Answer:
biology
Explanation:
Animals and plants
In high air pressure the molecules are
A-Warm and moving fast
b-Close together and moving slowly
c-far apart and moving slowly
d-hot and moving rapidly
A ball is sitting at the top of a ramp. As the ball rolls down the ramp, the potential energy of the ball decreases, what happens to the potential energy as the ball moves
Answer:
the potential energy decreases as it is converted to kinetic energy.
Explanation:
As things move, their potential energy converts to kinetic energy to power them along. When a ball rolls down the top of a ramp, all the potential energy it accumulated at the top of the ramp converts to kinetic energy to help it roll down. In other words, its potential energy decreases as its kinetic energy increases.
A large box slides across a frictionless surface with a velocity of 12 m/s and a mass of 4
kg, collides with a smaller box with a mass of 2 kg that is stationary. The boxes stick
together. What is the velocity of the two combined masses after collision?
8 m/s
O m/s
12 m/s
4 m/s
us 12:18
According to Newton's first law, an object at rest will _____.
never move
stay at rest forever
start moving
stay at rest unless moved by force
How fast were both runners traveling after 4 seconds?
40
Distance (in yards)
30
20
10
1
2.
3
0
Time in seconds
Answer:
they were fast ⛷⛷
Define Mechanical advantage
fe effort of 2125N is used to lift a Lead of 500N
through a Verticle high of 2.N using a buly System
if the distance Moved by the effort is 45m
Calculate 1. Work done on the load
2. work done by the effort
3. Efficiency of the System
Answer:
1) 1000Nm
2) 95,625Nm
3) 1.05%
Explanation:
Mechanical Advantage is the ratio of the load to the effort applied to an object.
MA = Load/Effort
1) Workdone on the load = Force(Load) * distance covered by the load
Workdone on the load = 500N * 2m
Workdone on the load = 1000Nm
2) work done by the effort = Effort * distance moves d by effort
work done by the effort = 2125 * 45
work done by the effort = 95,625Nm
3) Efficiency = Workdone on the load/ work done by the effort * 100
Efficiency = 1000/95625 * 100
Efficiency = 1.05%
Hence the efficiency of the system is 1.05%
A long, uninsulated steam line with a diameter of 100 mm and a surface emissivity of 0.8 transports steam at 150°C and is exposed to atmospheric air and large surroundings at an equivalent temperature of 20°C. (a) Calculate the rate of heat loss per unit length for a calm day. (b) Calculate the rate of heat loss on a breezy day when the wind speed is 8
Answer:
Heat loss per unit length = 642.358 W/m
The heat loss per unit length on a breezy day during 8 m/s speed is = 1760.205 W/m
Explanation:
From the information given:
Diameter D [tex]= 100 mm = 0.1 m[/tex]
Surface emissivity ε = 0.8
Temperature of steam [tex]T_s[/tex] = 150° C = 423K
Atmospheric air temperature [tex]T_{\infty} = 20^0 \ C = 293 \ K[/tex]
Velocity of wind V = 8 m/s
To calculate average film temperature:
[tex]T_f = \dfrac{T_s+T_{\infty}}{2}[/tex]
[tex]T_f = \dfrac{423+293}{2}[/tex]
[tex]T_f = \dfrac{716}{2}[/tex]
[tex]T_f = 358 \ K[/tex]
To calculate volume expansion coefficient
[tex]\beta= \dfrac{1}{T_f} \\ \\ \beta= \dfrac{1}{358} \\ \\ \beta= 2.79 \times 10^{-3} \ K^{-1}[/tex]
From the table of "Thermophysical properties of gases at atmospheric pressure" relating to 358 K of average film temperature; the following data are obtained;
Kinematic viscosity (v) = 21.7984 × 10⁻⁶ m²/s
Thermal conductivity k = 30.608 × 10⁻³ W/m.K
Thermal diffusivity ∝ = 31.244 × 10⁻⁶ m²/s
Prandtl no. Pr = 0.698
Rayleigh No. for the steam line is determined as follows:
