Answers
The angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec
The angular velocity (ω) of an object is the rate at which the object's angle position is changing in relation to time.
For a wheel attached to an incline angle, the angular velocity can be computed by considering the conservation of energy theorem.
As such the total kinetic energy (K.E) and rotational kinetic energy (R.K.E) at a point is equal to the total potential energy (P.E) at the other point.
i.e.
P.E = K.E + R.K.E
[tex]\mathbf{mgh = \dfrac{1}{2}m(r \times \omega)^2 + \dfrac{1}{2}\times I \times \omega^2}[/tex]
[tex]\mathbf{gh = \dfrac{1}{2}(r \times \omega)^2 + \dfrac{1}{2}\times r^2 \times \omega^2}[/tex]
[tex]\mathbf{2 \times \dfrac{gh}{r^2} =\omega^2 + \omega^2}[/tex]
[tex]\mathbf{2 \omega^2=2 \times \dfrac{9.81 \times 8 m }{2.0 ^2} }[/tex]
[tex]\mathbf{\omega^2=\dfrac{39.24 }{2}}[/tex]
[tex]\mathbf{\omega=\sqrt{19.62 } \ rad/sec}[/tex]
[tex]\mathbf{\omega=4.429 \ rad/sec}[/tex]
Therefore, we can conclude that the angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec
Learn more about angular velocity here:
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The angular velocity of the wheel depends on the mass, radius and the
mode of rotation of the wheel (with or without slipping).
The angle velocity at the bottom of the incline, ω ≈ 4.43 rad/secReasons:
The given parameters are;
Radius of the wheel, r = 2.0 m
Height of the incline, h = 8.0 m
Required:
Angular velocity of the wheel at the bottom of the incline.
Solution:
The potential energy of the wheel at the top of the hill, P.E. = m·g·h
[tex]Sum \ of \ the \ kinetic \ energy \ of \ the \ wheel, \ K.E. = \mathbf{\displaystyle \frac{1}{2} \cdot m \cdot v^2 + \frac{1}{2} \cdot I \cdot \omega ^2}[/tex]
Where;
v = The translational velocity of the wheel = ω·r
I = The moment of inertia of the wheel = m·r²
Therefore'
[tex]Sum \ of \ K.E. = \displaystyle \frac{1}{2} \cdot m \cdot (\omega \cdot r)^2 + \frac{1}{2} \cdot m \cdot r^2 \cdot \omega ^2 = \mathbf{m \cdot r^2 \cdot \omega^2}[/tex]
At the bottom of the hill, the potential energy is converted to kinetic energy
Therefore;
P.E. = Sum of K.E.
m·g·h = m·r²·ω²
g·h = r²·ω²
[tex]\displaystyle \omega = \sqrt{ \frac{g \cdot h}{r^2} } = \mathbf{ \frac{\sqrt{g \cdot h} }{r}}[/tex]
Where;
g = Acceleration due to gravity ≈ 9.81 m/s²
Therefore;
[tex]\displaystyle \omega = \frac{\sqrt{9.81 \times 8} }{2} \approx \mathbf{ 4.43}[/tex]
The angular velocity of the of the wheel at the bottom of the incline, ω ≈ 4.43 rad/secLearn more about the law of conservation of energy here:
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https://brainly.com/question/816294
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Decant it.
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a
A person throws a ball up into the air, and the ball falls back towar
would the kinetic energy be the lowest? (1 point)
at a point before the ball hits the ground
when the ball leaves the person's hand
o when the ball is at its highest point
at a point when the ball is still rising
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Answer:
when the ball is at its highest point
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hope this helps you!!
Write down the condition required for the thermonuclear fusion.
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The 0.01 kg marble is dropped from rest at A through the smooth glass tube and accumulate in the basket at C as shown in Figure Q2(b). Determine: i) the velocity of the marble at B ii) the horizontal distance R of the basket from the end of the tube, and iii) the speed at which the marble falls into the basket.
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Which is the independent variable in the graph of the U.S. population 1800-1900?
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Answers
Answer:
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Explanation:
The independent variable is the x axis of graph
plz help ASAP I'll mark as brainliest
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Hi there!
1.
Hooke's law states that:
F = -kx
k = Spring constant (N/m)
x = DISPLACEMENT from equilibrium (m)
Essentially, the force of a spring is PROPORTIONAL to its spring constant and its displacement from its equilibrium point.
2.
The force of the spring (T) is not proportional to the spring's length (l), but rather its DISPLACEMENT from its equilibrium length. (Δl)
3.
The equilibrium length is where the force of the spring (T) = 0N. Looking at the graph, the line intersects this value at l = 30cm.
