Answer:
conduction I believe if not its convection
Answer: Conduction transfers energy from the spoon to the hot chocolate.
Explanation: Heated water molecules and steam rise in the beaker, carrying heat by convection.
Two pieces of steel wire with identical cross sections have lengths of L and 2L. The wires are each fixed at both ends and stretched so that the tension in the longer wire is four times greater than in the shorter wire. If the fundamental frequency in the shorter wire is 60 Hz, what is the frequency of the second harmonic in the longer wire?
Answer:
Explanation:
Expression for fundamental frequency of tone produced in a wire under tension of T and length L is given as follows
[tex]f=\frac{1}{2L} \times \sqrt{\frac{T}{ m} }[/tex]
m is mass per unit length .
We shall apply this formula for given wires .
For shorter wire
[tex]60 =\frac{1}{2L} \times \sqrt{\frac{T}{ m} }[/tex]
For longer wire for second harmonic
length of wire is 2L , tension is 4T ,
[tex]f =\frac{2}{4L} \times \sqrt{\frac{4T}{ m} }[/tex]
[tex]f =\frac{2\times 2}{4L} \times \sqrt{\frac{T}{ m} }[/tex]
f = 2 x 60 = 120 Hz .
A receiver catches a football on the 50.0 yard line and is tackled 5.42 seconds later on the 12 yard line. What
was the runner's average speed?
Answer:
7.01yard/sec
Explanation:
Given parameters:
Initial position = 50yard
Final position = 12yard
Time = 5.42s
Unknown:
Average speed of runner = ?
Solution:
To solve this problem;
Speed = [tex]\frac{distance}{time}[/tex]
Distance covered = Initial position - final position = 50 - 12 = 38yards
So;
Speed = [tex]\frac{38}{5.42}[/tex] = 7.01yard/sec
When measuring wellness you must consider
A all components of health
B your physical fitness being in the top 10% of the population
C being free of diseases
D both physical and mental health
Answer:
The answer is A) all components of health
Explanation:
Got it right on edge
In the figure, if Q = 52 µC q =10 µC and d = 55 cm, what is the magnitude of the electrostatic force on q?
Answer:
F = 15.47 N
Explanation:
Given that,
Q = 52 µC
q = 10 µC
d = 55 cm = 0.55 m
We need to find the magnitude of the electrostatic force on q. The formula for the electrostatic force is given by :
[tex]F=k\dfrac{q_1q_2}{d^2}\\\\F=9\times 10^9\times \dfrac{52\times 10^{-6}\times 10\times 10^{-6}}{(0.55)^2}\\\\F=15.47\ N[/tex]
So, the magnitude of the electrostatic force is 15.47 N.
The magnitude of electrostatic force will be "15.47 N".
Electrostatic forceAccording to the question,
Charges, Q = 52 μC
q = 10 μC
Distance, d = 55 cm, or
= 0.55 m
Constant, k = 9 × 10⁹
We know the relation,
→ Electrostatic force, F = k [tex]\frac{q_1 q_2}{d^2}[/tex]
By substituting the values, we get
= 9 × 10⁹ × [tex]\frac{10\times 10^{-6}}{(0.55)^2}[/tex]
= 15.47 N
Thus the above answer is correct.
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I need this done by tonight!! Can anyone help me please? Answer these 4 questions
Answer:
1. 14 g of chocolate mixture.
2. 24 fl oz of chocolate milk
3. 10 cups of chocolate milk.
4. 12½ cups.
Explanation:
From the question given above, the following data were obtained:
1 TBSP = 7 g
1 Cup = 8 fl oz
2 Table spoons (TBSP) for 1 cup (8 fl oz) of milk.
1. Determination of the mass of chocolate mixture in 1 cup of chocolate milk.
From the question given above,
1 Cup required 2 Table spoons (TBSP)
But
1 TBSP = 7 g
Therefore,
2 TBSP = 2 × 7 = 14 g
Thus, 1 Cup required 14 g of chocolate mixture.
