Answers
Answer:
D) See the explanation below.
Explanation:
Heat transfer is always given from a body of higher temperature to a lower temperature. In this way the heat from the boiling water is transferred to the ice bars and the energy of the phase change from solid to liquid from the ice is transferred to the environment
Related Questions
What does the law of conservation of energy state? *
1. Total energy before transfer is equal to total energy after transfer
2. Total energy before transfer is less than total energy after transfer
3. Total energy before transfer is more than total energy after transfer
Answers
Answer:
total energy before transfer is equal to total energy after transfer
Answer:
1 is the correct one Answer
Which term is the product of force and distance? power work net force acceleration please help
Answers
Answer: WORK
Explanation:
Answer:
work
Explanation:
got it right on edge:p
what kind of mathematical relationship do you predict wll exist between velocity and number of filters
Answers
Answer:
[tex]PK_T = constant[/tex]
Explanation:
[tex]PK_T[/tex] = constant
terminal velocity is the maximum const. velocity to achieved by the partical when particle is free fall with the effect of gravity
so that here
eq of motion is
Fnet = mg - kv ................1
here a = 0
v = terminal velocity
0 = mg - kV
v = [tex]\frac{mg}{k}[/tex] m/s
prove principle of conversation of linear momentum in accordance to principle of classical relativity
Answers
Answer:
Cuz im batman hahahhahah
1. Sodium-24 has a half-life of 15 hours. How much sodium-24 will remain in an
26.0 g sample after 60 hours?
Answers
Answer:
1.625g
Explanation:
The half - life of a radioactive substance is the time taken for half of it to decay to half of its original composition.
For the given specie, the original composition is 26g and the half life is 15hrs:
At time 0hrs Mass = 26g
For the first half life at time 15hrs Mass = 13g
30hrs Mass = 6.5g
45hrs Mass = 3.25g
60hrs Mass = 1.625g
What do you think causes the changes in sunrise and sunset times over the course of the year?
Answers
Answer:
The combination of Earth's elliptical orbit and the tilt of its axis results in the Sun taking different paths across the sky at slightly different speeds each day. This gives us different sunrise and sunset times each day.
Explanation:
A 105 kg football player runs at 8.5 m/s and plows into the back of an 85 kg referee running at 3.5 m/s on the field causing the referee to fly forward at 6.0 m/s.
What must the momentum of the football player be after the collision?
Answers
Answer:
680 Kg.m/s
Explanation:
Mass of player; m_p = 105 kg.
Speed of player before Collision; v_pi = 8.5 m/s
Mass of referee; m_r = 85 kg
Speed of referee before collision; v_ri = 3.5 m/s
Speed of referee after collision; v_rf = 6 m/s
From conservation of momentum,
Initial momentum = final momentum
Thus;
(m_p × v_pi) + (m_r × v_ri) = (m_p × v_pf) + (m_r × v_rf)
Where (m_p × v_pf) is the momentum of the player after collision.
Thus, Plugging in the relevant values, we have;
(105 × 8.5) + (85 × 3.5) = (m_p × v_pf) + (85 × 6)
(m_p × v_pf) = (105 × 8.5) + (85 × 3.5) - (85 × 6)
(m_p × v_pf) = 680 Kg.m/s
The resistance of a wire depends on its length i and on its cross sectional area A the resistance is
Answers
Answer:
The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area
Explanation:
A 1250-kg car moves at 20.0 m/s. How much work must be done on the car to increase its speed to 30.0 m/s.
Answers
Answer:
312,497.5Joules
Explanation:
Work done = force × distance
W = FS
Get the force
F = ma
F = 1250×9.8
F = 12250N
Get the distances using the equation of motion
v² = u² +2gS
30² =20²+2(9.8)S
900 =400+19.6S
900-400 =19.6S
500 = 19.6S
S = 500/19.6
S = 25.51m
Get the work done
Work done = 12250×25.51
Workdone = 312,497.5Joules
Answer:
3.13 x 10^5 J
Explanation:
The work done increases the car's kinetic energy so that its new speed is 30 m/s. So, we set t, the net work required to increase the car's speed equal to the change in the car's kinetic energy.
