When heat travels between two objects that are touching,
it is moving by conduction.
The slope of a displacement time graph for a uniform motion represent what
Answer:
Velocity.
Explanation:
The slope of a displacement time graph for a uniform motion represent the gradient of the line i.e. the velocity of the object.
The velocity of an object is given by :
v = d/t
Where
d is displacement
t is time
Hence, the slope of the displacement-time graph gives the velocity of the object.
An elevator suspended by a vertical cable is moving downward at a constant speed. The tension in the cable must be A) greater than the weight of the elevator. B) less than the weight of the elevator. C) equal to the weight of the elevator.
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
5
Select the correct answer.
What is the current in a parallel circuit which has two resistors (17.2 ohms and
22.4 ohms) and a power source of 6.0 volts?
ОА.
0.30 amps
OB.
9.8 amps
OC.
0.61 amps
D.
1.2 amps
Reset
Next
Answer:
Current in a parallel circuit = 0.61 amps (Approx)
Explanation:
Given:
Voltage V = 6 volt
Two resistors = 17.2 , 22.4 in parallel circuit
Find:
Current in a parallel circuit
Computation:
1/R = 1/r1 + 1 / r2
1/R = 1/17.2 + 1 / 22.4
R = 9.73 ohms (Approx)
Current in a parallel circuit = V / R
Current in a parallel circuit = 6 / 9.73
Current in a parallel circuit = 0.61 amps (Approx)
can someone please help me I am so behind I neee to catch up but I need it to be correct both of them
Answer:
1.B, 2.A
Explanation:
If the object on the Moon were raised to a height of 30.0 m, what would be the potential energy? PE=mgh (g on the Moon is 1.62m/s)
What Kind of clothing would you most likely wear at 15 latitude?
A. Fleece or jacket and long pants
B. cotton shirts, light jacket, long pants
C. parka, gloves,insulated pants, heavy boots
D. shorts , sandals and T-shirts
Answer:
I'd say A, but I'm not 100% what kind of latitude you're referring to
Answer:
D. shorts, sandals and T-shirts
Explanation:
The 15th parallel north is a circle of latitude that is 15 degrees north of the Earth's equatorial plane. It crosses Africa, Asia, the Indian Ocean, the Pacific Ocean, Central America, the Caribbean and Atlantic Ocean. All these areas are all warm and weather for wearing shorts, sandals and T-shirts!
(Hope this helped! If so please mark this answer as brainliest!)
1. A silicon BJT is connected as shown in Fig 1, where RC = 3.6 k 2. VBE = 0.8 V. (10%)
(a) Predict Ic and specify Rp to establish Vce at 5 V.(5%)
(b) The BJT is said to be in forward-reverse bias. Explain what is meant by this. (5%)
Answer:
The circuit is missing attached below is the required circuit
answer :
a) Ic = 1.944 mA
Rp = 288.66 kΩ
b) The Emitter-base Junction of the BJT is forward biased while its collector-base junction is reverse biased
Explanation:
Rc = 3.6 kΩ
VBE = 0.8 v
1) predict Ic and specify Rp to establish Vce at 5 V
we will apply Kirchhoff's voltage law to resolve this
solution attached below
b ) The BJT is said to be in Forward reverse bias because The Emitter-base Junction of the BJT is forward biased while its collector-base junction is reverse biased
6. As distance increases, gravitational force *
(10 Points)
increases
decreases
In 2014, physicists from FOM Foundation at the University of Amsterdam introduced a new hypothesis of how the Pyramids at Giza were built. The group of physicists suggestedthat ancient Egyptians wetted sand in an effort toreduce friction and then pulled the 3000 kg stoneblocks to their final resting place. 15 men couldmove a block at a rate of 0.5m/sby pulling a largerope angled at 30owith respect to the plane anda tension of 7,200 N.
Required:
a. What is net work done on block?
b. What is speed of blck after it moved .25m?
c. What is work done by block if kinetic friction coefficient is 0.3?
d. What is net work including friction?
Answer:
The correct answer is:
(a) 0
(b) 0.5 m/s
(c) 7740 N
(d) 0
Explanation:
The given values are:
mass,
m = 3000 kg
Tension,
T = 7,200 N
Angle,
= 30°
(a)
Even as the block speed becomes unchanged, the kinetic energy (KE) will adjust as well:
⇒ [tex]\Delta K =0[/tex]
By using the theorem of energy, the net work done will be:
⇒ [tex]\Delta K =0[/tex]
(b)
According to the question, After 0.25 m the block is moving with the constant speed
= 0.5 m/s.
