The dependent variable is academic performance while the independent variable is presence/absence of tutorial support.
The correct results are:
The dependent variable is academic performanceThe independent variable is the presence/absence of tutorial supportThe control group are students who did not get the tutorial support.The experimental group were students that got the tutorial supportIn every experiment, there is a dependent and independent variable as well as an experimental and a control group.
The experimental group receive the treatment while the control group do not receive the treatment. The independent variable is manipulated and its impact on the dependent variable is evaluated.
The control group are students who did not receive the tutorial support while the experimental group are students that received the tutorial support.
The dependent variable in this case is academic performance. Its outcome depends on the presence or absence of tutorial support (independent variable).
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Interactive Solution 8.29 offers a model for this problem. The drive propeller of a ship starts from rest and accelerates at 2.38 x 10-3 rad/s2 for 2.04 x 103 s. For the next 1.48 x 103 s the propeller rotates at a constant angular speed. Then it decelerates at 2.63 x 10-3 rad/s2 until it slows (without reversing direction) to an angular speed of 2.42 rad/s. Find the total angular displacement of the propeller.
Answer:
Δθ = 15747.37 rad.
Explanation:
The total angular displacement is the sum of three partial displacements: one while accelerating from rest to a certain angular speed, a second one rotating at this same angular speed, and a third one while decelerating to a final angular speed.Applying the definition of angular acceleration, we can find the final angular speed for this first part as follows:[tex]\omega_{f1} = \alpha * \Delta t = 2.38*e-3rad/s2*2.04e3s = 4.9 rad/sec (1)[/tex]
Since the angular acceleration is constant, and the propeller starts from rest, we can use the following kinematic equation in order to find the first angular displacement θ₁:[tex]\omega_{f1}^{2} = 2* \alpha *\Delta\theta (2)[/tex]
Solving for Δθ in (2):[tex]\theta_{1} = \frac{\omega_{f1}^{2}}{2*\alpha } = \frac{(4.9rad/sec)^{2}}{2*2.38*e-3rad/sec2} = 5044.12 rad (3)[/tex]
The second displacement θ₂, (since along it the propeller rotates at a constant angular speed equal to (1), can be found just applying the definition of average angular velocity, as follows:[tex]\theta_{2} =\omega_{f1} * \Delta_{t2} = 4.9 rad/s * 1.48*e3 s = 7252 rad (4)[/tex]
Finally we can find the third displacement θ₃, applying the same kinematic equation as in (2), taking into account that the angular initial speed is not zero anymore:[tex]\omega_{f2}^{2} - \omega_{o2}^{2} = 2* \alpha *\Delta\theta (5)[/tex]
Replacing by the givens (α, ωf₂) and ω₀₂ from (1) we can solve for Δθ as follows:[tex]\theta_{3} = \frac{(\omega_{f2})^{2}- (\omega_{f1}) ^{2} }{2*\alpha } = \frac{(2.42rad/s^{2}) -(4.9rad/sec)^{2}}{2*(-2.63*e-3rad/sec2)} = 3451.25 rad (6)[/tex]
The total angular displacement is just the sum of (3), (4) and (6):Δθ = θ₁ + θ₂ + θ₃ = 5044.12 rad + 7252 rad + 3451.25 rad ⇒ Δθ = 15747.37 rad.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
How much work is done when 100 N of force is applied to a rock to move it 20 m
Answer: 2000 J
Explanation: work W = F s
Is there a way to see moon and the sun at once?
३.रात में घूमने वाला write one word substitute
Explanation:
रात में घूमने वाला arthaarat निशाचर
Define Mechanical advantage
fe effort of 2125N is used to lift a Lead of 500N
through a Verticle high of 2.N using a buly System
if the distance Moved by the effort is 45m
Calculate 1. Work done on the load
2. work done by the effort
3. Efficiency of the System
Answer:
1) 1000Nm
2) 95,625Nm
3) 1.05%
Explanation:
Mechanical Advantage is the ratio of the load to the effort applied to an object.
MA = Load/Effort
1) Workdone on the load = Force(Load) * distance covered by the load
Workdone on the load = 500N * 2m
Workdone on the load = 1000Nm
2) work done by the effort = Effort * distance moves d by effort
work done by the effort = 2125 * 45
work done by the effort = 95,625Nm
3) Efficiency = Workdone on the load/ work done by the effort * 100
Efficiency = 1000/95625 * 100
Efficiency = 1.05%
Hence the efficiency of the system is 1.05%
a pendulum clock having Copper keeps time at 20 degree Celsius it gains 15 second per day if cooled to 0°C celsius calculate the coefficient of linear expansion of copper.
