Answer:
The solution is given below:
Explanation:
The computation is shown below:
The acceleration is
= v^2 ÷ r
= 19^2 ÷ 35 m
= 10.3 m/s^2
According to the newton second law
F = mac
= 1240 kg (10.3 m/s^2)
= 12772 N
And,
frictional force (f) = \mu N
f = \mu (mg) = mac
\mu = ac ÷ g
= 10.3 m/s^2 ÷ 9.80 m/s^2
= 1.0510
In these ways it can be determined
An object of mass 45 kg is observed to accelerate at the rate of 6 m/s2. Calculate the force required to produce this acceleration
On a distance-time graph, what is shown when the curve is flat going from left to the right?
A. a negative speed
B. no speed
C. a positive speed
D. It does not mean anything.
Please help me !!im on a test
when two capacitor 3muF and 6muF are connected in a parallel and combination is charged to a potential of 120 volt the potential difference across the 3muF capacitor is
Answer:
V₁ = V = 120 V
Explanation:
Such a combination of capacitors in which;
1- Potential difference across each capacitor is the same
2- Total charge is distributed amongst the capacitors
; is called Parallel Combination.
Therefore, in this case, the potential difference across each capacitor will also be the same. Because the capacitors are connected in parallel here. So the voltage across 3 μF capacitor will be the same as the voltage across the 6 μF capacitor and they both will be equal to the total potential difference.
V₁ = V = 120 V
You are playing in a volley ball game Your team has 12 and the other team has 18.
How many points does your team needs to win?
How many points does the other team needs to win?
Answer:
you need 7 points and the other team just needs to stop you from scoring
Explanation:
Which is the best way to become familiar with your company's policies and procedures?
O
A. ask the person who hired you
O
B. look in the employee handbook
C. tell your supervisor you need help
D. visit the company's website
An ocean thermal energy conversion system is being proposed for electric power generation. Such a system is based on the standard power cycle for which the working fluid is evaporated, passed through a turbine, and subsequently condensed. The system is to be used in very special locations for which the oceanic water temperature near the surface is approximately 300 K, while the temperature at reasonable depths is approximately 280 K. The warmer water is
Answer:
Explanation:
Dear Student, this question is incomplete, and to attempt this question, we have attached the complete copy of the question in the image below. Please, Kindly refer to it when going through the solution to the question.
To objective is to find the:
(i) required heat exchanger area.
(ii) flow rate to be maintained in the evaporator.
Given that:
water temperature = 300 K
At a reasonable depth, the water is cold and its temperature = 280 K
The power output W = 2 MW
Efficiency [tex]\zeta[/tex] = 3%
where;
[tex]\zeta = \dfrac{W_{out}}{Q_{supplied }}[/tex]
[tex]Q_{supplied } = \dfrac{2}{0.03} \ MW[/tex]
[tex]Q_{supplied } = 66.66 \ MW[/tex]
However, from the evaporator, the heat transfer Q can be determined by using the formula:
Q = UA(L MTD)
where;
[tex]LMTD = \dfrac{\Delta T_1 - \Delta T_2}{In (\dfrac{\Delta T_1}{\Delta T_2} )}[/tex]
Also;
[tex]\Delta T_1 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_1 = 300 -290 \\ \\ \Delta T_1 = 10 \ K[/tex]
[tex]\Delta T_2 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_2 = 292 -290 \\ \\ \Delta T_2 = 2\ K[/tex]
[tex]LMTD = \dfrac{10 -2}{In (\dfrac{10}{2} )}[/tex]
[tex]LMTD = \dfrac{8}{In (5)}[/tex]
LMTD = 4.97
Thus, the required heat exchanger area A is calculated by using the formula:
[tex]Q_H = UA (LMTD)[/tex]
where;
U = overall heat coefficient given as 1200 W/m².K
[tex]66.667 \times 10^6 = 1200 \times A \times 4.97 \\ \\ A= \dfrac{66.667 \times 10^6}{1200 \times 4.97} \\ \\ \mathbf{A = 11178.236 \ m^2}[/tex]
The mass flow rate:
[tex]Q_{H} = mC_p(T_{in} -T_{out} ) \\ \\ 66.667 \times 10^6= m \times 4.18 (300 -292) \\ \\ m = \dfrac{ 66.667 \times 10^6}{4.18 \times 8} \\ \\ \mathbf{m = 1993630.383 \ kg/s}[/tex]
What does it mean when work is positive?
Answer:
When force and displacement are in the same direction, the work performed on an object is said to be positive work. Example: When a body moves on the horizontal surface, force and displacement act in the forward path. The work is done in this case known as Positive work.
