Answer:
Series
Cells in Series connection.In series, cells are joined end to end so that the same current flows through each cell. In case if the cells are connected in series the emf of the battery connected to the sum of the emf of the individual cell,If E is the overall emf of the battery combined with n number cells and E1, E2,......Em is the EMFs of individual cell.
Then
E= E1+E2+...............+Em.
The photo shows a pair of figure skaters performing a spin maneuver. The
axis of rotation goes through the left foot of the skater on the left. What
action could increase the pair's angular velocity?
An action which could increase the pair's angular velocity is the figure skater on the left pointing his right arm down instead of up.
What is angular velocity?Angular velocity simply refers to the rate of change of angular displacement of a physical object (body) with respect to time.
This ultimately implies that, angular velocity is a measure of how fast and quickly a physical object (body) rotates with respect to another point or how its angular displacement (position) changes with respect to time.
In this scenario, an action which could increase the pair's angular velocity is when the figure skater on the left points his right arm down instead of up.
Read more on angular velocity here: https://brainly.com/question/6860269
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the skater on the left pulls the skater on the right closer to him
Explanation:
apex bliches
What is the typical treatment for Hodgkin's disease?
desensitization treatments, bone marrow transplant, and biopsy
desensitization treatments, chemotherapy, and thymectomy
radiation, biopsy, and desensitization treatments
radiation, chemotherapy, and bone marrow transplant
Answer:
Pretty sure its D.
radiation, chemotherapy, and bone marrow transplant
Explanation:
Radiation, chemotherapy, and bone marrow transplant are the typical treatment for Hodgkin's disease. Therefore, option (D) is correct.
What is Hodgkin lymphoma?Hodgkin lymphoma can be described as a type of lymphoma, in which cancer forms from a specific type of white blood cell known as lymphocytes, where multinucleated Reed-Sternberg cells are present in the lymph nodes.
The most common symptom of Hodgkin's lymphoma is the painless enlargement of lymph nodes. Hodgkin lymphoma can be treated with radiation therapy, chemotherapy, and stem cell transplantation. The treatment generally depends on how advanced cancer has favorable features.
Radiation and chemotherapy drugs increase the risk of other cancers, lung disease, or heart disease over the subsequent decades.
Hodgkin lymphoma can be distinguished from non-cancerous causes of lymph node swelling and from other cancer. Definitive diagnosis is by lymph node biopsy. Blood tests are performed to assess the function of major organs and safety for chemotherapy.
Learn more about Hodgkin lymphoma, here:
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money is rare? or is it bad
Answer:
money is not rare. you can find it everywhere. what exactly is the question? whether it is bad or not is an opinion.
Explanation:
45
А ______ is a solid, liquid or gas that a wave travels through.
Answer:
medium
Explanation:
I am not really sure
Which wave has the smallest wave
period? What is its period?
Answer:
C
Explanation:
The said wave takes the shortest time to move/get transmitted
A string with a length of 6.2 m is vibrated at its natural frequency (first harmonic). What is the wavelength of the produced standing wave?
3.1 m
6.2 m
12.4 m
18.6 m
Answer:
12.4m
Explanation:
Given
Length of string = 6.2m
L = λ/2 (for natural frequency (first harmonic)
λ = 2L
λ = 2(6.2)
λ = 12.4m
Hence the wavelength of the produced standing wave is 12.4m
A long, uninsulated steam line with a diameter of 89 mm and a surface emissivity of 0.8 transports steam at 200C and is exposed to atmospheric air and large surroundings at an equivalent temperature of 20C. (a) Calculate the heat loss per unit length for a calm day. (b) Calculate the heat loss on a breezy day when the wind speed is 8 m/s. (c) For the conditions of part (a), calculate the heat loss with a 20-mm-thick layer of insulation (k 0.08 W/m K). Would the heat loss change significantly with an appreciable wind speed
Answer:
[tex]Q_net=534.67\frac{w}{m}[/tex]
Explanation:
From the question we are told that:
Steam line diameter [tex]D=89mm \approx 0.089m[/tex]
Surface emissivity [tex]\mu=0.8[/tex]
Steam temp [tex]T_s=200\textdegree C \approx 200+273=473k[/tex]
Surrounding temp [tex]T_a=20 \textdegree C \aprrox 20+273= 293k[/tex]
Generally the equation for heat loss per unit length due to radiation [tex]Q_{net}[/tex] is mathematically given by
[tex]Q_net=\sigma*\mu>(\pi *d)*(T_s^4-T_a^4)[/tex]
[tex]Q_net=5.6*10^8*0.8*(\pi *0.089)*(473^4-293^4)[/tex]
[tex]Q_net=534.67\frac{w}{m}[/tex]
PLEASE HELP
A ball, 1.8 kg, is attached to the end of a rope and spun in a horizontal circle above a student's head. As the student rotates the ball in a horizontal clockwise circle their lab partner counts 53 rotations in one minute. What is the ball's angular velocity in radians per second?