[tex]Ra_{D} = \dfrac{g \times \beta (T_s-T_{\infty}) \times D_b^3}{\alpha\times v}[/tex]
[tex]Ra_{D} = \dfrac{9.8 \times (2.79 *10^{-3})(150-20) \times (0.1)^3}{(31.244\times 10^{-6}) \times (21.7984\times 10^{-6})}[/tex]
[tex]Ra_{D} = 5.224 \times 10^6[/tex]
The average Nusselt number is:
[tex]Nu_D = \Big \{ 0.60 + \dfrac{0.387(Ra_D)^{1/6}}{[ 1+ (0.559/Pr)^{9/16}]^{8/27}} \Big \}^2[/tex]
[tex]Nu_D = \Big \{ 0.60 + \dfrac{0.387(5.224\times 10^6)^{1/6}}{[ 1+ (0.559/0.698)^{9/16}]^{8/27}} \Big \}^2[/tex]
[tex]Nu_D = \Big \{ 0.60 + \dfrac{5.0977}{[ 1.8826]^{8/27}}\Big \}^2[/tex]
[tex]Nu_D = \Big \{ 0.60 + 4.226 \Big \}^2[/tex]
[tex]Nu_D = 23.29[/tex]
However, for the heat transfer coefficient; we have:
[tex]h_D = \dfrac{Nu_D\times k}{D_b} \\ \\ h_D = \dfrac{(23.29) \times (30.608 \times 10^{-3} )}{0.1}[/tex]
[tex]h_D = 7.129 \ Wm^2 .K[/tex]
Hence, Stefan-Boltzmann constant [tex]\sigma = 5.67 \times 10^{-8} \ W/m^2.K^4[/tex]
Now;
To determine the heat loss using the formula:
[tex]q'_b = q'_{ev} + q'_{rad} \\ \\ q'_b = h_D (\pi D_o) (T_t-T_{\infty})+\varepsilon(\pi D_b)\sigma (T_t^4-T_{\infty }^4)[/tex]
[tex]q'_b = (7.129)(\pi*0.1) (423-293) + (0.8) (\pi*0.1) (5.67 *10^{-8}) (423^4-293^4) \\ \\ q'_b = 291.153 + 351.205 \\ \\ \mathbf{q'_b = 642.258 \ W/m}[/tex]
Now; here we need to determine the Reynold no and the average Nusselt number:
[tex]Re_D = \dfrac{VD_b}{v } \\ \\ Re_D = \dfrac{8 *0.1}{21.7984 \times 10^{-6}} \\ \\ Re_D = 3.6699 \times 10^4[/tex]
However, to determine the avg. Nusselt no by using Churchill-Bernstein correlation, we have;
[tex]Nu_D = 0.3 + \dfrac{0.62 \times Re_D^{1/2}* Pr^{1/3}}{[1+(0.4/Pr)^{2/3}]^{1/4}} [1+ (\dfrac{Re_D}{282000})^{5/8}]^{4/5}[/tex]
[tex]Nu_D = 0.3 + \dfrac{0.62 \times (3.6699*10^4)^{1/2}* (0.698)^{1/3}}{[1+(0.4/0.698)^{2/3}]^{1/4}} [1+ (\dfrac{3.669*10^4}{282000})^{5/8}]^{4/5}[/tex]
[tex]Nu_D = (0.3 +\dfrac{105.359}{1.140}\times 1.218) \\ \\ Nu_D = 112.86[/tex]
SO, the heat transfer coefficient for forced convection is determined as follows afterward:
[tex]h_D = \dfrac{Nu_{D}* k}{D_b} \\ \ h_D = \dfrac{112.86*30.608 *10^{-3}}{0.1} \\ \\ h_D = 34.5 \ W/m^2 .K[/tex]
Finally; The heat loss per unit length on a breezy day during 8 m/s speed is:
[tex]q'b = h_D (\pi D_b) (T_s-T_{\infty}) + \varepsilon (\pi D_b) \sigma (T_s^4-T_ {\infty}^4) \\ \\ q'b = (34.5) (\pi *0.1) (423-293) + (0.8) (\pi*0.1) (5.67*10^{-8}) (423^4 - 293^4) \\ \\ = 1409 +351.205 \\ \\ \mathbf{q'b = 1760.205 \ W/m}[/tex]
(will give brainliest to whoever is correct and shows reasoning) What is the acceleration of an object that has a velocity of 60m/s and is moving in a circle of radius 50m?
Answer:
5.0/s
Explanation:
Answer:
b and a it is this that abewsr
Calculate the momentum of a 10 kg bowling ball rolling at 2m/s towards north.
Answer:
momentum=mass x velocity= 10 x 2 = 20kgm/s
On a 10 kg cart (shown below), the cart is brought up to speed with 50N of force for 7m, horizontally. At this point (A), the cart begins to experience an average frictional force of 15N throughout the ride.
Find:
a) The total energy at (A)
b) The velocity at (B)
c) The velocity at (C)
d) Can the cart make it to Point (D)? Why or why not?
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].
A student's backpack has a mass of 9.6 kg. The student applies a force of 94.08 N [up] while walking through 1.4 km [E] to get to school. Calculate the work done by the student on the backpack
The student does zero work on the backpack because the upward force applied by the student is acting perpendicular to the backpack's displacement parallel to the ground.
