Answer:
97.3 MJ
Explanation:
The formula for the coefficient of Perfomance is given as
COE = Q/W, where
COE is the coefficient of Perfomance
Q is the heat provided
W serves as the work input.
Dividing both sides of the equation by a factor of time t, we get the coefficient of Perfomance in terms of heating power and input power, so we say
COE = P / P(i),
making heating power, P the subject of formula, we have
P = COE * P(i)
P = 3.85 * 7020 * 1 * 3600
P = 97297200 J
P = 97.3 MJ
What are conducting patterns?
Answer:
A conducting pattern is a pattern in which your dominant hand follows in order to establish beats and tempo to the choir. Conductors that are directing large orchestras and choirs will often times use a baton so that the entire group can clearly see the motions.
Explanation:
20 pts.
Which of the following statements is true?
O Electromagnets use electrlcity and magnets.
O Magnetic fields are strongest around the poles of a magnet.
O The south pole of a magnet will repel the south pole of another magnet.
O all of the above
Answer:
all are true so d is right
Explanation:
Electromagnets use electrlcity and magnets is true.
Magnetic fields are strongest around the poles of a magnet is true.
The south pole of a magnet will repel the south pole of another magnet is true
and since all of them is true the answer is d all of the above
Suppose a uniform solid sphere of mass M and radius R rolls without slipping down an inclined plane starting from rest. The linear velocity of the sphere at the bottom of the incline depends on?
Answer:
None of the mass or the radius of the sphere
Explanation:
When a uniform solid sphere of any given mass, say M and any given radius, say R, rolls without slipping downwards an inclined plane that starts from rest. The linear velocity of the sphere at about the bottom of the inclined happens not to depend on either of its mass or that of the radius of its sphere.
g You heard the sound of a distant explosion (3.50 A/10) seconds after you saw it happen. If the temperature of the air is (15.0 B) oC, how far were you from the site of the explosion
Answer:
The answer is "1557 meters".
Explanation:
speed of sound in ([tex]\frac{m}{s}[/tex]) [tex]= 331.5 + 0.60 \ T^{\circ}\ C\\\\[/tex]
[tex]\to V = 331.5 + 0.6 \times 24 = 346 \frac{m}{s}\\\\\to t = 4.5 \ seconds \\\\\to S = vt = 346 \times 4.5 = 1557 \ meters[/tex]
explain an experiment of the phenomenon of rainfall
Unclear/incomplete question. However, I inferred you need an explanation of the phenomenon of rainfall.
Explanation:
Basically, the phenomenon of rainfall follows a natural cycle called the water cycle. What we call 'rainfall' occurs when water condensed (in liquid form) in the atmosphere is made to fall down on the ground as tiny droplets as a result of the forces of gravity.
The water cycle makes rainfall possible:
First, water on the earth's surface is evaporated (or is absorbed into) the atmosphere.Next, it then condensed into liquid form; which later falls to the surface to the ground again. And the process continues.A major league pitcher can throw a baseball an excess of
A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion
[tex]{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{First \: penetrating \: length\:(s_{1}) = 3 \: cm}[/tex]
[tex]\\[/tex]
[tex]{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{Left \: Penetration \: length \: before \: it \: comes \: to \: rest \:( s_{2} )}[/tex]
[tex]\\[/tex]
[tex]{\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{Let \: Initial \: velocity = v\:m/s} \\\\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{Left \: velocity \: after \: s_{1} \: penetration = \dfrac{v}{2} \:m/s} \\\\ [/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{s_{1} = \dfrac{3}{100} = 0.03 \: m}[/tex]
[tex]\\[/tex]
☯ As we know that,
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ \bigg(\dfrac{v}{2} \bigg)^{2} = {v}^{2} + 2a s_{1}}[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} = {v}^{2} + 2 \times a \times 0.03 }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} - {v}^{2} = 0.06 \times a }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{\dfrac{ - 3{v}^{2} }{4} = 0.06 \times a }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{a = \dfrac{ - 3 {v}^{2} }{4 \times 0.06} }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ a = \dfrac{ - 25 {v}^{2} }{2}\:m/s^{2} ......(1) }[/tex]
[tex]\\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{ Initial\:velocity=v\:m/s} \\\\ [/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{ Final \: velocity = 0 \: m/s }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as}[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{{0}^{2} = {v}^{2} + 2 \times \dfrac{ - 25 {v}^{2} }{2} \times s }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ - {v}^{2} = - 25 {v}^{2} \times s }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ s = \dfrac{ - {v}^{2} }{ - 25 {v}^{2} }}[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ s = \dfrac{1}{25} }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ s = 0.04 \: m }[/tex]
[tex]\\[/tex]
☯ For left penetration (s₂)
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{s = s_{1} + s_{2} }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ 0.04 = 0.03 + s_{2}}[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ s_{2} = 0.04 - 0.03 }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{s_{2} = 0.01 \: m = {\boxed{\sf{\purple{1 \: cm }}} }}[/tex]
[tex]\\[/tex]
[tex]\star\:\sf{Left \: penetration \: before \: it \: come \: to \: rest \: is \:{\bf{ 1 \: cm}}} \\ [/tex]
James is planning on registering for a course in electrical engineering which of the following sub disciplines could he opt for instrumentation engineering prosthetics plant design or signal processing
Answer:
Instrmentation engineering
Explanation:
Hope this helps!