Answer:
The angular speed of the ball in radians per second is 5.55 rad/s.
Explanation:
Given;
mass of the ball, m = 1.8 kg
number of the ball's rotation per minute, n = 53 RPM
The angular speed of the ball in radians per second is calculated as follows;
[tex]\omega = 53\frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 \ s } \\\\\omega = 5.55 \ rad/s[/tex]
Therefore, the angular speed of the ball in radians per second is 5.55 rad/s.
A beaker is filled with water. A small plastic container contains a solid bar of aluminum, which has a mass of 40 g, and is placed on the water so that it floats. The water level reads 60 ml. Next, the bar of aluminum is taken out of the container and placed in the water so that it sinks to the bottom. By how much does the water level change?
Answer:
25.2 ml
Explanation:
When the aluminium block is inserted in the container, the overall amount of only the water in the beaker can equal V o.
The weight of the water expelled by the plastic container should be equal to the weight of the aluminium block, according to the buoyancy balance relation.
i.e.
[tex]\rho_w V_wg = m_{Al}g = \rho_{Al}V_{Al} g \\ \\ V_w = \dfrac{\rho_{Al}V_{AL}}{\rho_{w}}[/tex]
When the aluminium block is inserted into the plastic container, the initial volume of water = 60 ml
[tex]V_i = V_o + V_w[/tex]
[tex]V_i = V_o + \dfrac{\rho_{Al}V_{Al}}{\rho_{w}}---(1)[/tex]
When the aluminium block is placed outside the container, the volume of the water
[tex]V_f = V_o +V_{Al} ---(2)[/tex]
By subtracting equation (1) and (2)
[tex]V_i -V_f = V_o + \dfrac{\rho_{Al} V_{Al}}{\rho_w}- ( V_o + V_{Al}}) \\ \\ =\dfrac{\rho _{Al}V_{Al}}{\rho_w}-V_{Al} \\ \\ = V_{Al} \Big( \dfrac{\rho_{Al}}{\rho_{w}}-1 \Big)[/tex]
since;
[tex]m_{Al} = 40 g[/tex]
[tex]V _{Al} = \dfrac{40 \ g}{2.7 \ g/cm^3} \\ \\ V_{Al} = 14.815 \ cm^3[/tex]
Similarly;
[tex]\dfrac{\rho_{Al}}{\rho_{w}}= \dfrac{2.7 }{1.0}[/tex]
= 2.7
[tex]V_i -V_f =14.815\Big( 2.7-1 \Big) \\ \\ V_i -V_f = 25.1855 \ ml \\ \\ = \mathbf{25.2 \ ml}[/tex]
A typical electric refrigerator has a power rating of 500 Watts, which is the rate (J/s) at which electrical energy is supplied to do work needed to remove heat from the refrigerator. If the refrigerator releases heat to the room at a rate of 800 Watts, at what rate (in Watts) does it remove heat from inside of the refrigerator?
Answer:
The rate of heat removed from inside the refrigerator is 300 watts.
Explanation:
By the First Law of Thermodynamics and the definition of a Refrigeration Cycle, we have the following formula to determine the rate of heat removed from inside the refrigerator ([tex]\dot Q_{L}[/tex]), in watts:
[tex]\dot Q_{L} = \dot Q_{H}-\dot W[/tex] (1)
Where:
[tex]\dot Q_{H}[/tex] - Rate of heat released to the room, in watts.
[tex]\dot W[/tex] - Rate of electric energy needed by the refrigerator, in watts.
