraction of light?Refraction is the bending of a wave as it travels through different media. The two materials' different densities are what lead to the bending.
Refraction is defined as "the change in a wave's direction as it passes through a medium."
Although light refraction is one of the most frequently seen phenomena, refraction can also occur with sound and water waves. We can use optical tools like lenses, prisms, and magnifying glasses thanks to refraction. We can focus light on our retina because of the refraction of light, which is another benefit.
When light travels from water to air , it bends due to refractive property of light. So, he e will not probably success to thrust a spear into the fish.
Learn more about refraction here:
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NEED HELP ASAP, ILL GIVE YOU BRAINLIEST IF CORRECT (30POINTS)
Drag each label to the correct location on the image. Each label can be used more than once.
Identify the parts of the barred spiral galaxy.
SPIRAL ARM, NUCLEUS, BAR
NOTE I JUST FILLED IN THE SPOTS FOR YOU TO SEE, THEY ARE NOT CORRECT
Answer:
the bar is the top and bottem. the nucleas in the middle and the Spiral arm is the last space
Explanation:
Answer:
look pkch
Explanation:
ASAP Even though the force exerted on each object in a collision is the same strength, if the objects have different masses, their will be different. * O changes in velocity O amount of force O speed and direction
Answer:
it should be changes in velocity
Explanation:
I hope this helps!
Investigator Daniels is working on the scene of a suspected arson on a rainy evening. The site of the fire was a mechanic's garage, so there are plenty of accelerants
on the scene. According to the owner of the garage, there had been an argument with a customer the day before about his motorcycle. All the employees left by
6:00 pm. The office was unharmed but there is much damage in the mechanic's bay area and toolboxes. The flames look to have been most concentrated around a
bucket that the owner says was filled with oily rags. She is trying to sift through the evidence and find the relevant facts of the case so far. In order to effectively
Identify the relevant facts, Investigator Daniels creates a word bank. Which word would not fit into her word bank?
Argument
Rainy
Olly rags
Toolboxes
Answer:
Rainy
Explanation:
It is the only one that has nothing to do with the case
What are some possible factors that can be the X
and Y axis of a motion graph?
Answer:
x-Speed/velocity
y-time.
Explanation:
because Speed is a rate of change of distance while time how long it takes a a car to move to a specific point
Select the correct answer.
Which graph shows the correct relationship between kinetic energy and speed?
Answer:
c
Explanation:
It's c the last one u see