Answer:
Explanation:
From the information given;
mass of the crate m = 41 kg
constant horizontal force = 135 N
where;
[tex]s_1 = 15.0 \ m \\ \\ s_2 = 12.0 \ m[/tex]
coefficient of kinetic friction [tex]u_k[/tex] = 0.28
a)
To start with the work done by the applied force [tex](W_f)[/tex]
[tex]W_F = F\times (s_1 +s_2) \times cos(0) \ J[/tex]
[tex]W_F = 135 \times (12 +15) \times cos(0) \ J \\ \\ W_F = (135 \times 37 )J \\ \\ W_F =4995 \ J[/tex]
Work done by friction:
[tex]W_{ff} = -\mu\_k\times m \times g \times s_2 \\ \\ W_{ff} = -0.320 \times 41 \times 9.81 \times 12 \ J \\ \\ W_{ff} = -1544.49 \ J[/tex]
Work done by gravity:
[tex]W_g = mg \times (s_1+s_2) \times cos (90)} \ J \\ \\ W_g = 0 \ j[/tex]
Work done by normal force;
[tex]W_n = N \times (s_1 + s_2) \times cos (90) \ J[/tex]
[tex]W_n = 0 \ J[/tex]
b)
total work by all forces:
[tex]W = F \times (s_1 + s_2) + \mu_k \times m \times g \times s_2 \times 180 \\ \\ W = 135 \times (15+12) \ J - 0.320 \times 41 \times 9.81 \times 12[/tex]
W = 2100.5 J
c) By applying the work-energy theorem;
total work done = ΔK.E
[tex]W = \dfrac{1}{2}\times m \times (v^2 - u^2)[/tex]
[tex]2100.5 = 0.5 \times 41 \times v^2[/tex]
[tex]v^2 = \dfrac{2100.5}{ 0.5 \times 41 }[/tex]
[tex]v^2 = 102.46 \\ \\ v = \sqrt{102.46} \\ \\ \mathbf{v = 10.1 \ m/s}[/tex]
A hair dryer draws a current of 12.8 A.
(a)How many minutes does it take for
6.8 x 10° C of charge to pass through the
hair dryer? The fundamental charge is
1.602 x 10-19 C.
Answer in units of min.
(b)How many electrons does this amount of
charge represent?
Answer in units of electrons.
Answer:
(a) 8.85×10⁻³ minutes
(b) 4.24×10¹⁹ electrons
Explanation:
(a) Using,
Q = it............................. Equation 1
Where Q = quantity of charge, i = current, t = time.
Make t the subject of the equation
t = Q/i............................. Equation 2
Given: Q = 6.8×10⁰ C, i = 12.8 A
Substitute these values into equation 2
t = 6.8×10⁰/12.8
t = 8.85×10⁻³ minutes
(b) n = Q/(1.602×10⁻¹⁹)................. Equation 3
Where n = number of electrons.
Given: Q = 6.8×10⁰ C
Substitute into equation 2
n = 6.8×10⁰/1.602×10⁻¹⁹
n = 4.24×10¹⁹ electrons
(a) The time taken by the charge to flow from the drier will be [tex]\d8.85[/tex]×[tex]\d10^{-3}[/tex]minutes
(b) Amount of the electrons in the charge will be [tex]\d4.24[/tex]×[tex]\d10^{19}[/tex] electrons
What will be the time of the charge and number of the electrons in the charge ?As we know Q = IT
Where Q = quantity of charge, i = current, T = time.
From the above equation
T= Q/I.
Given: Q = [tex]6.8\times\d10^{0}[/tex] C, i = 12.8 A
Substitute these values
T= [tex]6.8[/tex]×[tex]\d10^{0}[/tex] /12.8
T = [tex]8.85[/tex]×[tex]\d10^{-3}[/tex] minutes
Now the number of the electrons present in the charge will be
n = Q/( [tex]1.602[/tex]×[tex]\d10^{-19}[/tex])
Where n = number of electrons.