2. Determination of the number fl oz of chocolate milk in 3 cups
1 Cup = 8 fl oz
Therefore,
3 Cups = 3 × 8
3 Cups = 24 fl oz
Thus, 24 fl oz of chocolate milk are in 3 cups.
3. Determination of the number of cups of chocolate milk produce from 20 TBSP.
2 TBSP is required to produce 1 cup.
Therefore,
20 TBSP will produce = 20/2 = 10 Cups.
Thus, 10 cups of chocolate milk produce from 20 TBSP.
4. Determination of the number of cups obtained from 100 fl oz chocolate milk.
8 fl oz is required to produce 1 cup.
Therefore,
100 fl oz will produce = 100 / 8 = 12½ cups.
Thus, 12½ cups is obtained from 100 fl oz chocolate milk.
true or false A person's speed around the Earth is faster at the poles than it is at the equator.
Answer:False
Explanation:The Earth rotates faster at the equator than at the poles.
a student lift a 25kg mass at vertical distance of 1.6m in a time of 2.0 seconds. a. Find the force needed to lift the mass (in N ). b. Find the work done by the student (in J). c. Find the power exerted by the student (in W)
Answer:
a. F = 245 Newton.
b. Workdone = 392 Joules.
c. Power = 196 Watts
Explanation:
Given the following data;
Mass = 25kg
Distance = 1.6m
Time = 2secs
a. To find the force needed to lift the mass (in N );
Force = mass * acceleration
We know that acceleration due to gravity is equal to 9.8
F = 25*9.8
F = 245N
b. To find the work done by the student (in J);
Workdone = force * distance
Workdone = 245 * 1.6
Workdone = 392 Joules.
c. To find the power exerted by the student (in W);
Power = workdone/time
Power = 392/2
Power = 196 Watts.
If two objects have different temperatures when they come in contact, heat will flow from the warmer
object to the cooler one UNTIL
Answer:
They both have an equal temperature.
Explanation:
Any material or object that allow the conduction (transfer) of electric charge or thermal energy is generally referred to as a conductor. Some examples of a conductor are metals, copper, aluminum, graphite, etc.
In the process of heat conduction, thermal energy is usually transferred from fast moving particles to slow moving particles during the collision of these particles. Also, thermal energy is typically transferred between objects that has different degrees of temperature and materials (particles) that are directly in contact with each other but differ in their ability to accept or give up electrons.
Hence, if two objects have different temperatures when they come in contact, heat will flow from the warmer object to the cooler one until they both have an equal temperature.
Which image shows the difference between the speed of molecules in hot and cold water? Explain your answer choice.
HELP ME,EVERYONE!!!!!!!! :(
Answer:
the answer is B
Explanation:
I think its B because on the top it shows the molecule speed and A looks like the water is cold, C shows that the hot
water is cooler, and D shows that both are cold
Various amplifier and load combinations are measured as listed below using rms values. For each, find the voltage, current, and power gains ( A v , Ai , and Ap, respectively) both as ratios and in dB:
(a) vI= 100 mV, iI = 100 μA, vO = 10 V, RL = 100Ω
(b) vI = 10 μV, iI = 100 nA, vO =1 V, RL= 10 kΩ
(c) vI =1 V, iI = 1 mA, vO =5 V, RL = 10Ω
Answer:
The solution to this question can be defined as follows:
Explanation:
In point (a):
[tex]v_i= 100 \ mV\\\\ i_I = 100 \mu \ A\\\\ v_O = 10 \ V\\\\ R_L = 100 \ \Omega \\\\i_L = \frac{V_0}{R_L} = \frac{10}{100} = 100 \ MA \\\\A_v = \frac{V_0}{V_i} = \frac{10}{100 \times 10^{-3}} =100 \\\\A_v(db) = 20 \lag (100) =40 \ db \\\\ A_i= \frac{i_L}{i_i} = \frac{100 \times 10^{-3}}{100 \times 10^{-6}} =1000 \\\\A_i(db) = 20 \lag (100) =60 \ db \\\\[/tex]
[tex]A_p= \frac{P_0}{p_i} =\frac{v_0 i_L}{v_i i_i} = \frac{ 10(100 \times 10^{-3})}{100 \times 10^{-6} \times 100 \times 10^{-3}} =100000\\\\ A_p(db) =10 \log (100000) =50 \ db \\\\[/tex]
In point (b):
[tex]v_i = 10 \mu V\\\\ i_i = 100 \ nA \\\\ v_O =1 \ V \\\\ R_L= 10 \ k \Omega \\\\i_0 = \frac{V_0}{R_L} = \frac{1}{10 \ K} = 100 \ \muA \\\\A_v = \frac{V_0}{V_i} = \frac{10}{10 \times 10^{-6}} =100000 \\\\A_v(db) = 20 \lag (100000) =100 \ db \\\\ A_i= \frac{i_0}{i_i} = \frac{100 \times 10^{-6}}{100 \times 10^{-9}} =1000 \\\\A_i(db) = 20 \lag (1000) =60 \ db \\\\[/tex]
[tex]A_p= \frac{P_0}{p_i} =\frac{v_0 i_0}{v_i i_i} = \frac{ 1 \times 100 \times 10^{-6})}{10 \times 10^{-6} \times 100 \times 10^{-9}} =100000000\\\\A_p(db) =10 \log (100000000) =80 \ db \\\\[/tex]
In point (C):
[tex]v_i =1\ V\\\\ i_I = 1 \ mA\\\\ v_O =5\ V \\\\ R_L = 10 \ \Omega \\\\i_0 = \frac{V_0}{R_L} = \frac{5}{10 } = 0.5 \ A \\\\A_v = \frac{V_0}{V_i} = \frac{5}{1} =5 \\\\A_v(db) = 20 \log 5 =13.97 \ db = 14 \db \\\\ A_i= \frac{i_0}{i_i} = \frac{0.5}{1\times 10^{-3}} =500 \\\\A_i(db) = 20 \log (500) =53.97 \ db = 54 \db \\\\[/tex]
[tex]A_p= \frac{P_0}{p_i} =\frac{v_0 i_0}{v_i i_i} = \frac{ 5 \times 0.5 }{1 \times 1 \times 10^{-3}} =2500\\\\A_p(db) =10 \log (2500) = 33.97 \ db = 34 \db\\\\[/tex]
what are ribosomes?
I'm tired. But I have insomnia. Big ugh moment. <.<.
Answer:
Ribosomes are organelles the make protein for the cell.
Some giant ocean waves have a wavelength of 25 m and travel at 6.5 m/s with a frequency of 0.26 HZ. What is the period of such a wave ?
Answer:
3.85s
Explanation:
Given parameters:
Wavelength = 25m
Velocity = 6.5m/s
Frequency = 0.26Hz
Unknown:
Period of the wave = ?
Solution:
The period of a wave is the inverse of the frequency of the wave.
Period = [tex]\frac{1}{frequency}[/tex]
Period = [tex]\frac{1}{0.26}[/tex] = 3.85s
I will give brainly
Defend Democritus' work on the atom and its contribution to the modern atomic model.
What is the function
of second plate in
parallel plate capacitor?
In which part of a lab report would be the following sentence most likely occur? “Since the data showed that the
Answer:
most likely be included in the analysis section of a lab report
Explanation:
This problem, a squid at rest suddenly sees a predator coming toward it and needs to escape. Assume the following:______.
(i) Including the water in its internal cavity, the squid has a total mass of 6.50 kg.
(ii) The mass of the water in its cavity is 1.75 kg.
(iii) In order to escape its predators, the squid needs to achieve an escape speed of 2.5 m/s.
Answer:
6.79 m/s
Explanation:
By applying the principle of conservation of momentum.
The total momentum = MV - mv = 0 (since the squid is beginning at rest)
the mass of the squid (M) in absence of water in its cavity = (6.5 - 1.75) kg
= 4.75 kg
speed of the squid (V) = 2.5 m/s
mass of the water expelled (m) = 1.75 kg
speed of the water (v) = ???