4.
We can begin by looking at the given graph.
When the spring force = 4N, the total length of the spring is 35 cm.
Now, the EQUILIBRIUM length is 30 cm, so the total elongation is:
35 - 30 = 5 cm.
5.1.
If the spring elongates by 10 cm, the total length of the spring is:
30 + 10 = 40 cm
According to the graph, a length of 40 cm corresponds to a force of 8N.
5.2.
We can solve for the weight of the ball using the following:
W (weight) = m (mass) · acceleration due to gravity (10N/kg)
Using a summation of forces:
∑F = T - W
The elongation that we are solving for occurs at the equilibrium point (net force = 0 N), so:
0 = T - W
T = W = 8 N
5.3.
0 = T - Mg
T = Mg
Use the prior value of T and gravity to solve:
8 = 10M
m = 0.8 kg
stuck with this one.
Answers
What is the half-life of the imaginary element Lokium? Show your work
Answers
Answer:What is the half life of the element Lokium?
The half-live of the element Lokium is 4.
Explanation:
sorry if im wrong, have good day
Answer:
The half-life of the element Lokium is 4
What is half-life?
Whether or not a given isotope is radioactive is a characteristic of that particular isotope. Some isotopes are stable indefinitely, while others are radioactive and decay through a characteristic form of emission. As time passes, less and less of the radioactive isotope will be present, and the level of radioactivity decreases. An interesting and useful aspect of radioactive decay is the half-life. The half-life of a radioactive isotope is the amount of time it takes for one half of the radioactive isotope to decay. The half-life of a specific radioactive isotope is constant; it is unaffected by conditions and is independent of the initial amount of that isotope.
#SPJ2
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Answer:
Explanation:
Objects want to continue doing what they're already doing because they are "lazy." This is called inertia.
Using the formula below, calculate the kinetic energy of the 6 gram stone going 10 Mph
going 10 mph
KE = 1/2 Mass x Speed?
=
Answers
Answer:
Explanation:
A wonderful (NOT) mixed unit problem
convert mph to m/s
v = 10 mi/hr(5280 ft/mi)(12 in/ft)(2.54 cm/in) / (100 cm/m) / (3600 s/hr)
v = 4.4704 m/s
KE = ½(0.006)4.4704²
KE = 0.05995342848 J
KE = 0.06 J
Suppose you want to make the demonstration more dramatic by attracting as much hair as possible with the balloon. Which of the following is the best choice to accomplish your goal? A. Decrease the electric force by holding the balloon closer to the hair. B. Increase the electric force by rubbing the balloon for a longer period of time. C. Decrease the electric force by using a larger balloon. D. Increase the electric force by holding the balloon farther away from the hair.
Answers
Answer:
B. Increase the electric force by rubbing the balloon for a longer period of time.
Explanation:
You are given two vectors A⃗ =−3.00i^+5.00j^ and B⃗ =7.00i^+2.00j^. Let counterclockwise angles be positive.
a)What angle θA, where 0∘≤θA<360∘, does A⃗ make with the +x-axis?
Express your answer in degrees.
b)What angle θB, where 0∘≤θB<360∘, does B⃗ make with the +x-axis?
Express your answer in degrees.
c)Vector C⃗ is the sum of A⃗ and B⃗ , so C⃗ =A⃗ +B⃗ . What angle θC, where 0∘≤θC<360∘, does C⃗ make with the +x-axis?
Express your answer in degrees.
Answers
Answer:
Explanation:
We can subtract directly the corresponding components and check using the parallelogram rule.
Explanation:
Have a look:
enter image source here
Where, graphically, I used the fact that:
→
A
−
→
B
=
→
A
+
(
−
→
B
)
For the magnitude we use Pythagoras (with the components) to get:
∣
∣
∣
→
A
−
→
B
∣
∣
∣
=
√
(
−
1
)
2
+
(
5
)
2
=
√
1
+
25
=
√
26
≈
5.1
For the direction I can see that will be
90
∘
from the
x
axis up to the
y
axis, plus the little bit passed the
y
axis given as:
θ
=
arctan
(
1
5
)
=
11.3
∘
giving in total: angle
=
90
∘
+
11.3
∘
=
101.3
∘
A 2.98-kg object oscillates on a spring with an amplitude of 8.05 cm. Its maximum acceleration is 3.55 m/s2. Calculate the total energy.