Given: Q = [tex]6.8\times\d10^{0}[/tex] C
Substitute Value of Q
n = [tex]6.8\times\d10^{0}[/tex]/ [tex]1.602\times\d10^{-19}[/tex]
n = [tex]4.24\times\d10^{19}[/tex] electrons
Thus
(a)The time taken by the charge to flow from the drier will be [tex]\d8.85[/tex]×[tex]\d10^{-3}[/tex]minutes
(b)Amount of the electrons in the charge will be [tex]\d4.24[/tex]×[tex]\d10^{19}[/tex] electrons
To know more about electric charge follow
https://brainly.com/question/14372859
why food cook faster with salt water than cook with pure water
Answer:
oil heats faster
Explanation:
23
In order for a 12 Volt power source
to produce a current of 0.085 amps,
a resistance of...
[?] Ohms is needed.
Enter
Haven't learned this yet.
Answer:
141.18 ohms
Explanation:
From the question given above, the following data were obtained:
Voltage (V) = 12
Current (I) = 0.085 A
Resistance (R) =?
The resistance needed can be obtained as follow:
V = IR
12 = 0.085 × R
Divide both side by 0.085
R = 12 / 0.085
R = 141.18 ohms
Therefore, a resistor of resistance 141.18 ohms is needed.
A light year is the amount of time it takes for light from the Sun to reach the Earth.
True
False
Which of the following changes when an unbalanced force acts on an object?
A. mass
B. motion
C. inertia
D. weight
The answer is Motion
If you blow across the open end of a soda bottle and produce a tone of 250 Hz, what will be the frequency of the next harmonic heard if you blow much harder?
___Hz
Answer:
Generally, the lowest overtone for a pipe open at one end and closed would be at y / 4 where y represents lambda, the wavelength.
Since F (frequency) = c / y Speed/wavelength
F2 / F1 = y1 / y2 because c is the same in both cases
F2 = y1/y2 * F1
F2 = 3 F1 = 750 /sec
Note that L = y1 / 4 = 3 y2 / 4 for these wavelengths to fit in the pipe
and y1 = 3 y2
The second harmonic will be three times the first harmonic. The answer is 750 Hz
VIBRATION OF WAVES IN PIPESClosed pipes have odd multiples of frequencies or harmonics. That is,
If [tex]F_{0}[/tex] = fundamental frequency = first harmonic
[tex]F_{1}[/tex] = 3[tex]F_{0}[/tex] = second harmonic
[tex]F_{2}[/tex] = 5[tex]F_{0}[/tex] = third harmonic
[tex]F_{3}[/tex] = 7[tex]F_{0}[/tex] = fourth harmonic
Let assume that the first harmonic is 250 Hz, If you blow it much harder, second, third or fourth harmonic can be produced.
By using the formula above,
second harmonic will be 3 x 250 = 750Hz
Therefore, the frequency of the next harmonic heard if you blow much harder will be 750 Hz
Learn more about Sound waves here: https://brainly.com/question/1199084
The gravitational force between two objects with masses 1kg and 28kg separated by a distance 7m is ____________10-11 N.
a.
3.81
b.
26.68
c.
9151.24
d.
1307.32
Hhhhhellllppp fastt
Answer:
a. 3.81
Explanation:
F = GMm/r^2
F = (6.67 x 10^-11 x 28 x 1) / 7^2
F = 3.81 x 10^-11 N
If you could help me please.
1) Does a 1 kg object weight 9.8 newtons on the moon? why?
2) How much does a 3-kg object weigh (on earth) in newtons?
3) How much does a 20-kg object weigh (on earth) in newton?
4) What must happen for the mass of an object to change?
5) What are 2 ways the weight of an object can change?
1) Does a 1 kg object weight 9.8 newtons on the moon? why?
No. 1kg of mass does not weigh 9.8N on the moon.
Weight = (mass) x (gravity).
Gravity is 9.8 m/s² on Earth, but gravity is only 1.62 m/s² on the moon.
2) How much does a 3-kg object weigh (on earth) in newtons?
Weight = (mass) x (gravity)
Gravity = 9.8 m/s² on Earth.
Weight = (3 kg) x (9.8 m/s² )
Weight = 29.4 N
3) How much does a 20-kg object weigh (on earth) in newton?
Weight = (mass) x (gravity)
Gravity = 9.8 m/s² on Earth.
Weight = (20 kg) x (9.8 m/s² )
Weight = 196 N
4) What must happen for the mass of an object to change?