∴
4.75 × 2.5 = 1.75 × v
[tex]v = \dfrac{4.75 \times 2.5}{1.75 }[/tex]
v = 6.79 m/s
In one type of Combustion reaction, _______ is combined with oxygen to create heat and light.
A) nitrogen
B) carbon
C) hydrogen
Answer:
carbon.
Explanation:
the reaction would then create c02 as a product
David is driving a steady 28.0 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2.90 m/s^2 at the instant when David passes.
Required:
a. How far does Tina drive before passing David?
b. What is her speed as she passes him?
Answer:
Explanation:
Let t represent the time for Tina to catch David.
Hence, considering the equation of linear motion S = ut + 1/2at^2..... 1
For David u = 28.0 m/s where 'a' is set to nought
S = ut
S = 28t.......2
For Tina consider equation 1
Where acceleration = 2.90m/s^2 and u is set at nought
S = 1/2×2.90 m/s×t^2.......3
Equate 2 and 3
28t = 1.45t^2
Divide through by t
28 = 1.45t
t = 28/1.45
t = 19.31seconds
Now put the value of t into equation 3
S = 1/2×2.90 m/s×t^2.......3
= 1.45×20×20
= 580m
Tina must have driven 580meters before passing David
Considering the equation of linear motion : V^2 = U^2+2as
Where u is set at nought
V^2 = 2as
V^2 = 2×2.9×580
V^2 = 3364
V = √3364
V = 58m/s
Her speed will be 58m/s
(a) Tina should drive for 580 m, before passing the David.
(b) The speed of Tina during her passage through the David is 58 m/s.
Given data:
The initial velocity of the David is, u = 28.0 m/s.
The magnitude of acceleration is, [tex]a = 2.90 \;\rm m/s^{2}[/tex].
(a)
We can use the second kinematic equations of motion to obtain the distance covered by Tina, before passing the David. As per the second kinematic equation of motion,
[tex]s= u't + \dfrac{1}{2}at^{2}[/tex]
Here, u' is the initial speed of Tina and t is the time interval. Then,
Let t represent the time for Tina to catch David.
Hence, considering the equation of linear motion as,
S = ut + 1/2at²...............................................................(1)
Also,
S = ut
S = 28t ...........................................................................(2)
For Tina consider equation 1
S = 1/2×2.90t²................................................................(3)
Equate 2 and 3
28t = 1.45t²
28 = 1.45t
t = 28/1.45
t = 19.31 seconds
Now put the value of t into equation (3)
S = 1/2×2.90 t².
= 1.45×20×20
= 580m
Thus, we can conclude that Tina should drive for 580 m, before passing the David.
(b)
Now, using the third kinematic equation of motion to obtain the speed of Tina during her passage through David as,
v² = u²+2as
Solving as,
v² = 28.0² + 2(2.90)(580)
v = √3364
v = 58m/s
Thus, we can conclude that the speed of Tina during her passage through the David is 58 m/s.
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Question 1 of 10
What might happen to personal information when it is transferred using
digital signals?
A. Some information might be changed when the data are copied.
B. It might be accessed by someone who was not the intended
recipient.
C. The information might change while being transmitted because of
noise.
D. The information might change to analog, making it less reliable.
Answer:
its b for sure
Explanation:
Answer:
B. It might be accessed by someone who was not the intended
recipient
Of the charge Q initially on a tiny sphere, a portion q is to be transferred to a second, nearby sphere. Both spheres can be treated as particles. For what value of q/Q>0.5 will the electrostatic force between the two parts have 1/3 of the maximum possible value?