(c)
The given kinetic friction coefficient is:
u = 0.3
The friction force will be:
= [tex]u(mg-Tsin30^{\circ})[/tex]
On substituting the values, we get,
= [tex]0.3[(3000\times 9.8)-(7200\times 0.5)][/tex]
= [tex]0.3[29400-3600][/tex]
= [tex]0.3\times 25800[/tex]
= [tex]7,740 \ N[/tex]
(d)
On including the friction,
The net work will be:
⇒ [tex]\Delta K=0[/tex]
A frictionless pendulum is made with a bob of mass 12.6 kg. The bob is held at height = 0.650 meter above the bottom of its trajectory, and then pushed
forward with an initial speed of 4.22 m/s. What amount of mechanical energy does the bob have when it reaches the bottom?
g A rigid air cylinder with a volume of 100 cm3 is punctured with a hole having a crosssectional area of 0.3 mm2 . The original pressure and temperature of the air inside the cylinder are 800 kPa and 35 C. As the air leaves the hole in the cylinder, it reaches a pressure of 100 kPa and a temperature of 5 C. The velocity of the air as it escapes through the hole is 100 m/s. Calculate the original mass of air inside the tank and the mass in the tank 5 seconds after it is punctured, assuming the exit conditions of the air remain independent of time g
Solution :
The volume of the rigid cylinder = [tex]$100 \ cm^3 = 100 \times 10^{-6} \ m^3$[/tex]
Initial pressure inside the cylinder, [tex]$P_i = 800 \ kPa$[/tex]
Initial temperature inside the cylinder, [tex]$T_i = 35^\circ C= 308 \ K$[/tex]
Final temperature inside the cylinder, [tex]$T_f = 5^\circ C= 278 \ K$[/tex]
Final pressure inside the cylinder, [tex]$P_f = 100 \ kPa$[/tex]
Area of the hole, A = [tex]$0.3 \ mm^2 = 3 \times 10^{-7} \ m^2$[/tex]
Velocity of the air through the hole, V = 100 m/s
The final pressure and the temperature inside the cylinder will be the condition same as the ambient conditions.
At initial state, from the equation of state,
PV = mRT, where R = 287 J/kg-K for air
[tex]$800 \times 10^3 \times100 \times10^{-6} = m_1 \times 287 \times 308$[/tex]
∴ [tex]$m_1=9.1 \times 10^{-4} \ kg$[/tex]
Since the exit condition does not change with time, we have ,
At ambient condition, [tex]$P_f = 100 \kPa$[/tex] and [tex]$T_f= 278 \ K$[/tex].
Therefore, we can find the density of the air
[tex]$P=\rho R T$[/tex]
[tex]$\rho = \frac{P}{RT}$[/tex]
[tex]$=\frac{100 \times 10^3}{287 \times 278}$[/tex]
[tex]$= 1.25 \ kg/m^3$[/tex]
Mass flow rate of air from the cylinder = [tex]$\dot m$[/tex]
[tex]$\dot m$[/tex] can be written as [tex]$\dot m$[/tex] [tex]$=\rho_{f} \times A \times v$[/tex]
[tex]$\dot m$[/tex] = [tex]$1.25 \times 3 \times 10^{-7} \times 100$[/tex]
[tex]$\dot m$[/tex] = [tex]$3.75 \times 10^{-5}$[/tex] kg/s
Mass escaped from the cylinder in 5 seconds
[tex]$m=3.75 \times 10^{-5} \times 5$[/tex]
[tex]$= 1.875 \times 10^{-4} \ kg$[/tex]
Mass of air remaining in the cylinder after 5 seconds :
[tex]$m_2 = m_1 - m$[/tex]
[tex]$m_2 = 9.1 \times 10^{-4} - 1.875 \times 10^{-4}$[/tex]
[tex]$m_2 = 7.225 \times 10^{-4} \ kg$[/tex]
= 0.7225 grams
how does temperature affect brownian motion
Answer: This (random) thermal motion of the particles due to the temperature is also called Brownian motion. ... The higher the temperature, the faster the diffusion will be, because the stronger the molecule movement and thus the “mixing”.
Explanation:
A student picks up two spheres that are the same size. One is made of steel and the other is made of aluminum. The student notices that the steel sphere feels much heavier than the aluminum sphere. He then holds one sphere in each hand at eye level and lets go of them at the same time. They fall to the floor. Which ball, if any, will hit the ground first and for what reason
Answer:
They will fall at the same time. This is because gravity accelerates all objects at the same speed, Earth's gravity being approximately 9.8m/s²
They'll both fall at the same time. This is because gravity accelerates everything at the same rate, with Earth's gravity being approximately 9.8 m/s2.