If we know that [tex]\dot Q_{H} = 800\,W[/tex] and [tex]\dot W = 500\,W[/tex], then the rate of heat removed from inside the refrigerator is:
[tex]\dot Q_{L} = \dot Q_{H}-\dot W[/tex]
[tex]\dot Q_{L} = 300\,W[/tex]
The rate of heat removed from inside the refrigerator is 300 watts.
B
С
17. A car moves at constant speed of 40 kmh-' along the
road shown in Fig. 17.1. The radius of curvature at
A is 350 m and the total acceleration of the car at B is
1.0 ms?
(a) Find the total acceleration of the car at
A
5 p
5
Fig. 17.1
[2]
i. A
[1]
ii. C
2
(b) Find the radius of curvature at B.
Answer:
a
Explanation:
a
What is a centripetal acceleration of a greyhound running on a circular track with a radius of 50 m at 12.5 m/s
Answer:
Centripetal acceleration = 3.125 m/s²
Explanation:
Given the following data;
Radius, r = 50 m
Velocity, V = 12.5 m/s
To find the centripetal acceleration;
Centripetal acceleration = Velocity²/radius
Centripetal acceleration = 12.5²/50
Centripetal acceleration = 156.25/50
Centripetal acceleration = 3.125 m/s²
Therefore, the centripetal acceleration of the greyhound running on a circular track is 3.125 meters per seconds square.
Calculate the height of a cliff if it takes 2.35s for a rock to hit the ground when it is thrown straight up from the cliff with a initial velocity 8m/s.
Answer:
y₀ = 10.625 m
Explanation:
For this exercise we will use the kinematic relations, where the upward direction is positive.
y = y₀ + v₀ t - ½ g t²
in the exercise they indicate the initial velocity v₀ = 8 m / s.
when the rock reaches the ground its height is zero
0 = y₀ + v₀ t - ½ g t²
y₀i = -v₀ t + ½ g t²
let's calculate
y₀ = - 8 2.5 + ½ 9.8 2.5²
y₀ = 10.625 m
How efficient are the small and large scale solar-power systems used in individual homes and industrial settings?
Plz someone help me
Answer:
I study physics too, want me to study with you?
Wind power is considered a renewable energy source.
O True
O False
Answer:
True
Explanation:
Answer:
your answer is true hope this helps
the electrostatic force between two objects is 40N. if the charge of one object is cut in half, and the distance is doubled, what is the new force?
Answer:
F1 = K Q1 Q2 / R1^2
F2 = K Q1 / 2 * Q2 / (2 R1)^2
F2 / F1 = 1/2 / 4 = 1/8
The new force is 5N (1/2 due to charge and 1/4 due to distance)
Jupiter's Great Red Spot is
1.slowly shrinking
2.a large storm
3.smaller than Saturn's spot
4.a large area of liquid hydrogen
Answer:
2 : A large storm
Explanation: I hope this helps
1. There are 15 boys and girls 25 in Grade VI- Jacinto . Name the ratio of girls to boys in
fractional form?
A. 15/20
B. 15/25
C. 20/25
D. 25230
A water pipe having a 4.00 cm inside diameter carries water into the basement of a house at a speed of 1.00 m/s and a pressure of 167 kPa. The pipe tapers to 1.4 cm and rises to the second floor 7.8 m above the input point. What is the speed at the second floor
Answer:
[tex]8.16\ \text{m/s}[/tex]
Explanation:
[tex]d_1[/tex] = Initial diameter = 4 cm
[tex]v_1[/tex] = Initial velocity = 1 m/s
[tex]d_2[/tex] = Final diameter = 7.8 m
[tex]v_2[/tex] = Final velocity
[tex]A[/tex] = Area = [tex]\pi\dfrac{d^2}{4}[/tex]
From the continuity equation we get
[tex]A_1v_1=A_2v_2\\\Rightarrow \pi\dfrac{d_1^2}{4}v_1=\pi\dfrac{d_2^2}{4}v_2\\\Rightarrow v_2=\dfrac{d_1^2}{d_2^2}v_1\\\Rightarrow v_2=\dfrac{4^2}{1.4^2}\times 1\\\Rightarrow v_2=8.16\ \text{m/s}[/tex]
The speed of water at the second floor is [tex]8.16\ \text{m/s}[/tex].