Answer:
Explanation:
Maximum value of force will be possible when both the sphere will have same charge . In that case charge on each sphere = Q / 2 =.5Q
F( max ) = k .5Q x .5Q / R²
=.25kQ² /R²
For the second case
F = k q ( Q-q)/ R²
F = .25kQ² /3R²
.25kQ² /3R² = k q ( Q-q)/ R²
.25 Q² = 3qQ - 3q²
3q² - 3qQ + .25 Q² = 0
q =
What is the instantaneous velocity of a freely falling object 11 s after it is released from a position of rest
Answer:
v= -107.8 m/s
Explanation:
Since the object is in free fall, this means that is moving at an accceleration equal to the one due to gravity.Since it starts at rest, we can apply the definition of acceleration, rearranging terms as follows:[tex]v_{f} = v_{o} + a*t = a*t = -g*t = 9.8m/s2*11s = -107.8 m/s (1)[/tex]
(Assuming as positive the upward direction)You perform nine (identical) measurements of the acceleration of gravity (units of m/s2): 10.1,9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, and 9.90. The true value is 9.81. Calculate the standard error of your results to ONE significant digit.
Answer:
0.01
Explanation:
Given the data:
10.1,9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, 9.90
True value = 9.81
Mean value :
Σx / n
Sample size, n = 9
(10.1 + 9.87 + 9.76 + 9.91 + 9.75 + 9.88 + 9.69 + 9.83 + 9.90) / 9
= 88.69 / 9
= 9.854
Standard deviation (σ) :
Sqrt (Σ(X - m)² / n)
[(10.1 - 9.854)^2 + (9.87 - 9.854)^2 + (9.76 - 9.854)^2 + (9.91 - 9.854)^2 + (9.75 - 9.854)^2 + (9.88 - 9.854)^2 + (9.69 - 9.854)^2 + (9.83 - 9.854)^2 + (9.90 - 9.854)^2] / 9
Sqrt(0.113824 / 9)
Sqrt(0.0126471)
σ = 0.1124593
Standard Error = σ / sqrt(n)
Standard Error = 0.1124593 / 9
Standard Error = 0.0124954
Standard Error = 0.01 ( 1 significant digit)
Particle A with charge q and mass ma and particle B with charge 2q and mass
mb, are accelerated from rest by a potential difference AV and subsequently
deflected by a uniform magnetic field into semicircular paths. The radii of the
trajectories by particle A and B are R and 3R, respectively. The direction of
the magnetic field is perpendicular to the velocity of the particle. Determine
their mass ratio?
An object is released from rest at a height H near and above the surface of Earth. As the object falls toward the surface, Earth’s atmosphere exerts a resistive force on the object such that it reaches a terminal velocity before it reaches the ground. Which of the following claims is true? Select two answers.
The system consisting of only the object is an open system.
The system consisting of only the object is an open system.
A
Earth’s atmosphere does negative work on the object as it falls toward the surface.
Earth’s atmosphere does negative work on the object as it falls toward the surface.
B
The change in the object’s kinetic energy from the instant it is released from rest, to the instant it reaches terminal velocity, is zero.
The change in the object’s kinetic energy from the instant it is released from rest, to the instant it reaches terminal velocity, is zero.
C
The total mechanical energy of the object-Earth system remains constant at all times in which the object is in motion.
Answer:
Second and Last Option Are Correct
Explanation:
How does heat from the sun get to earth?
describe measurement in our daily life
Sultan walks for 15 km at 35° south of east.
Which of the following journeys would result in the same displacement?
Answer:
☝
Explanation:
the diameter of the wheels on your car ( including the tires) is 25 inches. you are going to drive 250 miles today. each of your wheels is goingnto turn by an angle of
1. An engine absorbs 600 J of heat while doing 650 J of work. What is the change
in internal energy of the enginge? *
1250 J
-50 J
-1250 J
50 J
The change in internal energy of the engine is -50 joule. Hence, option (B) is correct.
What is law of conservation of energy?Energy cannot be created or destroyed, according to the law of conservation of energy. However, it is capable of change from one form to another. An isolated system's total energy is constant regardless of the types of energy present.
The absorb energy: Q = 600 Joule
Work done: W = 650 Joule.
Let, the change in internal energy of the engine= dU.
According to conservation of energy:
The absorb energy = change in internal energy + Work done
Q = dU + W
dU = Q - W
= 600 joule - 650 joule
= - 50 joule.
Hence, the change in internal energy of the engine is -50 joule.
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