Wnet = △K = Kf - Ki = 1/2mvf^2 - 1/2mvi^2 = 1/2m(vf^2-vi^2) = 1/2(1250kg)((30m/s)^2 - (20m/s)^2) = 3.13x10^5 J
h. If the mass of an object is 390 g and its volume is 50 cm?, then its density is ....... g/cm' or ....... kg/m”.
- The relative density of the object in part "h" is
Edit: nvm I'm an idiot I remembered this but thanks anyway
Answers
Answer: See explanation
Explanation:
The density would like be calculated by dividing the mass by volume. This will be:
Density = Mass / Volume
Density = 390 / 50
Density = 7.8g/cm
what can you conclude about the force of gravity?
Answers
Answer:
All objects, irrespective of their mass, experience the same acceleration g when falling freely under the influence of gravity at the same point on the Earth. If gravity is the only force acting on an object, the sum of kinetic energy and gravitational energy is constant. ...
Answer:
Gravity, also called gravitation, in mechanics, the universal force of attraction acting between all matter. ... On Earth all bodies have a weight, or downward force of gravity, proportional to their mass, which Earth's mass exerts on them. Gravity is measured by the acceleration that it gives to freely falling objects.The Earth's gravitational force accelerates objects when they fall. It constantly pulls, and the objects constantly speed up.Gravity is the force of attraction between two objects with mass and is dependent on the distance between these objects. Gravity is the force that repels two objects that have opposite charges. It is dependent upon the charges of the objects.
Explanation:
explain why work is not considered a form of energy.
Answers
Hope this helps.
If force c is 30n and force d us 20 N what is the size of the horizontal force
Answers
Hope this helped
(Are these statements about the spring constant true or
false?
a. The spring constant is a measure of the stiffness of
the spring.
b. The spring constant tells you how many newtons of
force it takes to stretch the spring one meter.
c. If a spring stretches easily, it will have a high spring
constant.
d. The spring constant of a spring varies with the
amount of stretch or compression of the spring.
Answers
Answer:
true
true
false
true
pls follow me and thank you
What are the three longest wavelengths for standing waves on a270-cm-long string that is fixed at both ends
Answers
Answer:
The answer is below
Explanation:
a) What are the three longest wavelengths for standing waves on a 270-cm-long string that is fixed at both ends? b. If the frequency of the second-largest wavelength is 50.0 Hz, what is the frequency of the third-longest wave length?
Solution:
a) The wavelengths (λ) for standing waves is given by the formula:
[tex]\lambda_m=\frac{2*length\ of\ string}{m}\\\\Where\ m=1,2,3,.\ .\ .\\\\Given\ that\ length\ of\ string = 270\ cm=2.7\ m,\ m=1,2,3(three\ longest\ wavelengths)\\\\Hence:\\\\\lambda_1=\frac{2(2.7)}{1}=5.4\ m\\\\\lambda_2=\frac{2(2.7)}{2}=2.7\ m \\\\\lambda_3=\frac{2(2.7)}{3}=1.8\ m[/tex]
b) The frequency (f) and wavelength (λ) is given by:
fλ = constant
Hence:
[tex]f_2\lambda_2=f_3\lambda_3\\\\f_2=50\ Hz\\\\2.7*50=f_3(1.8)\\\\f_3=\frac{2.7*50}{1.8} \\\\f_3=75\ Hz[/tex]
The three longest wavelengths for the standing waves on a 270-cm long string that is fixed at both ends are:
1. 5.4 meters.
2. 2.7 meters.
3. 1.8 meters.
Given the following data:
Length of string = 270 cm to m = [tex]\frac{270}{100} =2.7\;m[/tex]To determine the three (3) longest wavelengths for these standing waves:
Mathematically, the wavelength for standing waves is given by the formula:
[tex]\lambda_n = \frac{2L}{n}[/tex]
Where:
[tex]\lambda_n[/tex] is the wavelength for standing waves.L is the length of string.Note: n = 1, 2, and 3.