What is the maximum speed your car can go around a flat 300m radius turn. Assume a coefficient of static friction of 0.9.
Answer:
vmax = 51.4 m/s
Explanation:
If the turn is flat, this means that the car is not accelerated in the vertical direction, so the normal force and the force of gravity are equal each other in magnitude, as follows:[tex]F_{n} = F_{g} = m*g (1)[/tex]
At the same time there exists a single force that keeps the car turning, which is the centripetal force, aiming to the center of the turn.This force is not a different force, is just the net force acting on the car in the horizontal direction, which is the friction force.This friction reaches to a maximum value, when the friction force is equal to the static friction force.This force can be expressed as follows:[tex]F_{fs} = \mu_{s} *F_{n} = \mu_{s} * m*g (2)[/tex]
This force must be equal to the maximum centripetal force, which can be written as follows:[tex]F_{c} = m*\frac{v^{2}}{r} (3)[/tex]
Since the left sides of (2) and (3) are equal each other, right sides must be equal too:[tex]\mu_{s} * m*g = m*\frac{v^{2}}{r} (4)[/tex]
Rearranging, and solving for v, we get the maximum speed that the car can go around the turn without sliding:[tex]v_{max} =\sqrt{\mu_{s} *g*r} =\sqrt{0.9*9.8m/s2*300m} = 51.4 m/s (5)[/tex]
A horizontal spring has a spring constant of 2.00 N/m. One end of the spring is attached to a fixed wall. A mass of 0.200 kilograms is attached to the free end of the spring and the spring is compressed by 0.500 meters from its natural length. The mass is released. What is the speed of the mass when the spring returns to its natural length?
a. 2.24 m/s
b. 2.50 m/s
c. 5.00 m/s
d. 1.58 m/s
Answer:
(D) the speed of the mass when the spring returns to its natural length is 1.58 m/s.
Explanation:
Given;
spring constant of the spring, k = 2 N/m
mass attached to the spring, m = 0.2 kg
compression of the spring, x = 0.5 m
Apply the principle of conservation of mechanical energy;
K.E = P.E
¹/₂m(v² - u²) = ¹/₂kx²
where;
u is the initial speed of the mass = 0
¹/₂mv² = ¹/₂kx²
mv² = kx²
[tex]v^2= \frac{kx^2}{m} \\\\v= \sqrt{\frac{kx^2}{m}} \\\\v = \sqrt{\frac{(2)(0.5)^2}{0.2}} \\\\v = 1.58 \ m/s[/tex]
Therefore, the speed of the mass when the spring returns to its natural length is 1.58 m/s.
Bonnie and Clyde are trying to steal the world's largest diamond from a 10 story
building. Bonnie needs to make a landing spot for when Clyde jumps with the
diamond. She knows they will be jumping from 58 feet. Neglecting air resistance,
what is the speed Clyde will be falling at right before hitting the ground?
Answer:
The speed Clyde will be falling at is 33.72.
g A wave on a string has a speed of 13.3 m/s and a period of 0.3 s. What is the frequency of the wave
Answer:
3.33 Hz
Explanation:
The first step is to calculate the wavelength
= speed × period
= 13.3 × 0.3
= 3.99
Therefore the frequency of the wave can be calculated as follows
= speed/wavelength
= 13.3/3.99
= 3.33 Hz
A rock is at the top of a 20 meter tall hill. The rock has a mass of 10 kg. How much potential energy does it have?
Answer:
1960 JExplanation:
The potential energy of a body can be found by using the formula
PE = mgh
where
m is the mass
h is the height
g is the acceleration due to gravity which is 9.8 m/s²
PE = 10 × 9.8 × 20
We have the final answer as
1960 JHope this helps you
(6.9 * 10-6)(770 * 102)
Answer:
4948020
Explanation:
(6.9*10-6)(770*102)
Multiply 6.9 by 10 .
( 69 − 6 ) ( 770 ⋅ 102 )
Subtract 6 from 69 .
63 ( 770 ⋅ 102 )
Multiply 770 by 102 .
63 ⋅ 78540
Multiply 63 by 78540 .