Part C
Now vary the mass of the skateboarder. Change the mass slider to the lowest mass. Repeat the investigation using the
steepest ramp. How does changing the mass of the skateboarder affect his kinetic energy on a given ramp? How does it
affect his speed? Remember to use the pause button to clearly see his energy and speed. Repeat your trials as many
times as needed.
Answer:
When mass increases, kinetic energy also increases. Changing mass doesn’t affect his speed on a given ramp.
Explanation:
Answer:
When mass increases, kinetic energy also increases. Changing mass doesn’t affect his speed on a given ramp.
Explanation:
pluto
A car of mass, 1200 kg is pulled with constant force of 3000N in 20s. Work out A) The acceleration resulted: B) Distance travelled by the car, in the given time? C) Work done by the engine of the car D) The kinetic energy of the Car
Answer:
D
Explanation:
the energy is coming up and down
The answer is:
D) The kinetic energy of the car
Does an infrared wave or an x-ray travel faster in the vacuum of space?
Answer:
All electromagnetic radiation, of which radio waves and X-rays are examples, travels at the speed c in a vacuum. The only difference between the two is that the frequency of X-rays is very much higher than radio waves
Roland Incorporated manufactures and sells portable hair dryers. If the price of a hair dryer is $ 127.00, then the company sells 116 hair dryers per week, and the elasticity of demand is −0.463. Assume that the demand function is differentiable and that the only time the company sells exactly 116 hair dryers per week is when the price of a hair dryer is $ 127.00. Given this situation, which of the following do you know MUST be true?
a. The marginal revenue is negative when Roland sells 144 driers per week.
b. The marginal profit is negative when Roland sells 144 driers per week.
c. The marginal profit is positive when Roland sells 144 driers per week.
d. The marginal revenue is positive when Roland sells 144 driers per week.
Answer:
The marginal revenue is positive when Roland sells 144 driers per week ( D
Explanation:
Elasticity of demand = -0.463
P = $127
116 hair dryers sold per week ( q )
demand function is differentiable
prove the true statement below
Revenue = price per unit * quantity
determine the Marginal revenue = dr / dq
∴ MR = P + q dp/dq
price elasticity of demand E(p) = - dq / dp * p/q
where : E ( p ) = - 0.463 , q = 116 , p = $127
q * dp/dq = P / 0.463
∴ MR = P * 1.463 / 0.463
= 127 * 3.16 = $401.32
since the Marginal revenue > 0 HENCE The marginal revenue will be positive when Roland sells 144 driers per week.
how much work is done in accelerating a 2000 kg car from rest to a speed of 30 m/s?
Give me four reasons pluto is a cool planet / dwarf planet
How does a wind turbine make electricity?
1) The wind turns a generator inside.
2) Magic
3) Waves turn the generator.
4) Hamster turns the generator.
Answer:
1
Explanation:
Water is to be heated from 10°C to 80°C as it flows through a 2-cm-internal-diameter, 13-m-long tube. The tube is equipped with an electric resistance heater, which provides uniform heating throughout the surface of the tube. The outer surface of the heater is well insulated, so that in steady operation all the heat generated in the heater is transferred to the water in the tube. If the system is to provide hot water at a rate of 5 L/min, determine the power rating of the resistance heater. Also, estimate the inner surface temperature of the pipe at the exit.