Explanation:
Hope this helps you
A solenoid that is 66.2 cm long has a cross-sectional area of 18.0 cm2. There are 1300 turns of wire carrying a current of 8.15 A. (a) Calculate the energy density of the magnetic field inside the solenoid. (b) Find the total energy in joules stored in the magnetic field there (neglect end effects).
Answer:
(a) Energy Density = 160.94 J/m³
(b) Energy Stored = 0.192 J
Explanation:
(a)
The energy density of the magnetic field inside the solenoid is given by the following formula:
[tex]Energy\ Denisty = \frac{B^2}{2\mu_o}\\[/tex]
where,
B = magnetic field strength of solenoid = [tex]\frac{\mu_oNI}{l}[/tex]
Therefore,
[tex]Energy\ Density = \frac{\mu_oN^2I^2}{2l^2}[/tex]
where,
μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²
N = No. of turns = 1300
I = current = 8.15 A
L = length = 66.2 cm = 0.662 m
Therefore,
[tex]Energy\ Density = \frac{(4\pi\ x\ 10^{-7}\ N/A^2)(1300)^2(8.15\ A)^2}{2(0.662\ m)^2}[/tex]
Energy Density = 160.94 J/m³
(b)
Energy Stored = (Energy Density)(Volume)
Energy Stored = (Energy Density)(Area)(L)
Energy Stored = (160.94 J/m³)(0.0018 m²)(0.662 m)
Energy Stored = 0.192 J
An 800 kg charging bull rams through a wooden fence. It was travelling at
5 m/s, now it's travelling at 3 m/s. How much impulse did the bull
experience by smashing the fence?
Answer:
J = 1600 kg-m/s
Explanation:
Given that,
The mass of charging bull rams, m = 800 kg
Initial speed, u = 5 m/s
Final speed, v = 3 m/s
We need to find the impulse the bull experience by smashing the fence. Let it is J. We know that, impulse is equal to the change in momentum such that,
J = m(v-u)
Put all the values,
J = 800(3-5)
= 800(-2)
= -1600 kg-m/s
Hence, the magnitude of impulse is equal to 1600 kg-m/s.
Write about
a time you had to ride a bicycle on a difficult
surface. What did you have to do to adjust your
riding?
what is the maximum distance we can shoot a dart,from ground level provided our toy dart gun gives a maximum initial velocity of 2.7m/s and air resistance is negligible
Answer:
R = v^2 sin 2 theta / g
The range provides the distance a projectile can travel
R(max) = v^2 / g if theta = 45 deg
R = 2.7^2 / 9.8 = .74 m
Which of the following is a category of mechanical wave?
O A. Transverse
B. Frictional
C. Parallel
D. Perpendicular
Answer:
a
because the mechanical wave is when it goes over and over again
Answer:
The answer is a like i said 3hrs ago i dont know if this guy copied me tbh
Explanation:
When a particular hanging mass is suspended from the string, a standing wave with two segments is formed. When the weight is reduced by 2.2 kg, a standing wave with five segments is formed. What is the linear density of the string
Solution :
Mass is varied keeping frequency constant.
Wavelength, λ [tex]$=\frac{2l}{n}$[/tex]
where length of spring = l
number of segments = n
Velocity, v = λ x f
= [tex]$\sqrt{\frac{T}{\mu}}$[/tex]
[tex]$\mu $[/tex] = mass density, T = tension in string
[tex]$T=\frac{4 \mu l^2f^2}{n^2}$[/tex]
[tex]$T=mg = \frac{4 \mu l^2f^2}{n^2}$[/tex] , n = 2
[tex]$T = (m-2.2)g = \frac{4 \mu l^2f^2}{n^2}, n = 5$[/tex]
[tex]$\Rightarrow \frac{m}{m-2.2}=\frac{25}{4}$[/tex]
[tex]$\Rightarrow m = 2.619\ kg$[/tex]
Therefore, μ = 0.002785 kg/ m
Frequency is varied keeping T constant
[tex]$T=\frac{4 \mu l^2f^2}{n^2}, f=60 , \ \ n = 2$[/tex]
[tex]$T=\frac{4 \mu l^2f^2}{n^2}, f=? , \ \ n = 7$[/tex]
[tex]$\Rightarrow \frac{60^2}{4}=\frac{f^2}{49}$[/tex]
f = 210 Hz
The density of table sugar is 1.59g/cm3 what is the volume of 7.85g of sugar?