4948020
A 1430 kg car speeds up from rest
to 14.0 m/s in 7.00 s. lgnoring
friction, how much power did that
require?
The power required by the car if a 1430 kg car speeds up from rest to 14.0 m/s in 7.00 s is 20020 Joules.
What is power?In science and engineering, power is the rate at which work is completed or energy is delivered. It can be expressed as the product of the work completed (W) or the energy transferred (E) divided by the time interval (t), or W/t.
A high-powered motor can complete a given amount of work quickly, or a low-powered motor can accomplish it slowly. Foot-pounds per minute, joules per second (or watts), and ergs per second are examples of units of power, which measure work (or energy) per unit of time.
Given:
A 1430 kg car speeds up from rest to 14.0 m/s in 7.00 s,
Calculate the acceleration and displacement as shown below,
[tex]v = u + at[/tex]
14 = 0 + a × 7
a = 2 m/s²
[tex]s = ut + 1/2at^2[/tex]
s = 0 + 1 / 2 × 2 × 7²
s = 49 m,
Calculate the power as shown below,
Power = 1430 × 2 × 49 / 7
Power = 20020 Joules
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I
2800000 kg
Fig. 2.1
The total mass of the rocket and its full load of fuel is 2.8 106 kg. The constant force provided by
the rocket's motors is 3.2 x 107N.
32000000 N.
(a) Calculate
(i) the total weight of the rocket and the fuel.
weight =
Answer:
W = 27468000 [N] or 27.47 [kN]
Explanation:
The weight of a body is defined as the product of mass by gravitational acceleration.
[tex]W = m*g[/tex]
where:
W = weight of the rocket with fuel [kg]
m = mass = 2800000 [kg]
g = gravity acceleration = 9.81 [m/s²]
Now we can solve:
[tex]W = 2800000*9.81\\W = 27468000 [N][/tex]
A disk rotates at a constant angular velocity of 30 degrees per second. Consider a point on the edge of the disk. Through how many degrees has it rotated after 3 seconds?
Answer:
The disk covers a rotation of 90º after 3 seconds.
Explanation:
Since the disk rotates at constant angular speed, we can determine the change in angular position ([tex]\Delta \theta[/tex]), measured in sexagesimal degrees, by the following kinematic formula:
[tex]\Delta \theta = \omega\cdot \Delta t[/tex] (1)
Where:
[tex]\omega[/tex] - Angular velocity, measured in sexagesimal degrees per second.
[tex]\Delta t[/tex] - Time, measured in seconds.
If we know that [tex]\omega= 30\,\frac{\circ}{s}[/tex] and [tex]\Delta t = 3\,s[/tex], then the change in angular position is:
[tex]\Delta \theta = \left(30\,\frac{\circ}{s} \right)\cdot (3\,s)[/tex]
[tex]\Delta \theta = 90^{\circ}[/tex]
The disk covers a rotation of 90º after 3 seconds.
Need ASAP!! An object lies motionless on a rough inclined surface .
Answer:
yes
Explanation:
yes
Answer:
that's true
Explanation:
if the rough inclined plane was rough enough than it would be true
6. Drawing reliable conclusions from scientific evidence is important. Which of the following is a question that you can ask yourself to evaluate the reliability of conclusions and claims made by other scientists? How much money did they spend on their experiment? What evidence supports the claim? Do I like their results? Do I agree with their claim?
Answer:
What evidence supports their claim.
Explanation:
In the end, all that really matters in an experiment, is what scientific evidence supports the claim being made.
Answer:
What evidence supports their claim.
Explanation:
Help me please..
When the slope of a velocity vs. time graph is negative and constant,
what type of motion is occurring?
A) No motion
B) Constant speed
C) Acceleration
D) Constant Velocity
E) Going in circles
Which factor listed below has the greatest effect on the period of a pendulum?
a
Mass of the pendulum
b
Angle of the pendulum
c
Length of the string
d
Amplitude of the oscillations
Answer:
C
Explanation:
equation for pendulum period is
[tex]\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}[/tex]
and it depends only on its length (and gravity which is not on the list of answers)
A force of 64 N gives an object an acceleration of 23.53 m/s^2. What is the mass of the object?
A-2.00 kg
B-1505 kg
C-2.72 kg
D-736 kg
PS Final Exam
What type of friction exists between a moving car's tires and the road?