Answer:
- the power rating of the resistance heater is 24139.5 W
- the inner surface temperature of the pipe at the exit is 96.34°C
Explanation:
Given the data in the question;
Flow rate of water in the tube V" = 5L/min = 8.333 × 10⁻⁵ m³/s
The water is to be heated from 10°C to 80°C;
so Average or mean temperature [tex]T_{avg[/tex] will be;
[tex]T_{avg[/tex] = (T₁ + T₂) / 2 = (10 + 80) / 2 = 90/2 = 45°C
Now, from the Table " Properties of Water " at average temperature;
at [tex]T_{avg[/tex] = 45°C
density p = 990.1 kg/m³
specific heat [tex]C_p[/tex] = 4180 J/kg-k
thermal conductivity k = 0.637 W/m-°C
Now, we determine the mass flow;
m" = pV"
we substitute
m" = 990.1 × 8.333 × 10⁻⁵
m" = 0.08250 kg/s
we know that the power rating of the resistance heater is equal to the heat transfer rate to the water;
Q' = m"[tex]C_p[/tex]( T₂ - T₁ )
we substitute
Q' = (0.08250 × 4180 ) ( 80 - 10 )
Q' = 344.85 × 70
Q' = 24139.5 W
Hence, the power rating of the resistance heater is 24139.5 W
Next, we determine the average velocity of water in the tube;
[tex]V_{avg[/tex] = V" / [tex]A_c[/tex]
[tex]V_{avg[/tex] = V" / ( [tex]\frac{1}{4}[/tex]πD² )
given that; flows through a 2-cm-internal-diameter; D = 0.02 m
we substitute
[tex]V_{avg[/tex] = (8.333 × 10⁻⁵) / ( [tex]\frac{1}{4}[/tex]π × (0.02)² )
[tex]V_{avg[/tex] = (8.333 × 10⁻⁵) / ( 3.14159 × 10⁻⁴ )
[tex]V_{avg[/tex] = 0.265 m/s
Also, from table " saturated water property table "
At 45°C
viscosity μ = 0.596 × 10⁻³ kg/m-s
Prandtl number Pr = 3.91
Now, we determine the kinematic viscosity
v = μ / p
we substitute
v = ( 0.596 × 10⁻³ ) / 990.1
v = 6.01959 × 10⁻⁷ m²/s
so, Reynolds number in the flow region will be;
Re = ([tex]V_{avg[/tex] × D) / v
we substitute
Re = ( 0.265 × 0.02) / (6.01959 × 10⁻⁷)
Re = 8804.586
we can see that our Reynolds number ( 8804.586 ) more than 2300 and less than 10,000.
Hydraulic and thermal entry length are equal in this flow region,
such that;
[tex]L_h[/tex] = [tex]L_t[/tex]
⇒ 10 × D = 10 × 0.02 = 0.2 m
we can see that the entry length ( 0.2 m ) is smaller than the given length ( 13 m ) in the question; the flow is a turbulent flow.
So we the Nuddelt number
Nu = [tex]0.023Re^{0.8} Pr^{0.4[/tex]
Nu = 0.023 × [tex]8804.586^{0.8[/tex] × [tex]3.91^{0.4[/tex]
Nu = 56.8
Hence, the heat transfer coefficient h will be;
h = [tex]\frac{k}{D}[/tex] × Nu
we substitute
h = [tex]\frac{0.637}{0.02}[/tex] × 56.8
h = 31.85 × 56.8
h = 1809.1 W/m²-°C
Now, area of the heat transfer will be
A[tex]_s[/tex] = πDL
we substitute
A[tex]_s[/tex] = π × 0.02 × 13
A[tex]_s[/tex] = 0.8168 m²
Finally we determine the inner temperature of the pipe at exit. using the relation;
Q' = hA[tex]_s[/tex]( T₃ - T₂ )
we substitute
24139.5 = 1809.1 × 0.8168( T₃ - 10 )
24139.5 = 1477.67288( T₃ - 80 )
24139.5 = 1477.67288T₃ - 118213.8304
24139.5 + 118213.8304 = 1477.67288T₃
1477.67288T₃ = 142353.3304
T₃ = 142353.3304 / 1477.67288T
T₃ = 96.34°C
Therefore, the inner surface temperature of the pipe at the exit is 96.34°C
A long conducting cylinder of radius a carrying a total charge +q is surrounded by a
concentric thin conducting cylindrical shell of radius b carrying a total charge -2q.
Draw the cross section of this arrangement. Use Gauss’s law to find the electric field
strength at a point r away from the axis,
Answer:
I don't know this, sorry
name the three major types of clouds
Answer:
Cumulus, Stratus, and Cirrus. There are three main cloud types.
Explanation:
hopes it help^_^
Wind power produces a lot of air pollution.
O True
O False
Answer:
Wind is a renewable energy source. Overall, using wind to produce energy has fewer effects on the environment than many other energy sources. Wind turbines do not release emissions that can pollute the air or water (with rare exceptions), and they do not require water for cooling.
Explanation: