An inventor develops a stationary cycling device by which an individual, while pedaling, can convert all of the energy expended into heat for warming water. How much mechanical energy is required to increase the temperature of 300 g of water (enough for 1 cup of coffee) from 20°C to 95°C? (1 cal = 4.186 J, the specific heat of water is 4186 J/kg⋅°C)
Answer:
Emec = 94050 [J]
Explanation:
In order to solve this problem, we must understand that all thermal energy is converted into mechanical energy.
The thermal energy can be calculated by means of the following expression.
[tex]Q=m*C_{p}*(T_{final}-T_{initial})[/tex]
where:
Q = heat [J]
Cp = specific heat of water = 4186 [J/kg*°C]
m = mass = 300 [g] = 0.3 [kg]
T_final = 95 [°C]
T_initial = 20 [°C]
Now we can calculate the heat, replacing the given values:
[tex]Q=0.3*4180*(95-20)\\Q= 94050[J][/tex]
Since all this energy must come from the mechanical energy delivered by the exercise bike, and no energy is lost during the process, the mechanical energy must be equal to the thermal energy.
[tex]Q=E_{mec}\\E_{mec}=94050[J][/tex]
A 2 kg toy car moves at a speed of 5 m/s. If a child applies a 5N force for 2 m in the same direction the car is already moving, what is the change in kinetic energy of the car?
Answer:
[tex]10\: \mathrm{J}[/tex]
Explanation:
The kinetic energy of an object is [tex]KE=\frac{1}{2}mv^2[/tex], where [tex]m[/tex] is the mass of the object and [tex]v[/tex] is the velocity of the object.
The toy car's initial kinetic energy is [tex]KE_{i}=\frac{1}{2}\cdot 2\cdot 5^2=25\: \mathrm{J}[/tex].
After the child applies a 5N force on it in the same direction, its velocity will increase but its mass will stay the same.
To find the final velocity of the toy car, we can use kinematic equation [tex]v_f^2=v_i^2+2a\Delta x, \\ v_f=\sqrt{v_i^2+2a\Delta x}[/tex]
We are given [tex]v_i=5\: \mathrm{m/s}[/tex] and [tex]\Delta x = 2\: \mathrm{m}[/tex].
To find acceleration:
[tex]F=ma, a=\frac{F}{m}=\frac{5}{2}=2.5\: \mathrm{m/s^2}[/tex].
Now substitute [tex]v_i=5\: \mathrm{m/s}, \: a=2\: \mathrm{m/s^2}, \: \Delta x = 2\: \mathrm{m}[/tex] into [tex]v_f=\sqrt{v_i^2+2a\Delta x}[/tex] to get [tex]v_f\approx 5.92\: \mathrm{m/s}[/tex].
Using this, we can find the final kinetic energy of the toy car is [tex]KE_f=\frac{1}{2}\cdot 2\cdot 5.92^2[/tex].
Thus, the change in kinetic energy is [tex]KE_f-KE_i=\frac{1}{2}\cdot2\cdot 5.92^2-\frac{1}{2}\cdot 2\cdot 5^2=\fbox{$10\: \mathrm{J}$}[/tex] (one significant figure).
The change in the kinetic energy of the car is 10 J.
The given parameters;
mass of the car, m = 2 kginitial velocity of the car, u = 5 m/sforce applied by the child, F = 5 Ndistance traveled, s = 2 mThe acceleration of the car is calculated as follows;
[tex]F = ma\\\\a = \frac{F}{m} \\\\a = \frac{5}{2} \\\\a = 2.5 \ m/s^2[/tex]
The final velocity of the car is calculated as follows;
[tex]v^2 = u^2 + 2as\\\\v = \sqrt{u^2 + 2as} \\\\v = \sqrt{5^2 \ + \ 2(2.5)(2)} \\\\v = 5.92 \ m/s[/tex]
The change in the kinetic energy of the car is calculated as follows;
[tex]\Delta K.E = \frac{1}{2} m(v^2 - u^2)\\\\\Delta K.E = \frac{1}{2} \times 2 \times (5.92^2\ - \ 5^2)\\\\\Delta K.E = 10 \ J[/tex]
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What statement is not an example of Newton’s first law of motion
Answer:
c
Explanation:
im smart....................... i think
This law is about inertia, and the law displayed in A is Newton's third law of equal and opposite reactions, so option A is correct.
What is Newton’s first law of motion?The basis of classical mechanics is laid out in three assertions known as Newton's laws of motion, which were first articulated by English physicist and mathematician Isaac Newton. These laws describe the relationships between forces acting on a body and its motion.
Unless a force acts on a body that is at rest or moving in a straight line at a constant speed, Newton's first law asserts that it will continue to be at rest or move in that direction.
This law is about inertia (an object wanting to stay in its state of motion) and the law displayed in A is Newton's third law of equal and opposite reactions, therefore, it is not an example of Newton’s first law of motion.
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At a certain instant, a ball is thrown downward with a velocity of 8.0 m/s from a height of 40 m. At the same instant, another ball is thrown upward from ground level directly in line with the first ball with a velocity of 12 m/s. Find (a) the time when the balls collide and (b) the height at which they collide. Take g = 10 m/s2
Answer:
(a) The two balls collide [tex]2\; \rm s[/tex] after launch.
(b) The height of the collision is [tex]4\; \rm m[/tex].
(Assuming that air resistance is negligible.)
Explanation:
Let vector quantities (displacements, velocities, acceleration, etc.) that point upward be positive. Conversely, let vector quantities that point downward be negative.
The gravitational acceleration of the earth points dowards (towards the ground.) Therefore, the sign of [tex]g[/tex] should be negative. The question states that the magnitude of [tex]g\![/tex] is [tex]10\; \rm m \cdot s^{-2}[/tex]. Hence, the signed value of [tex]\! g[/tex] should be [tex]\left(-10\; \rm m \cdot s^{-2}\right)[/tex].
Similarly, the initial velocity of the ball thrown downwards should also be negative: [tex]\left(-8.0\; \rm m \cdot s^{-1}\right)[/tex].
On the other hand, the initial velocity of the ball thrown upwards should be positive: [tex]\left(12\; \rm m \cdot s^{-1}\right)[/tex].
Let [tex]v_0[/tex] and [tex]h_0[/tex] denote the initial velocity and height of one such ball. The following SUVAT equation gives the height of that ball at time [tex]t[/tex]:
[tex]\displaystyle h(t) = \frac{1}{2}\, g \cdot {t}^{2} + v_0 \cdot t + h_0[/tex].
For both balls, [tex]g = \left(-10\; \rm m \cdot s^{-2}\right)[/tex].
For the ball thrown downwards:
Initial velocity: [tex]v_0 = \left(-8.0\; \rm m \cdot s^{-1}\right)[/tex].Initial height: [tex]h_0 = 40\; \rm m[/tex].[tex]\displaystyle h(t) = -5\, t^{2} + (-8.0)\, t + 40[/tex] (where [tex]h[/tex] is in meters and [tex]t[/tex] is in seconds.)
Similarly, for the ball thrown upwards:
Initial velocity: [tex]v_0 = \left(12\; \rm m \cdot s^{-1}\right)[/tex].Initial height: [tex]h_0 = 0\; \rm m[/tex].[tex]\displaystyle h(t) = -5\, t^{2} + 12\, t[/tex] (where [tex]h[/tex] is in meters and [tex]t[/tex] is in seconds.)
Equate the two expressions and solve for [tex]t[/tex]:
[tex]-5\, t^{2} + (-8.0)\, t + 40 = -5\, t^{2} + 12\, t[/tex].
[tex]t = 2[/tex].
Therefore, the collision takes place [tex]2\, \rm s[/tex] after launch.
Substitute [tex]t = 2[/tex] into either of the two original expressions to find the height of collision:
[tex]h = -5\times 2^{2} + 12 \times 2 = 4\; \rm m[/tex].
In other words, the two balls collide when their height was [tex]4\; \rm m[/tex].
The time the two balls collide is 0.4 seconds while the height at which they collide is 4m
The given parameters are :
Initial Velocity U = 8m/s
Height H = 40m
For the second ball, the initial velocity = 12m/s
a.) For the first ball, the height attained at the point of collision will be
h = ut + 1/2gt^2
h = 8t + 1/2 x 10t^2 ........ (1)
For the second ball, the height attained at the point of collision will be
h = 12t - 1/2 x 10t^2 .........(2)
Since the height will be the same for the two balls, equate the two equations
8t + 10t^2 = 12t - 10t^2
Collect the like term
8t - 12t = -5t^2 - 5t^2
-4t = -10^2
10t = 4
t = 4/10
t = 0.4s
b.) Substitute time t in any of the equation to find the height
h = 12(0.4) - 0.5 x 10(0.4)^2
h = 4.8 - 0.8
h = 4m
Therefore, the time the two balls collide is 0.4 seconds while the height at which they collide is 4m
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A block of mass m = 4.4 kg slides from left to right across a frictionless surface with a speed 9.2 m/s It collides in a perfectly elastic collision with a second block of mass M that is at rest. After the collision, the 4.4-kg block reverses direction, and its new speed is 2.5 m/s The block of mass M travels to the right at a speed V of 6.7 m/s what is M
Answer:
[tex]m_2=6.3\:\mathrm{kg}[/tex]
Explanation:
In a perfectly elastic collision, the total kinetic energy of the system is maintained. Therefore, we can set up the following equation:
[tex]\frac{1}{2}m_1{v_1}^2+\frac{1}{2}m_2{v_2}^2=\frac{1}{2}m_1{v_{1'}}^2+\frac{1}{2}m_2{v_{2'}}^2[/tex]
Since the second block was initially at rest, [tex]\frac{1}{2}m_2{v_2}^2=0[/tex].
Plugging in all given values, we have:
[tex]\frac{1}{2}m_1{v_1}^2=\frac{1}{2}m_1{v_{1'}}^2+\frac{1}{2}m_2{v_{2'}}^2,\\\\\frac{1}{2}\cdot4.4\cdot9.2^2=\frac{1}{2}\cdot 4.4 \cdot (-2.5)^2+\frac{1}{2}\cdot m_2\cdot 6.7^2,\\m_2=\fbox{$6.3\:\mathrm{kg}$}[/tex].
The force of gravity acting on an object is directed through this
center of gravity and toward the center of the
Explanation:
Every object has a center of gravity. ... The force of gravity acting on an object is directed through this center of gravity and toward the center of the earth. The object's weight, W, can be represented by a vector directed down (along the line the object would fall if it were dropped).
It should be towards the center of the earth.
The following information should be considered:
Each and every object contains the center of gravity. The force of activity acted on an object that director via his gravity center & towards the center of the earth.Learn more: brainly.com/question/17429689
Two identical 0.25 kg balls are involved in a head-on collision. Ball A is initially travelling at 3.5 m/s, and ball B is initally at rest. Determine the velocity of each ball after the collision.
Answer:
a) mv(final):<0,0,0> minus mv(initial):<25,0,0> = <-25,0,0>
b) mv(final):<25,0,0> minus mv(initial):<0,0,0> = <25,0,0>
c) conservation of momentum makes it <0,0,0>
for a-b-c, momentum_system + momentum_surroundings = 0
Explanation:
Hope this helps
The velocity of each ball after the collision is 1.75 m/s
Law of conservation of momentum states that:
Total momentum before collision = Total momentum after collision
m₁u₁ + m₂u₂ = (m₁ + m₂)v
Where m₁, m₂ is the mass of object, u₁, u₂ is the initial velocity before collision and v is the final velocity after collision
Given that: m₁ = m₂ = 0.25 kg, u₁ = 3.5 m/s, u₂ = 0, hence:
0.25(3.5) + 0.25(0) = (0.25 + 0.25)v
v = 1.75 m/s
The velocity of each ball after the collision is 1.75 m/s
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A ray of monochromatic light is incident on a plane mirror at and angle of 30. The angle of reflection for the light is
1)15
2)30
3)60
4)90
Answer:
30 degrees
Explanation: reflection, same angle
For a ray of monochromatic light is incident on a plane mirror at and angle of 30°. The angle of reflection for the light is 30°.
Reflection occurs when radiation bounces off from a surface. Light is an electromagnetic wave and it can be reflected. According to the laws of reflection, the angle of incidence is equal to the law of reflection.
Hence, for a ray of monochromatic light is incident on a plane mirror at and angle of 30°. The angle of reflection for the light is 30°.
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10) A soccer player kicks a soccer ball (m = 0.42 kg) accelerating from rest to 32.5m/s in 0.21s. Determine the force that sends soccer ball towards the goal.
G
U
E
S
S
Formula
11) Small rockets are fired to make small adjustments in the speed of a satellite. A certain small rocket can change the velocity of a 72,000kg satellite from 0m/s to 0.63m/s in 1296s. What force is exerted by the rocket on the satellite?
G
U
E
S
S
Formula
please I need help I don't understand it and I had to deliver it yesterday helpp:(
Answer:
10. 65 N
11. 35 N
Explanation:
10. Determination the force that sends the soccer ball towards the goal.
We'll begin by calculating the acceleration of the ball. This can be obtained as follow:
Initial velocity (u) of the ball = 0 m/s
Final velocity (v) of the ball = 32.5m/s time (t) = 0.21 s
Acceleration (a) of the ball =?
a = (v – u) /t
a = (32.5 – 0) / 0.21
a = 32.5 / 0.21
a = 154.76 m/s²
Finally, we shall determine the force that sends the soccer ball towards the goal. This can be obtained as follow:
Mass (m) of the ball = 0.42 kg
Acceleration (a) of the ball = 154.76 m/s²
Force (F) =?
F = ma
F = 0.42 × 154.76
F = 65 N
Thus, the force that sends soccer ball towards the goal is 65 N
11. Determination of the force exerted by the rocket on the satellite.
We'll begin by calculating the acceleration of the satellite. This can be obtained as follow:
Initial velocity (u) of satellite = 0 m/s
Final velocity (v) of satellite = 0.63 m/s
Time (t) = 1296 s
Acceleration (a) of the satellite =?
a = (v – u) /t
a = (0.63 – 0) / 1296
a = 0.63 / 1296
a = 4.861×10¯⁴ m/s²
Finally, we shall determine the force exerted by the rocket on the satellite. This can be obtained as follow:
Mass (m) of the satellite = 72000 Kg
Acceleration (a) of the satellite = 4.861×10¯⁴ m/s²
Force (F) =?
F = ma
F = 72000 × 4.861×10¯⁴
F = 35 N
Thus, the force exerted by the rocket on the satellite is 35 N
In a parallel circuit, if one connection is broken
A. all of the connections stop working. B. B only broken one will stop working
C. Nothing happens to circuit.
Explain:
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
Our Sun’s mass is 1.0 and our Earth’s mass is 2.0. The distance is standard as given on the simulation. Describe the path of the Earth.
Answer:
Earth orbits the Sun at an average distance of 149.60 million km (92.96 million mi), and one complete orbit takes 365.256 days (1 sidereal year), during which time Earth has traveled 940 million km (584 million mi).
Explanation:
a group of students working in a high school chemistry lab believe they have discovered a new element! how exciting! upon further testing by scientists (with better equipment),it is found that the element contains 74 protons and 110 neutrons.
Explanation:
From the experiment:
Number of protons = 74
Number of neutrons = 110
Number of protons in an element is the atomic number of the element. It is used to locate and position and element on the periodic table.
For a neutral or uncharged atom, the number of protons is the same as the number of electrons.
The element whose number of protons or atomic number if 74 is Tungsten
Mass number = 74 + 110 = 184g/mol
Explain how momentum is determined and conserved.
ASAP!!
Answer: momentum is determined and conserved.
Answer:
monument is determined
Which term describes the relatively constant internal conditions of an organism?
Single Cellular
Uniformity
Homeostasis
Organ System
Answer:
The correct answer is C. Homeostasis
Explanation:
Homeostasis describes the relatively constant internal physical conditions of an organism.
A volcano erupts spewing ash into the air and sending lava flowing down the side of the mountain. Looking at the image explain how the eruption of the volcano involves each of the Earth’s four spheres
Answer:
A lateral eruptions or lateral blast is a volcanic eruption which is directed laterally from a volcano rather than upwards from the summit. Lateral eruptions are caused by the outward expansion of flanks due to rising magma. Breaking occurs at the flanks of volcanoes making it easier for magma to flow outward.
Explanation:
Question 12 of 25
What type of energy is thermal energy?
O A. Chemical energy
O B. Nuclear energy
O C. Kinetic energy
O D. Gravitational potential energy
It’s C kinetic energy
In the diagram at the right, each grid square is 1.0μmx 1.0 μm. If Mass A = Mass C = 10kg and Mass B = 28 kg, determine the gravitational field at the location marked by the red dot.
Answer:
B
Explanation:
8-2 is 6
times that times 8 u get B
Consider a simple pendulum consisting of a massive bob suspended from a fixed point by a string. Let T denote the time (the period of the pendulum) that it takes the bob to complete one cycle of oscillation (the time it takes for the pendulum to swing back and forth one time). How does the period of the swing of the simple pendulum depend on the quantities that define the pendulum and the quantities that determine the motion
Answer:
The period of the swing depends on only the length of the string and not on the mass of the bob and the period of the pendulum depends on only the horizontal component of g.
Explanation:
The period of the swing depends on only the length of the string and not on the mass of the bob. Since the length of the string and the mass of the bob define the pendulum.
Also, the properties that define the motion are the component of the weight of the bob in the horizontal direction which determines the to and fro movement of the bob. So, the period of the pendulum depends on only the horizontal component of g.
So, T = 2π√(l/g) where l = length of pendulum and g = acceleration due to gravity.
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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Can someone help me out please I got it wrong
Answer:
3 maybe since protons=atomic
What is the strength of an electric field that will put a force of
1.28 x 10-15 N on a proton?
Answer: E = 7,490.6 N/C
Explanation:
If we have a field E, and a particle with a charge q, the force that the particle experiences is:
F = E*q
In this case, we know that the force is:
F = 1.2*10^(-15) N
And we know that the particle is a proton, where the charge of a proton is:
q = 1.602*10^(-19) C
Then we can replace these two values in the equation to get:
1.2*10^(-15) N = E*1.602*10^(-19) C
We just need to isolate E.
(1.2*10^(-15) N)/(1.602*10^(-19) C) = E
7,490.6 N/C = E
That is the strength of the electric field.
Two 13.0-cm-diameter electrodes 0.59 cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 14 V battery. What is the charge on each electrode after insulating handles are used to pull the electrodes away from each other until they are 1.7 cm apart? The electrodes remain connected to the battery during this process. What is the electric field strength inside the capacitor after insulating handles are used to pull the electrodes away from each other until they are 1.7 cm apart? The electrodes remain connected to the battery during this process. What is the potential difference between the electrodes after insulating handles are used to pull the electrodes away from each other until they are 1.7 cm apart? The electrodes remain connected to the battery during this process.
Answer:
Explanation:
The capacitor is of parallel plate capacitor type
Capacitance C = ε₀ A / 4π d
ε₀ is 8.85 x 10⁻¹² , A is plate area and d is distance between plate .
d = 1.7 cm
C = 8.85 x 10⁻¹² x π x (6.5 x 10⁻² )² / 4π x 1.7 x 10⁻²
= 55 x 10⁻¹⁴F .
Charge on each electrode = C x V , V is voltage of battery .
= 55 x 10⁻¹⁴ x 14
= 770 x 10⁻¹⁴ C
Electric field strength = V / d where V is potential difference of battery , d is distance between plate .
= 14 / 1.7 x 10⁻²
= 8.23 x 10² V / m
The potential difference between plate
= potential difference of the battery
= 14 V .
A subway train starts from rest at a station and accelerates at a rate of 1.60 m/s 2 for 14.0 s. It runs at constant speed for 70.0 s and slows down at a rate of 3.50 m/s 2 until it stops at the next station. Find the total distance covered.