?.............................
it takes 560s for a runner to complete one circular lap, moving at a speed of 6.00 m/s. what is the radius of a track?
Answer:
534.8 meters
Explanation:
Use T=(2*pi*r)/v
560=(2*pi*r)/6
3360=2*pi*r
1680=pi*r
534.8 meters=radius
It takes 560s for a runner to complete one circular lap, moving at a speed of 6.00 m/s. The radius of a track is 534.7 m.
What is Distance?The distance covered by a body is equal to the sum of total path covered. It is equal to the total path traveled by an object during its entire journey.This quantity is always positive. It can't be 0 or a negative number.It is defined as a scalar quantity.
Mathematically, it can be calculated as follows :
distance = speed × time
The formula relating distance (d), speed (s), and time (t) is
d = st
First, Calculating the distance,
d = 560 s × 6 m·s⁻¹
= 3360 m
When, Calculating the track radius,
The distance travelled is the circumference of a circle,
C = 2пr
r = 3360/2п
= 534.7 m
The radius of the track is 534.7 m.
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. Assume that the batter does hit the ball. If the bat's instantaneous angular velocity is 30 rad/s at the instant of contact, and the distance from the sweet spot on the bat to the axis of rotation is 1.25 m, what is the instantaneous linear velocity of the sweet spot at the instant of ball contact
Answer:
37.5 m/s
Explanation:
Using,
Formula
v = ωr....................... Equation 1
Where ω = instantaneous angular velocity, v = instantaneous linear velocity, r = radius or distance from the sweet spot of the bat to the axis of rotation.
From the question,
Given: ω = 30 rad/s, r = 1.25 m
Substitute these values into equation 1
v = 30(1.25)
v = 37.5 m/s.
Hence the instantaneous linear velocity of the sweet spot at the instant of ball contact is 37.5 m/s
According to Newton's first law, an object at rest will _____.
never move
stay at rest forever
start moving
stay at rest unless moved by force
A planet of mass M has a moon of mass m in a circular orbit of radius R. An object is placed between the planet and the moon on the line joining the center of the planet to the center of the moon so that the net gravitational force on the object is zero. How far is the object placed from the center of the planet
Answer:
r =[tex]\frac{ 1 \pm \sqrt{ \frac{m}{M} } }{1 - \frac{m}{M} }[/tex]
Explanation:
Let's apply the universal gravitation law to the body (c), we use the indications 1 for the planet and 2 for the moon
∑ F = 0
-F_{1c} + F_{2c} = 0
F_{1c} = F_{2c}
let's write the force equations
[tex]G \frac{m_c M}{r^2} = G \frac{m_c m}{(d-r)^2}[/tex]
where d is the distance between the planet and the moon.
[tex]\frac{M}{r^2} = \frac{m}{(d-r)^2}[/tex]
(d-r)² = [tex]\frac{m}{M} \ \ r^2[/tex]
d² - 2rd + r² = \frac{m}{M} \ \ r^2
d² - 2rd + r² (1 - [tex]\frac{m}{M}[/tex]) = 0
(1 - [tex]\frac{m}{M}[/tex]) r² - 2d r + d² = 0
we solve the second degree equation
r = [2d ± [tex]\sqrt{ 4d^2 - 4 ( 1 - \frac{m}{M} ) }[/tex] ] / 2 (1- [tex]\frac{m}{M}[/tex])
r = [2d ± 2d [tex]\sqrt{ \frac{m}{M} }[/tex]] / 2d (1- [tex]\frac{m}{M}[/tex])
r =[tex]\frac{ 1 \pm \sqrt{ \frac{m}{M} } }{1 - \frac{m}{M} }[/tex]
there are two points for which the gravitational force is zero
The distance between object from planet will be "[tex]\frac{R}{[1+\sqrt{\frac{m}{M} } ]}[/tex]".
According to the question,
Let,
Object is "x" m from planet center = R - xGravitational force = 0Mass of object = m₁As we know,
→ [tex]Prerequisites-Gravitational \ force = \frac{GMm}{r^2}[/tex]
Now,
→ [tex]\frac{GMm_1}{x^2} = \frac{Gmm_1}{(R-x)^2}[/tex]
→ [tex]\frac{(R-x)^2}{x^2} = \frac{m}{M}[/tex]
→ [tex]\frac{R-x}{x} =\sqrt{\frac{m}{M} }[/tex]
→ [tex]x = \frac{R}{[1+ \sqrt{\frac{m}{M} } ]}[/tex]
Thus the answer above is appropriate.
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How fast were both runners traveling after 4 seconds?
40
Distance (in yards)
30
20
10
1
2.
3
0
Time in seconds
Answer:
they were fast ⛷⛷
On a 10 kg cart (shown below), the cart is brought up to speed with 50N of force for 7m, horizontally. At this point (A), the cart begins to experience an average frictional force of 15N throughout the ride.
Find:
a) The total energy at (A)
b) The velocity at (B)
c) The velocity at (C)
d) Can the cart make it to Point (D)? Why or why not?
Highest density of electrostatic charges in a metal is found where
I don't know the answer but I just want points sorry
(1) Which appliance is designed to transfer electrical energy to kinetic energy?
D)
A food mbuer
BB kettle
Clamp
D radio
Answer:
bb kettle
Explanation:
it transfres electricsl to kinetic
which one is odd copper,plastic,rubber
Answer:
It's plastic.
trust me it's plastic, i've rad it somewhere.
All of them have something that's not like the others.
-- Rubber is the only one on the list that has two repeated letters.
-- Plastic is the only one on the list thagt has no repeated letters.
-- Plastic is the only one on the list that has no 'r' in its name.
-- Copper is the only one on the list that is an element, not a compound.
-- Copper is the only good electrical conductor on the list.
-- Plastic is the only one on the list with more than six letters in its name.
-- Rubber is the only one on the list with no 'p' in its name.
-- Plastic is the only one on the list that doesn't end in "-er".
a sharp image is formed when light reflects from a
Answer:
Regular reflection
Explanation:
Regular reflection occurs when light reflects off a very smooth surface and forms a clear image.
i hope this helps a bit.
According to the context, a sharp image is formed when light reflects from a regular reflection.
What is regular reflection?It is reflection without diffusion that obeys the laws of geometrical optics, as in mirrors.
This reflection of light happens when the angles that the two rays determine with the surface are equal.
Therefore, we can conclude that according to the context, a sharp image is formed when light reflects from a regular reflection.
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why doesn't a radio operating with two batteries function when one of the batteries is reversed?
Answer:
If you have two batteries and they have precisely the same voltage then placing one backwards will effectively cancel out the voltages and no current will flow. However, batteries aren't like that. The slightest difference in voltages mean that current will flow.
Explanation:
A cylindrical body has 6 m height and its radius is 2 metre calculate its volume. Ans :75.428m3
Answer:
75.4
Explanation:
r= 2
h= 6
v= 22/7 *r*r*h
v= 75.42
Carl works hard to get a grades on his report card because his mother pays him 25 dollars for each semester he earns straight as Carl’s behavior is being influenced by
A girl weighing 45kg is standing on the floor, exerting a downward force of 200N on the floor. The force exerted on her by the floor is ..............
Select one:
a.
No force exerted
b.
Less than 2000N
c.
Equal to 200 N
d.
Greater than 200 N
Answer:
c.
Equal to 200 N..........
What is it called when the moon passes through the penumbra of Earth’s shadow?(1 point)
total lunar eclipse
total solar eclipse
partial lunar eclipse
partial solar eclipse
Answer: I'm not sure, but I think it would be a total lunar eclipse
When the moon passes through the penumbra of Earth’s shadow it is referred to as partial lunar eclipse. The correct option is C.
What is partial lunar eclipse?A partial lunar eclipse occurs when the moon is not completely immersed in the umbra of the earth's shadow.
During a partial solar eclipse, the Moon, Sun, and Earth do not align perfectly straight, and the Moon casts only the penumbra of its shadow on Earth. From our vantage point, it appears that the Moon has eaten the Sun.
A shadow's penumbra is the lighter outer edge. Partial solar eclipses are caused by the Moon's penumbra, while penumbral lunar eclipses are caused by the Earth's penumbra. The penumbra is a type of lighter shadow.
The Penumbra is a half-shadow region that occurs when an object only partially covers a light source.
Thus, the correct option is C.
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1. A perspex box has a 10 cm square base and contains water to a height of 10 cm. A piece of rock of mass 600g is lowered into the water and the level rises to 12 cm.
(a) What is the volume of water displaced by the rock?
(b) What is the volume of the rock?