When n = 1:
[tex]\lambda_1 = \frac{2\times 2.7}{1} \\\\\lambda_1 = 5.4 \;meters[/tex]
When n = 2:
[tex]\lambda_2 = \frac{2\times 2.7}{2} \\\\\lambda_2 = 2.7 \;meters[/tex]
When n = 3:
[tex]\lambda_3 = \frac{2\times 2.7}{3} \\\\\lambda_3 =\frac{5.4}{3} \\\\\lambda_3 = 1.8 \;meters[/tex]
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A 0.015 kg marble moving to the right at 0.225 m/s makes an elastic head-on collision with a 0.03 kg shooter marble moving to the left at 0.18 m/s. After the collision, the smaller marble moves to the left at 0.315 m/s. What is the velocity of the 0.03 kg marble after the collision
Answers
Answer:
0.135m/s
Explanation:
Using the law of conservation of momentum
m1u1+m2u2 = m1v1 +m2v2
m1 and m2 are the masses
u1 and u2 are the initial velocities
v1 and v2 are thee final velocities
Substitute the given parameters and get v2
0.015(0.225)+0.03(0.18) = 0.015(0.315)+0.03v2
0.003375+0.0054 = 0.004725 + 0.03v2
0.008775 - 0.004725 = 0.03v2
0.00405 = 0.03v2
v2 = 0.00405/0.03
v2 = 0.135m/s
Hence the final velocity of the 0.03kg marble after collision is 0.135m/s
In a chemical reaction, the mass of the reactants is 13,3 grams. Which of the
following could represent the mass of the products?
Answers
Answer:
my assuytyyhyyyyyy gftdrrdtiifyb tvhyvth rv b yy
A 10 kg weight is suspended in the air by a strong cable. How much work is done, per unit time, in suspending the weight
Answers
Answer:
There is no work done.
Explanation:
Given the following data;
Mass = 10 kg
To find the work done?
In Physics, work done can be defined as the amount of energy transfered when an object or body is moved over a distance due to the action of an external force.
Mathematically, work done is given by the formula;
Work done = force * distance
[tex] W = F * d[/tex]
Where,
W is the work doneF represents the force acting on a body. d represents the distance covered by the body.However, the weight suspended in the air by a strong cable does not move or experience any form of displacement. Therefore, there is no work done.
A car and a truck are both traveling at the constant speed of 35 mi/h; the car is 40 ft behind the truck. The driver of the car wants to pass the truck, i.e., he wishes to place his car at B, 40 ft in front of the truck, and then resume the speed of 35 mi/h; The maximum acceleration of the car is 5 ft/s2 and the maximum deceleration obtained by applying the brakes is 20 ft/s2. What is the shortest time in which the driver of the car can complete the passing operation? (assuming that the driver of the car does not pay any attention to the speed limit
Answers
Answer:
The shortest time in which the driver can complete the operation is approximately 6.32456 seconds
Explanation:
The given parameters are;
The speed of the truck and the car = 35 mi/h ≈ 51.33 ft./s
Let "t" represent the time it takes the car to pass the truck by 40 ft., we have;
The distance covered by the truck = 35 mi/h × t
The distance covered by the car = 35 mi/h × t + 80 ft = 51.33
Let the distance over which the car accelerates = d
We have;
d = 51.33×t₁ + 1/2×5×t₁²
51.33·t + 80 - d = (51.33 + 5·t₁)·(t - t₁) - 1/2·20·(t - t₁)²
51.33 = 51.33 + 5·t₁ - 20·(t - t₁)
∴ 5·t₁ = 20·(t - t₁)
5·t₁ = 20·t - 20·t₁
25·t₁ = 20·t
t₁ = 4·t/5
(51.33 + 5·t₁)·(t - t₁) - 1/2·20·(t - t₁)² + 51.33×t₁ + 1/2×5×t₁² = 51.33·t + 80
We get;
(51.33 + 5·(4·t/5))·(t - (4·t/5)) - 1/2·20·(t - (4·t/5))² + 51.33×(4·t/5) + 1/2×5×(4·t/5)² = 51.33·t + 80
(51.33 + 4·t)·(t/5) - 2·t²/5 + 51.33 × (4·t/5) + 8·t²/5 = 51.33·t + 80
51.33×t/5 + 51.33×4·t/5 + 4·t²/5 - 2·t²/5 + 8·t²/5 = 51.33·t + 80
51.33·t + 10·t²/5 = 51.33·t + 80
2·t² = 51.33·t + 80 - 51.33·t = 80
t² = 80/2 = 40
t = √40 = 2·√10 ≈ 6.32456
The shortest time in which the driver can complete the operation is "t" ≈ 6.32456 seconds
please help me this is timed .
Find x if a = 3.0 m/s^2
Answers
Answer:
x = 50 N
Explanation:
Given that we have a net force, a mass, and acceleration, we can use the fundamental formula for force found in newton's second law which is F = m × a.
Given a mass of 150 kg, and an acceleration 3.0m/s². We can substitute these two values in our formula to calculate the magnitude of these forces or it's net force to identify the unknown force acting on our known force for this situation to work.
_______
F (Net force) = F2 (Second force which we are given) - F1 (First force) = m × a
m (mass which we are given) = 150 kg
a (acceleration which we are given) = 3.0m/s
________
So F = m × a → F2 - F1 = m × a →
500 - F1 = 150 × 3.0 → 500 - F1 = 450 →
-F1 = -50 → F1 = 50
A low rider truck accelerates from 0 m/s to 30.0 m/s with a displacement of 50m. What is the vehicle's acceleration?
a. 2.00m/s
b. 5.00m/s
c. 9.0m/s
d. 10.0m/s
Answers
Answer:
a = 9 [m/s²]
Explanation:
To solve this problem we must use the following equation of kinematics.
[tex]v_{f}^{2}=v_{o}^{2} +2*a*x[/tex]
where:
Vf = final velocity = 30 [m/s]
Vo = initial velocity = 0
a = acceleration [m/s²]
x = displacement = 50 [m]
Now replacing:
[tex]30^{2}=0 +2*a*50\\100*a=900\\a=9[m/s^{2} ][/tex]
2- A car on a straight highway goes in the positive direction for 8 km and then backs up for 3.6 km. What are the distance and displacement covered by the train?
Answers
Answer:
11.6km
4.4km in the negative direction
Explanation:
Distance is the total length of path covered and traveled by a body.
So, for this car on a straight line;
Total distance = 8km + 3.6km = 11.6km
Displacement is the distance traveled along a path and the direction it takes.
It is a vector quantity with magnitude and directional attributes.
For this journey;
Displacement = 8km - 3.6km = 4.4km in the negative direction.
What is the distance of separation between objects of masses 5.6 x 10 5 kg and 8.8 x 10 6 kg if the force of gravity between them is 440 N
Answers
Answer:
2.87m
Explanation:
Using the law of gravitation to solve this question
F = GMm/r²
G is the gravitational constant
M and m are the masses
r is the distance between the masses
Substitute the given values
G = 6.67×10^-11 m³/kgs²
M =8.8 x 10^6 kg
m = 5.6 x 10^5 kg
F =440N
400 = 6.67×10^-11×8.8 x 10^6 ×5.6 x 10^5/r²
400r² = 328.698×10
400r² = 3286.98
r² = 3286.98/400
r² = 8.21745
r = √8.21745
r = 2.87m
Hence the distance of separation is 2.87m
The distance of separation will be "0.86 m".
Given:
Mass,
[tex]m_1 = 5.6\times 10^5 \ kg[/tex][tex]m_2 = 8.8\times 10^6 \ kg[/tex]Force,
[tex]F = 440 \ N[/tex]As we know,
→ [tex]F = \frac{Gm_1 m_2}{r^2}[/tex]
By putting the values, we get
→ [tex]440=\frac{6.67\times 10^{-11}\times 5.6\times 10^5\times 8.8\times 10^6}{r^2}[/tex]
→ [tex]r^2 = \frac{6.67\times 5.6\times 8.8}{440}[/tex]
→ [tex]r^2 = 0.74704[/tex]
→ [tex]r = \sqrt{0.74704}[/tex]
→ [tex]= 0.86 \ m[/tex]
Thus above approach is correct.
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A helicopter flies with an air speed of 175 km/h, heading south. The wind is blowing at 85 km/h to the east relative to the ground. Calculate the speed and direction of the helicopter.
Answers
Answer:
154° at 195 km/h
Explanation:
The helicopter is moving south at 175 km/h, relative to the wind.
But the wind is moving east at 85 km/h, relative to the ground.
This means that the helicopter is moving south east relative to the ground.
Every hour, the helicopter will move 175 km to the south and 85 km to the east, relative to the ground.