Water enter the horizontal, circular cross-sectional, sudden-contraction nozzle sketched below at section (1) with a uniformly distributed velocity of 30 ft/s and a pressure of 80 psi. The water exits from the nozzle into the atmosphere at section (2) where the uniformly distributed velocity is 100 ft/s. Determinethe axial component of the anchoring force required to hold the contraction in place.
This question is incomplete, the missing image is uploaded along this answer.
Answer:
the axial component of the anchoring force required to hold the contraction in place is 365.6 lb
Explanation:
Given the data in the question and as illustrated in the image below;
first we calculate the area at section 1
A₁ = (πD²)/4
we substitute
A₁ = (π(3 in)²)/4
A₁ = 7.06858 in²
we know that; 1 ft = 12 in
A₁ = ( 7.06858 / (12²) ) ft²
A₁ = ( 7.06858 / 144 ) ft²
A₁ = 0.0491 ft²
now, we write the elation for area at section 2
A₂ = πd²/4
here, d is the diameter at section 2
next, we use the conservation of mass equation between the two section;
m" = pV₁A₁ = pV₂A₂
we calculate the mass flow rate;
m" = pV₁A₁
= (1.94[tex]\frac{slug}{ft^2}[/tex]) × 30[tex]\frac{ft}{s}[/tex] × 0.0491 ft²
= 2.8576 slug/s
Now, Apply the linear momentum along the horizontal direction for the control volume between 1 - 2
-pV₁A₁V₁ = pV₂A₂V₂ = P₁A₁ - F[tex]_A[/tex] - P₂A₂
m"( V₂ - V₁ ) = P₁A₁ - F[tex]_A[/tex] - P₂A₂
F[tex]_A[/tex] = P₁A₁ - P₂A₂ - m"( V₂ - V₁ )
we substitute
F[tex]_A[/tex] = ((80×[tex]\frac{144 in^2}{1 ft^2}[/tex])×0.0491 ft²) - (0×(πd²/4)) - 2.8576( 100 - 30 )ft/s
F[tex]_A[/tex] = 565.632 - 0 - 200.032
F[tex]_A[/tex] = 565.632 - 200.032
F[tex]_A[/tex] = 365.6 lb
Therefore, the axial component of the anchoring force required to hold the contraction in place is 365.6 lb
३.रात में घूमने वाला write one word substitute
Explanation:
रात में घूमने वाला arthaarat निशाचर
Which of the following types of mirrors has an optical axis?
a convex mirror
a concave mirror
a double convex mirror
All of the choices are correct.
Answer:
its B
Explanation:
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:
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.
if the water measures -5 feet at low tide and 3ft at high tide what is the tidal range
Answer:
8 feet
................
How much work is done when 100 N of force is applied to a rock to move it 20 m
Answer: 2000 J
Explanation: work W = F s
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
Mechanical energy is the most concentrated form of energy.
a. true
b. false
If it takes 560 joules of work to move an object 10 meters what force was needed
Answer:
F = 56 N
Explanation:
Assuming that the applied force is parallel to the displacement, according to the definition of work, we can write the following expression for the force required:[tex]F = \frac{W}{\Delta x} =\frac{560J}{10m} = 56 N (1)[/tex]
3. Materials that lets electricity to pass
Answer:
materials that allow electricity to pass through them are called conductors, some examples of conductors are many metals, such as copper, iron and steel.
At the base of a hill, a 90 kg cart drives at 13 m/s toward it then lifts off the accelerator pedal). If the cart just barely makes it to the top of this hill and stops, how high must the hill be?
Answer:
8.45 m
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 90 Kg
Initial velocity (u) = 13 m/s
Final velocity (v) = 0 m/s
Height (h) =?
NOTE: Acceleration due to gravity (g) = 10 m/s²
The height of the hill can be obtained as follow:
v² = u² – 2gh (since the cart is going against gravity)
0² = 13² – (2 × 10 × h)
0 = 169 – 20h
Rearrange
20h = 169
Divide both side by 20
h = 169/20
h = 8.45 m
Therefore, the height of the hill is 8.45 m
The knee extensors insert on the tibia at an angle of 30 degrees (from the longitudinal axis of the tibia), at a distance of 3 cm from the axis of rotation at the knee. How much force must the knee extensors exert to produce an angular acceleration at the knee of 1 rad/s2 , given a mass of the lower leg and foot of 4.5 kg, and a radius of gyration of 23 cm
Answer:
the knee extensors must exert 15.87 N
Explanation:
Given the data in the question;
mass m = 4.5 kg
radius of gyration k = 23 cm = 0.23 m
angle ∅ = 30°
∝ = 1 rad/s²
distance of 3 cm from the axis of rotation at the knee r = 3 cm = 0.03 m
using the expression;
ζ = I∝
ζ = mk²∝
we substitute
ζ = 4.5 × (0.23)² × 1
ζ = 0.23805 N-m
so
from; ζ = rFsin∅
F = ζ / rsin∅
we substitute
F = 0.23805 / (0.03 × sin( 30 ° )
F = 0.23805 / (0.03 × 0.5)
F F = 0.23805 / 0.015
F = 15.87 N
Therefore, the knee extensors must exert 15.87 N
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
why doesn't a radio operating with two batteries function when one of the batteries is reversed?