Answers
Answer:
a = ω^2 A formula for max acceleration (ignoring sign)
V = ω A formula for max velocity
V^2 = ω^2 A^2 = a A from first equation
E = 1/2 M V^2 = 1/2 * 2.98 * 3.55 * .0805 = .426 J
(kg * m/sec^2 * m = kg m^2 / sec^2 = Joule
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Answers
Answer:
The removal of heat energy slows the speed of particles
Explanation:
When you add heat to a substance, the heat energy gets transferred to kinetic energy, and the molecules began to move a greater distance at a greater speed. When you remove heat, the opposite happens.
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Diffusion
Facilitated transport
Osmotic pressure
Endocytosis
Answers
Question
what is the force exerted on a selectively permeable membrane beacuse water has moved from an area of higher concentration to an area of Lower concentration of water?
Answer
OSMOSIS
Explanation
Osmosis: This is the hydrostatic force acting to equalize the concentration of water on both sides of the membrane that is impermeable to substances dissolved in that water. Water will move along its concentration gradient.
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A guitar string 63.6 cm long vibrates with a standing wave that has five antinodes. Which harmonic is this
Answers
Answer:
fifth harmonic
Explanation:
A device is rated at 1.3kW when connected to a 120 V source. The equivalent resistance of this device in ohm is:
a- 18.3
b- 12.0
c- 11.1
d- 14.4
Answers
Answer:
D.
correct me if im wrong
brainlest plsss<333
(c) It is suggested that one side of the copper sheet cools to a lower temperature than the
other side.
Explain why this does not happen.
[2]
Answers
Answer:
Explanation:
The word "sheet" implies that the copper is quite thin.
Copper is also a very good conductor of heat.
Therefore, with a very short heat flow distance to cover and a high rate of heat transmission, temperature differences on either side of the sheet are almost instantaneously eliminated by heat flow.
25.Figure 22.22 shows a plot
of potential versus position
along the x-axis. Make a plot
of the x-component of the
electric field for this situation.
Answers
The relationship between the potential and the electric field allows to find the results for the value of the electric field as a function of the distance is:
In the attachment we see the graph of the electric field as a function of distance.
Electric potential is defined by the change in potential energy of a test charge between two points, between the value of the test charge.
dV = - E . ds
E = [tex]- \frac{dV}{ds} \ \hat s[/tex]
Where the bold letters indicate vectors, V is the potential difference, E the electric field and s the path.
Let's apply this expression for each section of the given graph:
1) section from x₀ = 0 to x_f = 2 m, the potential is V₀ = 2 V is constant.
The derivative of a constant is zero.
E = 0
2) Section between x₀ = 2 and x_f = 4 m, the potential varies linearly from V₀ = 2 v to V_f = -2 V.
We look for the equation of the line.
V-V₀ = m (x- x₀)
We carry out the derivative.
E = - m i ^
The slope (m) is:
[tex]m= \frac{V_f - V_o}{x_f- x_o}[/tex]
Let's calculate.
[tex]m= \frac{-2 -2}{4-2} = \ -2 \ V/m[/tex]
Let's substitute.
E = [tex]2 \hat i \ V/m[/tex]
3) From x₀ = 4 to x_f = 4.5 m, the potential varies from V₀ = -2 to V_f = 0.
We look for the equation of the line and we derive.
E = - m i ^
Let's substitute.
[tex]m = \frac{0-(-2)}{4.5-4} = \ 4 V/m[/tex]
E = - 4 [tex]\hat i[/tex] V / m
4) From x₀ = 4.5 m to x_f = 6m. The potential is constant and the derivative of a constant is zero.
E = 0
5) From x₀ = 6m to x_f = 8 m, the potential changes linearly from v₀ = 0 to V_f = 1 V
We look for the equation of the line and we derive.
E = - m i ^
[tex]m = \frac{1-0}{8-6} = \ 0.5 \ V/m[/tex]
E = - 0.5 [tex]\hat i[/tex] V/m
6) From x₀ = 8m to x_f = 9m, the potential changes linearly from V₀ = 1 V to V_f = -1.
We look for the equation of the line and we derive.
E = - m i ^
[tex]m = \frac{-1-1}{9-8} = \ -2 \ V/m[/tex]
Let's substitute.
E = 2 [tex]\hat i[/tex] V/m
7) From x₀ = 9m to x_f = 10 m, the potential changes linearly from V₀ = -1 V to V_f = -2 V
We look for the equation of the line and we derive.
E = - m i ^
[tex]m = \frac{-2+1}{10-9} = \ -1 \ V/m[/tex]
Let's substitute.
E = 1 [tex]\hat i[/tex] V/m
In the attachment we can see these Electric fields as a function of distance.
In conclusion, the relationship between the potential and the electric field we can find the results for the value of the electric field as a function of the distance is:
In the attachment we see the graph of the electric field as a function of distance.
Learn more about the electric field here: brainly.com/question/14306881
A small ball with a mass of 0.6 kg and a velocity of 12 m/s hits another ball with the same mass. The firs forward and hits a third ball with a mass of 0.2 kg. If the system is closed, what is the velocity of the third ball?
4 m/s
36 m/s
30 m/s
1.44 m/s
Answers
4. If a charged body is negatively charged, it has excess of _______
Answers
Answer:
Electrons
Explanation:
Electrons are negatively charged sub-atomic particles, therefore when a body's negatively charged, it means that there's more electrons than protons.
A crate is released on a frictionless plank inclined at angle q with
respect to the horizontal. Which of the following relationships is true?
(Assume that the x-axis is parallel to the surface of the incline.)
a. Fy = Fg
c. Fy = Fx
b. Fx = 0
d. none of the above
Answers
None of the given options is correct based on the relationship for the components of the crate's weight.
The given parameter:
Angle of inclination, = qThe vertical component of the force on the crate is calculated as follows;
[tex]F_y = W \times cos(q)\\\\F_y = F_gcos(q)[/tex]
The horizontal component of the crate is calculated as follows;
[tex]F_x = F_g \times sin(q)\\\\F_x = F_g sin(q)[/tex]
Thus, we can conclude that none of the given options is correct based on the relationship for the components of the crate's weight.
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i need help with the problem below
Answers
Answer:
Explanation:
a) F = ma
a = F/m
a = 9(800) / 1 x 10⁹ = 7.2 x 10⁻⁶ m/s
b) t = v/a
t = 200 / 7.2 x 10⁻⁶
t = 2.8 x 10⁷ s about 10½ months
c) v² = u² + 2as
s = (v² - u²) / 2a
s = (200² - 0²) / (2( 7.2 x 10⁻⁶))
s = 2.8 x 10⁹ m nearly 7 times around the earth
And all this assumes NO FRICTION.
A 2.2 kg model rocket is shot straight up in the air from the ground, with an initial velocity of 36.4 m/s. The rocket reaches its maximum height, and falls back to the ground. What is the maximum height of the rocket? Round your answer to 2 decimal places.
Answers
Answer:
Explanation:
Ignoring friction, the initial kinetic energy will convert to maximum potential energy at its highest point.
PE = KE
mgh = ½mv²
h = v²/2g
h = 36.4²/ (2(9.81))
h = 67.53109...
h = 67.53 m
The block in the figure below has a mass of 5.1 kg and it rests on an incline of angle . You pull on the rope with a force F = 34 N. Assume the incline is smooth and determine the angle of the incline if the block moves with constant speed.
Answers
42.9°
Explanation:
Let's assume that the x-axis is aligned with the incline and the positive direction is up the incline. We can then apply Newton's 2nd law as follows:
[tex]x:\;\;\;\;F - mg\sin{\theta} = 0\;\;\;\;[/tex]
[tex]\Rightarrow mg\sin{\theta} = F[/tex]
Note that the net force is zero because the block is moving with a constant speed when the angle of the incline is set at [tex]\theta.[/tex] Solving for the angle, we get
[tex]\sin{\theta} = \dfrac{F}{mg}[/tex]
or
[tex]\theta = \sin^{-1}\left(\dfrac{F}{mg}\right)[/tex]
[tex]\;\;\;= \sin^{-1}\left[\dfrac{34\:\text{N}}{(5.1\:\text{kg})(9.8\:\text{m/s}^2)}\right][/tex]
[tex]\;\;\;=42.9°[/tex]
explain how water erosion changes land forms
Answers
Answer:
Water erosion can slowly wash away dirt and rocks, wearing away landforms and sometimes forming rivers that can even carve out stone over time.
Andy Petite pitches a 0.8 kg baseball with a velocity of 67 m/s. Josh Hamilton
hits the ball back at a velocity of -44 m/s. Determine the impulse Josh
Hamilton's bat delivered to the baseball and enter to the nearest kg-m/s with
the correct sign.
Answers
The Impulse delivered to the baseball is 89 kgm/s.
To solve the problem above, we use the formula of impulse.
⇒ Formula:
I = m(v-u)................. Equation 1Where:
I = Impulse delivered to the baseballm = mass of the baseballv = Final velocity of the baseballu = initial speed of the baseballFrom the question,
⇒ Given:
m = 0.8 kgu = 67 m/sv = -44 m/s⇒ Substitute these values into equation 1
I = 0.8(-44-67)I = 0.8(-111)I = -88.8I ≈ -89 kgm/sNote: The negative tells that the impulse is in the same direction as the final velocity and therefore can be ignored.
Hence, The Impulse delivered to the baseball is 89 kgm/s.
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