Answer: 4.94cm³
Explanation:
Data;
ρ = 1.59g/cm³
mass = 7.85g
volume = ?
density = mass / volume
ρ = m / v
v = m / ρ
v = 7.85 / 1.59
v = 4.94cm³
Man-made climate change is an example
of...
Calculate the potential difference across the 8 ohm resistor
Explanation:
if the current is 1A
V=iR
V= 1 × 8
V = 8volts
Please help if you can!
Answer:
C, Red has the longest one
c red
red has longest wavelength
amnh dot org
The period of a sound wave coming from an instrument is 2 seconds. What 1 point
is the frequency of the sound? (f = 1/T) *
5 Hz
50 Hz
ОО
0.5 Hz
Answer:
b
Explanation:
How much heat is required to raise the temperature of 50 grams of water from 30 °C to 90 °C? C of water 4186 J / kg C.
12558 J
12558000 J
125580 J
1255800 J
Answer:
12558 J
Explanation:
Please do mark as brainliest. Hope this helps! :)
The length of the slope of a mountain is 2780 m, and it makes
its base?
angle of 14.1° with the horizontal. What is the height of the mountain, relative to
Additional Materials
Reading
Answer:
677 m
Explanation:
Using the definition of the sine of an angle, we can write
sin 14.1 = (height of mountain) / (slope length of mountain)
sin 14.1 = H / (2780 m) ---> H = (2780 m) x sin 14.1
= 677 m
The height of the mountain is 677.21 m
The given parameters;
length of the slope, L = 2780 m
angle of inclination, Ф = 14.1°
let the height of the mountain, = h
A simple sketch of the problem is given below;
↓P
↓
↓ h
↓ 14.1°
↓------------------------------------------------Q
A straight line joining PQ is the hypotenuse of the right triangle.
The height of the right triangle is calculated as follows;
[tex]sin(14.1) = \frac{h}{PQ} \\\\\h = PQ \times sin(14.1)\\\\h = 2780 \times sin(14.1)\\\\h = 677.21 \ m[/tex]
Thus, the height of the mountain is 677.21 m
Learn more here: https://brainly.com/question/4326804
True or false. When a girl walks the action of pushing and the equal amd opposite reaction is being projected forward
This is true I think
It applies to Newton's Laws
it's true because it's a part of newtons law
A 25.0 kg mass is traveling to the right with a speed of 2.80 m/s on a smooth horizontal surface when it collides with and sticks to a second 25.0 kg mass that is initially at rest but is attached to one end of a light, horizontal spring with force constant 170.0 N/m. The other end of the spring is fixed to a wall to the right of the second mass.
1) Find the frequency of the subsequent oscillations.2) Find the amplitude of the subsequent oscillations.3) Find the period of the subsequent oscillations.4) How long does it take the system to return the first time to the position it had immediately after the collision?
Answer:
Explanation:
1 ) angular frequency ω = √ ( k / m )
=√ ( 170 / 50 )
= 1.844 rad /s
2πn = 1.844 where n is frequency of oscillation
n = 1.844 / (2 x 3.14 )
= .294 per sec
= .294 x 60 = 18 approx. per minute .
Velocity just after collision of composite mass ( using law of conservation of momentum )
= 25 x 2.8 / 50
v = 1.4 m/s
If new amplitude be A
1/2 k A² = 1/2 m v²
m = 25 + 25 = 50 kg
170 x A² = 50 x 1.4²
A = 0.76 m
3 ) period of oscillation = 1 /n
= 1 / .294
= 3.4 s
4 ) It will take complete one period of oscillation ie 3.4 s to come to its original position.
g You drop a 3.6-kg ball from a height of 3.5 m above one end of a uniform bar that pivots at its center. The bar has mass 9.9 kg and is 4.2 m in length. At the other end of the bar sits another 3.6-kg ball, unattached to the bar. The dropped ball sticks to the bar after the collision. Assume that the bar is horizontal when the dropped ball hits it. How high (in meters) will the other ball go after the collision
Answer:
h = 3.5 m
Explanation:
First, we will calculate the final speed of the ball when it collides with a seesaw. Using the third equation of motion:
[tex]2gh = v_f^2 - v_i^2\\[/tex]
where,
g = acceleration due to gravity = 9.81 m/s²
h = height = 3.5 m
vf = final speed = ?
vi = initial speed = 0 m/s
Therefore,
[tex](2)(9.81\ m/s^2)(3.5\ m) = v_f^2 - (0\ m/s)^2\\v_f = \sqrt{68.67\ m^2/s^2}\\v_f = 8.3\ m/s[/tex]
Now, we will apply the law of conservation of momentum:
[tex]m_1v_1 = m_2v_2[/tex]
where,
m₁ = mass of colliding ball = 3.6 kg
m₂ = mass of ball on the other end = 3.6 kg
v₁ = vf = final velocity of ball while collision = 8.3 m/s
v₂ = vi = initial velocity of other end ball = ?