What is gravity with some instances?The energy that holds the gases inside the sun together. the force that causes a ball to fall after being thrown into the air the force that causes a car to coast downhill even when the gas pedal is not depressed the force that causes a glass to shatterGravity, also known as gravitation, affects all material objects in the universe. Gravity attracts any two objects or particles with nonzero mass toward one another. Gravity affects everything from subatomic particles to galaxy clusters. Gravity is the attraction force between two objects. It's what causes things to fall and keeps us from floating away into space. Gravity is a fundamental natural force.To learn more about gravity, refer to:
https://brainly.com/question/16275567
#SPJ2
A load of mass 120kg is raised
vertically through a height of 2m in
30s by a machine whose efficiency,
is 100% Calculate the power generated
by the machine
Answer: P = 120 kg·9.81 m/s² · 2 m / 30 s = 78 W
Explanation: power P = Work done / time
Work is lifting work = mgh in which g = 9.81 m/s²
Time 30 s
The Hall effect can be used to determine the density of mobile electrons in a conductor. A thin strip of the material being investigated is immersed in a magnetic field and oriented so that its surface is perpendicular to the field. In a particular measurement, the magnetic field strength was 0.685 T, the strip was 0.107 mm thick, the current along the strip was 2.25 A, and the Hall voltage between the strip's edges was 2.59 mV.Find the density nof mobile electrons in the material. The elementary charge is 1.602×10−19 C.
Answer:
the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³
Explanation:
Given the data in the question;
we make use of the following expression;
hall Voltage VH = IB / ned
where I = 2.25 A
B = 0.685 T
d = 0.107 mm = 0.107 × 10⁻³ m
e = 1.602×10⁻¹⁹ C
VH = 2.59 mV = 2.59 × 10⁻³ volt
n is the electron density
so from the form; VH = IB / ned
VHned = IB
n = IB / VHed
so we substitute
n = (2.25 × 0.685) / ( 2.59 × 10⁻³ × 1.602×10⁻¹⁹ × 0.107 × 10⁻³ )
n = 1.54125 / 4.4396226 × 10⁻²⁶
n = 3.4716 × 10²⁵ m⁻³
Therefore, the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³
What season is the Southern hemisphere in
Answer:
hope this answer helps this is what I understand
Answer:
In the southern hemisphere it is fall but almost winter
Explanation:
They are 6 months ahead of the northern hemisphere. at least thats
how I think about it
The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the terminal velocity (in meters per second and kilometers per hour) of an 86.0 kg skydiver falling in a pike (headfirst) position with a surface area of 0.145 m2. (Assume that the density of air is 1.21 kg/m3 and the drag coefficient of a skydiver in a pike position is 0.7. For each answer, enter a number.)
Answer:
terminal velocity is;
v = 117.54 m/s
v = 423.144 km/hr
Explanation:
Given the data in the question;
we know that, the force on a body due to gravity is;
[tex]F_g[/tex] = mg
where m is mass and g is acceleration due to gravity
Force of drag is;
[tex]F_d[/tex] = [tex]\frac{1}{2}[/tex]pCAv²
where p is the density of fluid, C is the drag coefficient, A is the area and v is the terminal velocity.
Terminal velocity is reach when the force of gravity is equal to the force of drag.
[tex]F_g = F_d[/tex]
mg = [tex]\frac{1}{2}[/tex]pCAv²
we solve for v
v = √( 2mg / pCA )
so we substitute in our values
v = √( [2×(86 kg)×9.8 m/s² ] / [ 1.21 kg/m³ × 0.7 × 0.145 m²] )
v = √( 1685.6 / 0.122015 )
v = √( 13814.6949 )
v = 117.54 m/s
v = ( 117.54 m/s × 3.6 ) = 423.144 km/hr
Therefore terminal velocity is;
v = 117.54 m/s
v = 423.144 km/hr
For the circuit in the previous part, the current flowing in the wire between the positive terminal of the battery and the resistor is ___________ the current flowing between the resistor and the negative terminal of the battery. For the circuit in the previous part, the current flowing in the wire between the positive terminal of the battery and the resistor is ___________ the current flowing between the resistor and the negative terminal of the battery. less than equal to greater than
Answer:
Explanation:
because it is godhood
Which type of biological molecule would control cell activities?
A) Amino Acids.
C) Nucleic Acids.
B) Lipids.
D) Carbohydrates.