Answer:
1796.48 m
Explanation:
Given that :
First part of journey :
Initial Velocity, u = 0
Acceleration, a, = 1.60 m/s²
Time, t = 14 s
Distance traveled, S = 0.5at²
S = 0.5 * 1.60 * 14²
S1 = 156.8m
2nd part :
Speed is constant
Time = 70 seconds
At constant speed ;
Distance = speed * time
Speed, V = u + at
V = 0 + 1.6*14
V = 22.4 m/s
Distance, S2 = 22.4 * 70 = 1568 m
3rd part :
Deceleration = - 3.50m/s²
Final velocity, v = 0
Time taken to attain rest
V = u + at
0 = 22.4 - 3.5(t)
3.5t = 22.4
t = 22.4/3.5
t = 6.4 seconds
S3 = ut - 0.5at² (deceleration)
S3 = (22.4*6.4) - 0.5(3.5)*6.4^2
S3 = 71.68m
S1 + S2 + S3
156.8m + 1568m + 71.68m
= 1796.48 m
A system experiences a change in internal energy of 36 kJ in a process that involves a transfer of 14 kJ of heat into the surroundings. Simultaneously, which of the following is true?
a. Q= 14 u=36
b. U=Q-W
c. 14-36=22
d. 22 kJ of work is done on the system.
Answer:
b. U = Q - W
Explanation:
Given;
change in internal energy, ΔU = 36 kJ
heat transferred to the surroundings Q = 14 kJ
Apply first law of thermodynamic; the change in internal energy is equal to heat added to the system minus work done by the system.
ΔU = Q - W
Since heat was lost to surroundings, Q = - Q
ΔU = (-Q) - W
36 kJ = -14 kJ - W
36 kJ + 14 kJ = - W
50 kJ = - W
W = - 50 kJ (the negative sign shows that work has been done on the system)
Thus, 50 kJ of work is done on the system.
The only correct answer in the given options is "b" U = Q-W
Before there is a transfer of charges between objects, they are uncharged. What does this mean? (pls answer by 7:35)
Answer: This means that the objects didn't hold or have any charges before it was charged by something or someone
Explanation:
Like a dead phone it dosn't have any "charge", but after you put the charger in the wall you can plug the phone in and get it charged It's an on going cycle of energy being turned into other energy.
Answer:
The answer is C
Explanation:
The positive and negative charges are equal.
Kiara starts at 4, walks 6 blocks left and 2 blocks right. What is her displacement?
A horse and a dog have same momentum. Which of them have greater kinetic energy.
horse
dog
both have same K.E
insufficient information
Answer:
C. both have same K.E
Explanation:
Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.
Mathematically, momentum is given by the formula;
[tex] Momentum = Mass * Velocity [/tex]
Kinetic energy can be defined as an energy possessed by an object or body due to its motion.
Mathematically, kinetic energy is given by the formula;
[tex] K.E = \frac{1}{2}MV^{2}[/tex]
Where, K.E represents kinetic energy measured in Joules.
M represents mass measured in kilograms.
V represents velocity measured in metres per seconds square.
Both momentum and kinetic energy are related to the velocity of an object or a body.
Since the horse and a dog have same momentum. Thus, they both have same kinetic energy.
how much power does it take to do 104 J of work in 8 sec?
Answer:
Given:
W=104J
t=8sec
Solve:
P=?
Equation:
P=W/t or Fd/t
P= 104J'8sec
P=13W
I’m confused how to start it and I just need help atleast doing one.
Answer:
CuCl₂ + H₂S → CuS + 2HCl
Explanation:
The unbalanced reaction expression is given as;
CuCl₂ + H₂S → CuS + HCl
The problem here involves balancing of chemical equations.
We use a mathematical approach to solve this problem. Here assign coefficients a, b, c and d as values that will effect the balance;
aCuCl₂ + bH₂S → cCuS + dHCl
Conserving Cu: a = c
Cl: 2a = d
H: 2b = d
S: b = c
let a = 1; c = 1, b = 1 and d = 2
CuCl₂ + H₂S → CuS + 2HCl