O rolling friction
O static friction
O sliding friction
During a hockey game, a puck is given an initial speed of 10 m/s. It slides 50 m on the horizontal ice before it stops due to friction. What is the coefficient of kinetic friction between the pick and the ice.A) 0.12B) 0.10C) 0.11D) 0.090
Answer:
The value is [tex]\mu_k = 0.102[/tex]
Explanation:
From the question we are told that
The initial speed of the pluck is [tex]u = 10 \ m/s[/tex]
The distance it slides on the horizontal ice is [tex]s = 50 \ m[/tex]
Generally from kinematic equation we have that
[tex]v^2 = u^2 + 2as[/tex]
Here v is is the final velocity and the value is 0 m/s given that the pluck came to rest, so
[tex]0^2 = 10 ^2 + 2* a * 50[/tex]
=> [tex]a = - 1 \ m/s^2[/tex]
Here the negative sign show that the pluck is decelerating
Generally the force applied on the pluck is equal to the frictional force experienced by the pluck
So
[tex]F = F_f[/tex]
=> [tex]m * a = m* g * \mu_k[/tex]
=> [tex]1 = 9.8 * \mu_k[/tex]
=> [tex]\mu_k = 0.102[/tex]
2 kg ball of clay moving at 40 m/s collides with a 5 kg ball of clay moving at 10 m/s directly toward the first ball. What is the final velocity if both balls stick together after the collision?
Answer:
vf = 4.3 m/s
Explanation:
Assuming no external forces acting during the collision, total momentum must be conserved:[tex]p_{o} = p_{f} (1)[/tex]
where p₀ = initial momentum, and pf = final momentum.
The initial momentum is just the sum (vector sum) of the initial momenta of both balls, as follows:[tex]p_{o} = m_{1} * v_{1o} + m_{2} * v_{2o} = 2 kg* 40 m/s - 5kg* 10m/s = 30 kg*m/s (2)[/tex]
The final momentum, assuming both balls stick together after the collision, can be expressed as follows:[tex]p_{f} = (m_{1} + m_{2} ) * v_{f} = 7 kg * v_{f} (3)[/tex]
From (2) and (3), solving for vf, we get:[tex]v_{f} =\frac{30 kg*m/s}{7 kg} = 4.3 m/s (4)[/tex]
A motorcycle skids for a distance of 2.0 m with the icy road pushing on its tires with force of 120 N as its
brakes are applied
What is the change in kinetic energy for the motorcycle?
Round the answer to two significant digits.
Answer:
-240
Explanation:
A motorcycle skids for a distance of 2.0 m on an icy road, then the change in kinetic energy for the motorcycle will be equal to -240 J.
What is kinetic energy?The force which a moving object has is referred to as kinetic energy in physics. It is defined as the number of effort required to propel a person of a specific mass from still to a specific velocity.
Aside from slight fluctuations in speed, your body holds onto the kinetic energy it obtains during acceleration.
When the body slows down from its present level to a condition of rest, the same quantity of energy is used.
Formally, kinetic energy is any quantity that has a gradient concerning time in the Lagrangian of a system.
As per the given information in the question,
Distance, d = 2.0 m
Friction, f = 120 N
The angle between displacement and friction force, θ = 180°
Now, the change in kinetic energy for the motorcycle = Work done by the friction.
K.E = f × d(cos θ)
= 120 (2.0 m)(cos 180°)
Δ K.E = -240 J
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Please help!!
A person is driving in a circle at 20m/s. Are they accelerating?
air is a gas and the particles can be pushed closer toegether which is called
why do feet smell and noses run?
Answer:
Nose has mucous glands with hairs which helps the body in trapping pollutants and infectants from entering inside the body. On the other hand,our feet is composed of millions of sweat pores when dirt and other things accumulate,it smells because of sweat mixed with the dirt and other dirty things of the ground.
Explanation:
Hope this helps
Bio-mimetic gel was used in the study of genetics and ultimately banded by the Federation. A Ferengi transport ship is carrying a 5-centimeter [cm] radius cylindrical tank of bio-mimetic gel to sell on the black market. A hydrostatic pressure probe located at the bottom of the tank reads 60 millimeters of mercury [mm Hg]. If the specific gravity of the gel is 8.9, what is the mass of gel contained in the tank in units of kilograms [kg]
Answer:
m = 6.4 kg
Explanation:
Now, we use the following equation to find the mass of the gel:
[tex]P = \rho gh[/tex]
where,
P = (60 mm of Hg)(133.3 Pa/1 mm of Hg) = 7999.2 Pa = 7.99 KPa
ρ = density of gel = [tex]\frac{m}{V} = \frac{m}{\pi r^2 h}[/tex]
m = mass of gel = ?
r = radius of cylindrical tank = 5 cm = 0.05 m
h = height of tank
Therefore,
[tex]7999.2 Pa = (\frac{m}{\pi(0.05\ m)^2\ h})(9.81\ m/s^2)h\\\\\frac{(7999.2\ Pa)(\pi)(0.05\ m)^2}{9.81\ m/s^2} = m\\[/tex]
m = 6.4 kg