When an object moves, its mass increases. The faster it moves, the greater its mass gets. But this is all part of Einstein's "Relativity". The object has to move at a significant fraction of the speed of light before any change can be noticed or measured. So as far as we are concerned, in everyday life, the mass of an object doesn't change, no matter where it is, or what you do to it.
5) What are 2 ways the weight of an object can change?
First, remember that the mass of an object doesn't change, no matter where it is, what you do to it, or what else is around it.
But its weight can change, because its weight depends on the strength of gravity in the place where the object is, and that gravity is the result of what else is around it in the neighborhood. So the weight can change even though the mass doesn't.
The weight of an object changes if you take it to a place where gravity is stronger or weaker.
Let's say we have an object whose mass is 90.72 kilograms. Like me !
As long as I stay on earth, where gravity is 9.8 m/s² , I weigh 889 Newtons (200 pounds).
. . . Fly me to the moon. Gravity = 1.62 m/s² Weight = 147 Newtons (33 lbs)
. . . Drag me to Jupiter. Gravity = 24.8 m/s² Weight = 2,249 N (506 pounds)
My mass never changed, but my weight sure did.
In this experiment, you will use a track and a toy car to explore the concept of movement. You will measure the time it takes the car to travel certain distances, and then complete some calculations. In the space below, write a scientific question that you will answer by doing this experiment.
Answer: if weight affects how fast they go?
Explanation:
Answer:
How can we change the speed of a toy car on a racetrack to describe the car’s motion?
Explanation:
thats the sample respond
The amount of light that enters the pupil is controlled by the:
retina.
lens.
inis.
Answer: The amount of light that enters the pupil is controlled by the Iris
Explanation:
a cohesive force between the liquids molecules is responsible for the fluids is called
Answer:
static force
Explanation:
mark me brainliest
A positive charge Q2 is uniformly distributed over a nonconducting disc of radius a which has a concentric circular hole of radius b. At the center of the hole there is another nonconducting disc of radius d where a charge Q1 is uniformly distributed.
a) Find the surface charge density of the disc with the hole σ2.
b) Find the surface charge density 01 of the disc of radius d.
c) Find the total charge enclosed by the circle of radius
Answer:
a) σ = [tex]\frac{Q_1}{ a^2 - b^2}[/tex] , b) σ = [tex]\frac{Q_2}{d^2}[/tex] , c) Q_ {total} = Q₁ + Q₂, σ_ {net} = [tex]\frac{Q_1 + Q_2}{\pi \ a^2}[/tex]
Explanation:
a) The very useful concept of charge density is defined by
σ = Q / A
In this case we have a circular disk
The are of a circle is
A = π r²
in this case we have a hole in the center of radius r = b, so
A_net = π r² - π r_ {hollow} ²
A_ {net} = π (a² - b²)
whereby the density is
σ = [tex]\frac{Q_1}{ a^2 - b^2}[/tex]
b) The density of the other disk is
σ = Q₂ / A₂
σ = [tex]\frac{Q_2}{d^2}[/tex]
c) The total waxed load is requested by the larger circle
Q_ {total} = Q₁ + Q₂
the net charge density, in the whole system is
σ = [tex]\frac{Q_{total} }{ A_{total} }[/tex]
the area is
A_{total} = π a²
since the other circle is inside, we are ignoring the space between the two circles
σ_ {net} = [tex]\frac{Q_1 + Q_2}{\pi \ a^2}[/tex]
What is happening in the graph shown below?
A.
The object moves away from the origin at a speed of 3 m/s, stands still 6 m away from the origin for 3 seconds, then moves toward the origin at a speed of 2 m/s.
B.
The object moves toward the origin at a speed of 3 m/s, stands still 6 m away from the origin for 3 seconds, then moves away from the origin at a speed of 2 m/s.
C.
The object moves toward the origin at a speed of 6 m/s, stands still 6 m away from the origin for 3 seconds, then moves away from the origin at a speed of 8 m/s.
D.
The object moves away from the origin at a speed of 6 m/s, stands still 6 m away from the origin for 3 seconds, then moves toward the origin at a speed of 8 m/s.
Answer:
D. The object moves away from the origin at a speed of 3 m/s, stands still 6 m away from the origin for 3 seconds, then moves toward the origin at a speed of 2 m/s.
Explanation:
I just got it right lol