Answer:
C. Nucleic Acids.
Explanation:
Nucleic acids are thus uniquely capable of directing their own self-replication, allowing them to function as the fundamental informational molecules of the cell. The information carried by DNA and RNA directs the synthesis of specific proteins, which control most cellular activities.
Did some simple research, if that's fine and I hope this is correct as it says on the sites. Have a good one :)
A trapeze artist swings in simple harmonic motion with a period of 3.8 s.
Calculate the length of the cables supporting the trapeze. (g=9.81 m/s2)
Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a large toy rocket to the back of a sled and take the modified sled to a large, flat, snowy field. You ignite the rocket and observe that the sled accelerates from rest in the forward direction at a rate of 11.5 m/s^2 for a time period of 3.30 s. After this time period, the rocket engine abruptly shuts off, and the sled subsequently undergoes a constant backward acceleration due to friction of 4.15 m/s^2.
Required:
a. After the rocket turns off, how much time does it take for the sled to come to a stop?
b. By the time the sled finally comes to a rest, how far has it traveled from its starting point?
Answer:
a) t = 9.2s
b) Δx = 242.2 m
Explanation:
a)
In order to find the time that the sled traveled since the rocket was turned off, we need to find the first the speed that it had at that moment.Applying the definition of accceleration, since we know that the sled started from rest, we can find the value of the final speed (for this part) as follows:[tex]v_{f1} = a_{1} * t_{1} = 11.5m/s2* 3.30 s = 38.0 m/s (1)[/tex]
This speed, is just the initial speed for the second part, so we can find the time traveled from the moment the rocket was turned off until it came to an stop, as follows:[tex]t_{2} = \frac{v_{f1}}{a_{2} } = \frac{38m/s}{4.15m/s} = 9.2 s (2)[/tex]
b)
We need to find find first the displacement when the sled was accelerating.Assuming the acceleration is constant, since it started from rest, we can use the following kinematic equation:[tex]v_{f1} ^{2} = 2* a_{1} * x_{1} (3)[/tex]
Solving for x₁:[tex]x_{1} =\frac{v_{f1}^{2} }{2*a_{1}} =\frac{(38m/s)^{2} }{2*11.5m/s2} =62.8 m (4)[/tex]
In the same way, we can use the same equation, replacing the values of the final speed (which becomes zero), initial speed (which is the same as vf1), and a, which becomes -4.15 m/s2 as it is backwards.[tex]-v_{f1} ^{2} = 2* a_{2} * x_{2} (5)[/tex]
Solving for x₂:[tex]x_{2} =\frac{-v_{f1}^{2} }{2*a_{2}} =\frac{-(38m/s)^{2} }{2*(-4.15m/s)^2} =174.0 m (6)[/tex]
Δx = x₁ + x₂ = 68.2 m + 174.0 m = 242.2 m (7)A 2.00-kg ball is moving at 2.20 m/s toward the right. It collides elastically with a 4.00-kg ball that is initially at rest. 1) Calculate the final velocity of the 2.00-kg ball. (Express your answer to three significant figures.)
Answer:
The final velocity of the 2kg ball is 1.270 m/s
Explanation:
According to Newton's second and third laws of motion
Newton's second law state that "the rate of change of momentum is proportional to the applied force and takes place in the direction of that force".
Newton's third law state that "for every action, there must be an equal and opposite reaction".
The combinations of these two laws resulted in an elastic collision
Given that:
m1 = 2kg
u1 = 2.20m/s
m2 = 4.00kg
u2 = 0m/s
An Elastic collision is when kinetic energy before = kinetic energy after
E.K before = [tex]1/2mv^{2}[/tex]
E.K before = 1/2 * 2 * (2.20)^2
E.K = 1/2 * 2 * 4.84
E.K before = 4.84j
E.K after = 1/2 x (4 + 2)v^2
E.K after = 1/2(6v^2)
E.K after = 3v^2
Since E.K before = E.K after
4.84 = 3v^2
Divide through by 3
4.84/3 = 3v^2/3
1.6133 = v^2
[tex]V = \sqrt{1.6133} \\V = 1.270 m/s[/tex]
Four bicyclists travel different distances and times along a straight path. Which cyclist traveled with the greatest average
speed?