(c) Calculate the density of the rock
Answer:
(a) The volume of water is 100 cm³
(b) The volume of the rock is 20 cm³
(c) The density of the rock is 30 g/cm³
Explanation:
The given parameters of the perspex box are;
The area of the base of the box, A = 10 cm²
The initial level of water in the box, h₁ = 10 cm
The mass of the rock placed in the box, m = 600 g
The final level of water in the box, h₂ = 12 cm
(a) The volume of water in the box, 'V', is given as follows;
V = A × h₁
∴ The volume of water in the box, V = 10 cm² × 10 cm = 100 cm³
The volume of water in the box, V = 100 cm³
(b) When the rock is placed in the box the total volume, [tex]V_T[/tex], is given by the sum of the rock, [tex]V_r[/tex], and the water, V, is given as follows;
[tex]V_T[/tex] = [tex]V_r[/tex] + V
[tex]V_T[/tex] = A × h₂
∴ [tex]V_T[/tex] = 10 cm² × 12 cm = 120 cm³
The total volume, [tex]V_T[/tex] = 120 cm³
The volume of the rock, [tex]V_r[/tex] = [tex]V_T[/tex] - V
∴ [tex]V_r[/tex] = 120 cm³ - 100 cm³ = 20 cm³
The volume of the rock, [tex]V_r[/tex] = 20 cm³
(c) The density of the rock, ρ = (Mass of the rock, m)/(The volume of the rock)
∴ The density of the rock, ρ = 600 g/(20 cm³) = 30 g/cm³
An object is accelerated by a net force in which direction?
A. at an angle to the force
B. in the direction of the force
C. in the direction opposite to the force
D. Any of these is possible.
Answer:
B. in the direction of the force
Explanation:
Sana nakatulong
Mechanical energy is the most concentrated form of energy.
a. true
b. false
Sam moves an 800 N wheelbarrow 5 meters in 15 seconds. How much work did he do?
Answer:
work done= force × displacement
=800×5
=4000J
Explanation:
The amount of work done is the result of the magnitude of force applied and the displacement of the body due to the force applied. Therefore, work done is defined as the product of the applied force and the displacement of the body.
How much energy would be required to move the earth into a circular orbit with a radius 2.0 kmkm larger than its current radius
Answer:
[tex]3.52\times 10^{25}\ \text{J}[/tex]
Explanation:
G = Gravitational constant = [tex]6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2[/tex]
M = Mass of Sun = [tex]1.989\times 10^{30}\ \text{kg}[/tex]
m = Mass of Earth = [tex]5.972\times 10^{24}\ \text{kg}[/tex]
[tex]r_i[/tex] = Initial radius of orbit = [tex]1.5\times 10^{11}\ \text{m}[/tex]
[tex]r_f[/tex] = Final radius of orbit = [tex]((1.5\times 10^{11})+2\times 10^3)\ \text{m}[/tex]
Energy required is given by
[tex]E=\dfrac{1}{2}\Delta U\\\Rightarrow E=\dfrac{GMm}{2}(\dfrac{1}{r_i}-\dfrac{1}{r_f})\\\Rightarrow E=\dfrac{6.674\times 10^{-11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{2}(\dfrac{1}{1.5\times 10^{11}}-\dfrac{1}{(1.5\times 10^{11})+2\times 10^3})\\\Rightarrow E=3.52\times 10^{25}\ \text{J}[/tex]
The energy required would be [tex]3.52\times 10^{25}\ \text{J}[/tex].
The gravitational potential energy of an object is defined as the energy it has due to its position in a gravitational field. A ball with a weight of 50 N is lifted to a height of 1 meter. Which graph correctly represents the change in gravitational potential energy (shaded in blue) as it is lifted to this height?
Answer:
athletic
Explanation:
because internet system has been down since we were in few days
During which phase is the moon not visible?
A) Full Moon
B) First quarter
C) New moon
D) Waxing crescent
Answer:
they are right it is a new moon
Explanation:
took the test
At the base of a hill, a 90 kg cart drives at 13 m/s toward it then lifts off the accelerator pedal). If the cart just barely makes it to the top of this hill and stops, how high must the hill be?
Answer:
8.45 m
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 90 Kg
Initial velocity (u) = 13 m/s
Final velocity (v) = 0 m/s
Height (h) =?
NOTE: Acceleration due to gravity (g) = 10 m/s²
The height of the hill can be obtained as follow:
v² = u² – 2gh (since the cart is going against gravity)
0² = 13² – (2 × 10 × h)
0 = 169 – 20h
Rearrange
20h = 169
Divide both side by 20
h = 169/20
h = 8.45 m
Therefore, the height of the hill is 8.45 m