Therefore,
[tex](3.6\ kg)(8.3\ m/s)=(3.6\ kg)(v_i)\\v_i = 8.3\ m/s[/tex]
Now, we again use the third equation of motion for the upward motion of the ball:
[tex]2gh = v_f^2 - v_i^2\\[/tex]
where,
g = acceleration due to gravity = -9.81 m/s² (negative for upward motion)
h = height = ?
vf = final speed = 0 m/s
vi = initial speed = 8.3 m/s
Therefore,
[tex](2)(9.81\ m/s^2)h = (0\ m/s)^2-(8.3\ m/s)^2\\[/tex]
h = 3.5 m
what is the definition of a moment of force?
Answer:
The Moment of a force is a measure of its tendency to cause a body to rotate about a specific point or axis.
Answer:
Torque
Explanation:
I serached it up sjdjbdjd
A 10kg block is Pulled along a horizontal
Surface by a force
of 50N at an angles
of 37° with the horizontal If the
coefficient of sliding friction b/n the
block and the surface is o.2
(g=10m/s^2 Sin 37=O.6 and cos 37 = 0.8)
A, what frictional forces acting on the block?
B,what is the acceleration of the block?
Answer:
hope u can understand the method
Two parallel copper rods supply power to a high-energy experiment, carrying the same current in opposite directions. The rods are held 8.0 cm apart by insulating blocks mounted every 1.5 m. If each block can tolerate a maximum tension force of 200 N, what is the maximum allowable current
Answer:
the maximum allowable current is 7302.967 amperl
Explanation:
The computation of the maximum allowable current is shown below;
Force F = mean ÷ 4π 2 I_1 I_2 ÷d × ΔL
200 N = (10)^-7 (2I × I) ÷ 0.08 × 1.5
200 = 3.75 × 10^-6 I^2
I = √200 ÷ √ 3.75 × 10^-6
= 7302.967 amperl
Hence, the maximum allowable current is 7302.967 amperl
Basically we applied the above formula
Electroconvulsive therapy would be done under the
supervision of a counseling psychologist, where high level
of electric shock would be admistered.
Select one:
True
False
Answer:
the answer of this question is true
PLS HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Answer:
A) pass the ball to a teammate
B) Smash the shuttlecock downward in your opponents court
C)Do a fake hit . . .
D) Do a fake hit . . .
(Best guess)
A soccer ball is released from rest at the top of a grassy incline. After 6.2 seconds, the ball travels 47 meters. One second later, the ball reaches the bottom of the incline.
(a) What was the balls acceleration?(assume that the acceleration was constant).
(b) How long was the incline?
Answer:
(a) a = 2.44 m/s²
(b) s = 63.24 m
Explanation:
(a)
We will use the second equation of motion here:
[tex]s = v_it+\frac{1}{2}at^2[/tex]
where,
s = distance covered = 47 m
vi = initial speed = 0 m/s
t = time taken = 6.2 s
a = acceleration = ?
Therefore,
[tex]47\ m = (0\ m/s)(6.2\ s)+\frac{1}{2}a(6.2\ s)^2\\\\a = \frac{2(47\ m)}{(6.2\ s)^2}[/tex]
a = 2.44 m/s²
(b)
Now, we will again use the second equation of motion for the complete length of the inclined plane:
[tex]s = v_it+\frac{1}{2}at^2[/tex]
where,
s = distance covered = ?
vi = initial speed = 0 m/s
t = time taken = 7.2 s
a = acceleration = 2.44 m/s²
Therefore,
[tex]s = (0\ m/s)(6.2\ s)+\frac{1}{2}(2.44\ m/s^2)(7.2\ s)^2\\\\[/tex]
s = 63.24 m
The Equipartition Theorem follows from the fundamental postulate of statistical mechanics--that every energetically accessible quantum state of a system has equal probability of being populated, which in turn leads to the Boltzmann distribution for a system in thermal equilibrium.
a. True
b. False
Answer:
Hello! Your answer would be, A) True
Explanation:
Hope I helped! Ask me anything if you have any questions. Brainiest plz!♥ Hope you make a 100%. Have a nice morning! -Amelia♥