This means that we can determine the speed and direction of the helicopter using a right triangle and simple trigonometry.
Refer to the triangle b1.
The distance traveled by the helicopter in 1 hour is denoted by d.
d is the hypotenuse of the right triangle.
Using the Pythagorean Theorem, we can calculate d to be 195 km (rounded to 3 s. f.)
Hence the helicopter is traveling at 195 km/h relative to the ground.
To calculate the direction we use,
tan (x) = opposite/adjacent = 85/175
So the angle x is,
[tex]x = arctan (\frac{85}{175} )[/tex] = 25.9°
Relative to the North, the helicopter is moving at 180° - 25.9° = 154° (rounded to 3 s. f.)
Two point charges, Q1 and Q2, are separated by a distance R. If the magnitudes of both charges are tripled and their separation is also tripled, what happens to the magnitude of the electrostatic force that each charge exerts on the other
Answers
Answer:
The force remains the same.
Explanation:
Let the magnitude of the forces in each case be F1 and F2 respectively.
The charges are Q1 and Q2.
The distance of separation between them is R.
Hence, for F1;
F1 = KQ1Q2/R^2
For F2:
F2 = K * 3Q1 * 3Q2/(3R)^2
F2 = 9KQ1Q2/9R^2
F2 =KQ1Q2/R^2
Hence F1=F2
The force is the same in both cases!
15 points! Answer FAST!
When heat is transferred, it flows_____.
A. From the warmer object to the cooler object only if the cooler object is an insulator
B. From the cooler object to the warmer object regardless of the materials
C. From the warmer object to the cooler object regardless of the materials
D. From the warmer object to the cooler object only if the warmer object is a conductor.
Answers
Answer:heat brings it up then down
Explanation:
How much extra water does a 147-lb concrete canoe displace compared to an ultra lightweight 36-lb Kevlar canoe of the same size carrying the same load
Answers
Answer:
1.7791 ft^3
Explanation:
The computation of the extra water is as follows;
Here we use the following formula
Volume of the water = ÷ Weight of concrete canoe ÷ value of the weight of the specific water
= 147 lb ÷ 62.4 lb /ft^3
= 2.356 ft^3
Now the volume of water occupied in ultra lightweight
= 36 lb ÷ 62.4 lb /ft^3
= 0.5769 ft^3
Now the water volume displaced is
= 2.356 - 0.5769
= 1.7791 ft^3
A constant force of 20. newtons applied to a box causes it to move at a constant speed of 4.0 meters per second. How much work is done in the boxin 6.0 seconds?
1) 240 joules
2) 120 joules
3) 480 joules
4) 80. joules
Answers
The work done is 480 Joules.
Option 3 is correct.
The work done is computed as multiplication of force and displacement.
Given that, Force = 20 N
Displacement = 4 meter per second
For 6 second,
Displacement [tex]=4*6=24m[/tex]
[tex]Workdone=Force*displacement\\\\Workdone=20*24=480J[/tex]
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Two objects are dropped from a bridge, an interval of 1.00 s apart. What is their separation 1.00 s after the second object is released
Answers
Answer:
171.5m
Explanation:
The velocity of sound in water = 343m/s
Time taken = 1.00secs
using the formula to calculate the distance
2x = vt
x is the distance
v is the speed of sound
t is the time
x = vt/2
x = 343(1)/2
x = 171.5m
hence their separation 1.00 s after the second object is released is 171.5m
Which three statements describe mechanical waves?
A. The waves can travel through empty space.
B. The waves transfer energy by causing particles of matter to
move.
C. The waves need matter to transfer energy.
D. The waves can transfer energy through solids, liquids, and gases.
Answers
The three statements that are correct about the mechanical waves are options B, C, and D.
Mechanical waves:Mechanical waves are produced due to disturbances in any matter. Which clearly indicates that the mechanical waves cannot travel through empty space.
The mechanical waves require matter to transfer energy through oscillation in the matter. So it is true that the waves transfer energy by causing particles of matter to move.
Also, the mechanical waves need matter to transfer energy, the waves can transfer energy through solids, liquids, and gases.
An example of a mechanical wave is the sound produced by the guitar strings.
Learn more about mechanical waves:
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