Answer:
If you have two batteries and they have precisely the same voltage then placing one backwards will effectively cancel out the voltages and no current will flow. However, batteries aren't like that. The slightest difference in voltages mean that current will flow.
Explanation:
g Two long parallel wires are a center-to-center distance of 2.50 cm apart and carry equal anti-parallel currents of 2.70 A. Find the magnitude of the magnetic field at the point P which is equidistant from the wires. (R
Answer:
864 mT
Explanation:
The magnetic field due to a long straight wire B = μ₀i/2πR where μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, i = current in wire, and R = distance from center of wire to point of magnetic field.
The magnitude of magnetic field due to the first wire carrying current i = 2.70 A at distance R which is mid-point between the wires is B = μ₀i/2πR.
Since the other wire also carries the same current at distance R, the magnitude of the magnetic field is B = μ₀i/2πR.
The resultant magnetic field at B is B' = B + B = 2B = 2(μ₀i/2πR) = μ₀i/πR
Now R = 2.50 cm/2 = 1.25 cm = 1.25 × 10⁻² m and i = 2.70 A.
Substituting these into B' = μ₀i/πR, we have
B' = 4π × 10⁻⁷ H/m × 2.70 A/π(1.25 × 10⁻² m)
B = 10.8/1.25 × 10⁻⁵ T
B = 8.64 × 10⁻⁵ T
B = 864 × 10⁻³ T
B = 864 mT
This question involves the concept of the magnetic field due to two current-carrying wires in the same direction, parallel to each other.
The magnitude of the magnetic field at the point P, which is equidistant from the wires is "8.64 x 10⁻⁵ T".
The following formula is used to find the magnetic field at the center distance between two parallel current-carrying wires in the same direction:
[tex]B = \frac{\mu_oI_1}{2\pi r}+\frac{\mu_oI_2}{2\pi r}\\\\But,\ I_1=I_2=I\\\\B = \frac{\mu_oI}{\pi r}[/tex]
where,
B = magnetic field at required point = ?
μ₀ = permeability of free space = 4π x 10⁻⁷ H/m
I = current = 2.7 A
r = distance from wires to the point = 2.5 cm/2 = 1.25 cm = 0.0125 m
Therefore,
[tex]B=\frac{(4\pi\ x\ 10^{-7}\ H/m)(2.7\ A)}{\pi (0.0125\ m)}[/tex]
B = 8.64 x 10⁻⁵ T
Learn more about the magnetic field here:
https://brainly.com/question/23096032?referrer=searchResults
Kevin decides to soup up his car by replacing the car's wheels with ones that have 1.4 times the diameter of the original wheels. Note that the speedometer in a car is calibrated based on the tire's diameter and on the distance the tire covers in each revolution. (a) Will the reading of the speedometer change
Answer:
No.
Explanation:
Given that Kevin decides to soup up his car by replacing the car's wheels with ones that have 1.4 times the diameter of the original wheels. Note that the speedometer in a car is calibrated based on the tire's diameter and on the distance the tire covers in each revolution. (a) Will the reading of the speedometer change ?
Considering the formula
V = wr
Where
V = linear speed
W = angular speed
r = radius of the wheel.
But W = 2πrf
Where the the 2 and pi are constant. The radius of the first wheel will be small but counter balance with the larger frequency.
While the radius of the second wheel may be large but it will be of a small frequency.
We can therefore conclude that the reading on the speedometer will not change. Because speedometer will read the linear speed V.
Four rods that obey Hooke's law are each put under tension.
a. A rod 50.0 cm long with cross-sectional area 1.00 mm2 and with a 200 N force applied on each end.
b. A rod 25.0 cm long with cross-sectional area 1.00 mm2 and with a 200 N force applied on each end.
c. A rod 20.0 cm long with cross-sectional area 2.00 mm2 and with a 100 N force applied on each end. Order the rods according to the tensile stress on each rod, from smallest to largest.
Order the rods according to the tensile stress on each rod, from smallest to largest.
1. c < b < a
2. a=b
3. a< b < c
4. c
Answer:
c < a<=b
Explanation:
Tensile stress = (force) /Area
for A:
Tensile stress = 200/1 =200N/mm²
For B:
Tensile stress = 200/1 =200N/mm2
For C:
Tensile stress = 100/2 =50N/mm²
Ranking from smallest to largest we have;
C<A<=B which is option 4