A
B
Cyclist 2 travels
87 min 22 s
Cyclist 4 travels
108 min 24 s
D
Cyclist 1 travels
95 m in 27 s
Cyclist 3 travels
106 m in 26 s
Answer:
The cyclist with the greatest average speed is Cyclist 4 with average speed of 4.5 m/s
Explanation:
Given;
Cyclist 1 travels 9 m in 27 s
Cyclist 2 travels 87 m in 22 s
Cyclist 3 travels 106 m in 26 s
Cyclist 4 travels 108 m in 24 s
Determine the average speed of the cyclists as follows;
Average speed of Cyclist 1: v = 9/27 = 0.33 m/s
Average speed of Cyclist 2: v = 87/22 = 3.96 m/s
Average speed of Cyclist 3: v = 106/26 = 4.08 m/s
Average speed of Cyclist 4: v = 108/24 = 4.5 m/s
Therefore, the cyclist with the greatest average speed is Cyclist 4 with average speed of 4.5 m/s
Question 11 of 15
Which of the following compounds is carbon tetrahydride?
A. CaH4
B. CH4
C. CaH
D. CAH
the correct answer would be "B"
A physics major is working to pay her college tuition by performing in a traveling carnival. She rides a motorcycle inside a hollow, transparent plastic sphere. After gaining sufficient speed, she travels in a vertical circle with a radius of 15.0 m. She has a mass of 80.0 kg and her motorcycle has mass 30.0 kg. What minimum speed must she have at the top of the circle for the motorcycle tires to remain in contact with the sphere
Answer:
v = 12.1 m/s
Explanation:
When at the top of the circle, there are two forces acting on the combined mass of the rider and the motorcycle.These are the force of gravity (downward) and the normal force, which is directed from the surface away from it, perpendicular to the surface.In this case, as the motorcycle runs in the interior of the circle, at the top point this force is completely vertical, and is also downward.Since the motorcycle is moving in a vertical circle, there must be a force, keeping the object moving around a circle.This force is the centripetal force, aims towards the center of the circle, and is just the net force aiming in this direction at any point.At the top point, this force is just the sum of the normal force and the weight of the mass of the rider and the motorcycle combined, as follows (we take the direction towards the center as positive):[tex]F_{c} = N + m*g (1)[/tex]
Now, we know that the centripetal force is related with the tangential speed at this point and the radius of the circle as follows:[tex]F_{c} = m*\frac{v^{2}}{r} (2)[/tex]
Since the normal force takes any value as needed to make (1) equal to (2), if the speed diminishes, it will be needed less force to keep the equality valid.In the limit, when the motorcyvle tires barely touch the surface, this normal force becomes zero.In this condition, from (1) and (2), we can find the minimum possible value of the speed that still keeps the motorcycle touching the surface, as follows:[tex]v_{min} =\sqrt{r*g} =\sqrt{15.0m*9.8m/s2} = 12.1 m/s (3)[/tex]1. A basket coffee filter (see below) is very light and has a large drag coefficient. It is possible to stack several filters together so they have the same drag coefficient as a single filter. Suppose you tried dropping one filter from a ladder, then tried dropping two stacked filters from the same height, then 3 and so on. For each try you measure the time taken for the filters to fall to the floor. How would you expect the time for the filters to fall to compare to the number of filters
Answer:
he times are getting closer as we use each filter, in the expression n would be the number of filters
t = [tex]\sqrt{ \frac{2(y_o - (n-1) h)}{g} }[/tex]
Explanation:
For this exercise let's use the kinematics relations
y = y₀ + v₀ t - ½ g t²
When the first filter reaches the ground, its height is y = 0, as they release its initial velocity is zero
for the 1st filter
0 = y₀ - ½ g t²
t² = 2y₀ / g
t = [tex]\sqrt{ \frac{2y_o}{g} }[/tex]
when we release the second filter upon arrival it has a height y = h where h is the height of each filter
h = y₀ - ½ g t²
t = [tex]\sqrt{ \frac{2(y_o- h)}{g} }[/tex]
when we release the third filter it reaches y = 2h
2h = y₀ - ½ g t²
t = [tex]\sqrt{ \frac{2(y_o -2h)}{g} }[/tex]
we can write the terms of this succession
(n-1) = y₀ - [tex]\frac{1}{2}[/tex] g t²
t = [tex]\sqrt{ \frac{2(y_o - (n-1) h)}{g} }[/tex]
therefore we see that the times are getting closer as we use each filter, in the expression n would be the number of filters
What happens to the energy of a rubber band when it is stretched?
Electromagnetic radiation is made up of wave components of electric and magnetic fields moving together. What form of energy is most closely associated with electromagnetic waves?
A.
light energy
B.
chemical energy
C.
electrical energy
D.
mechanical energy
Answer:
A) Because light is due to the electromagnetic fields (fields moving at the speed of light)
Note: Electrical energy is also used to describe the energy usage
in RLC circuits
What is heredity and how does it influence physical fitness?
2) How much work is required to pull a sled 15
meters if you use 30N of force?
2 people
Explanation: