Obtain a copy of the Student Guide for this lab. Your teacher may provide a copy, or you can click the link to access it. Be sure to read the entire Student Guide for this lab.
Did you read the Student Guide carefully?
yes
no
I cannot provide the information for the student guide as I do not have access to the specific lab or the guide. So, I cannot confirm whether I have read the student guide or not. Please provide me with the necessary information or a specific lab for me to provide accurate and helpful information. Instead of this I can give a brief about the student guide for a physics lab.
What is the Student Guide for the physics lab?A student guide for a physics lab is a document that provides students with information about the expectations, procedures, and safety guidelines for the lab. The guide usually includes the following information:
1. Introduction: A brief overview of the lab, including its purpose and objectives.
2. Materials and equipment: A list of the materials and equipment that students will use during the lab, as well as instructions on how to use them.
3. Procedure: A step-by-step guide on how to perform the experiment, including any calculations or data analysis that may be required.
4. Safety guidelines: A list of safety guidelines that students should follow during the lab, including proper handling of materials, use of protective equipment, and emergency procedures.
5. Pre-lab assignments: Assignments that students need to complete before the lab, such as reading the lab manual or reviewing relevant concepts.
6. Post-lab assignments: Assignments that students need to complete after the lab, such as analyzing data, writing lab reports, or answering questions.
Therefore, A student guide is an essential tool that helps students prepare for the lab, understand the concepts and procedures involved, and stay safe while conducting the experiment. It is important for students to read the entire guide carefully to ensure they have a clear understanding of the lab expectations and procedures.
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An object whose weight is 10kg is placed on smooth plane inclined at 30° to the horizontal. find the acceleration of the object as it moves down the plane
please i need help
Answer:
4.9 m/s²
Explanation:
Draw a free body diagram. There are two forces on the object:
Weight force mg pulling straight down,
and normal force N pushing perpendicular to the plane.
Sum the forces in the parallel direction.
∑F = ma
mg sin θ = ma
a = g sin θ
a = (9.8 m/s²) (sin 30°)
a = 4.9 m/s²
For the circuit shown in (Figure 1), find the potential difference between points a and b. Each resistor has
R = 160 N and each battery is 1.5 V
The potential difference between points a and b is zero.
Total emf of the series circuit
The total emf in the circuit is the sum of all the emf in the circuit.
emf(total) = 1.5 + 1.5 = 3.0 V
Potential differenceThe potential difference between two points, a and b is calculated as follows;
V(ab) = Va - Vb
V(ab) = 1.5 - 1.5
V(ab) = 0
Thus, the potential difference between points a and b is zero.
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One scientific investigation may result in
A.
further investigations.
B.
the generation of new ideas or phenomena to study.
C.
the development of new methods or procedures for investigation.
D.
all of these
A student conducts an investigation on electricity and magnetism. Which relationship will the student discover between the current and the magnetic field strength in a coiled wire?
Answer:
As current increases, the magnetic field strength increases.
So you're correct! :)
As the strength of the magnetic field increases, the current also increases.
From classical electromagnetism, we know that current is induced on conductor owing to relative motion between the conductor and the magnet. This is the principle upon which electromagnetic induction is based.
In the experiment, the student will find that the strength of the magnetic field is directly proportional to the induced current. Hence, as the strength of the magnetic field increases, the current also increases.
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An object moving with uniform acceleration has a velocity of 10.5 cm/s in the positive x-direction when its x-coordinate is 2.72 cm. If its x-coordinate 2.30 s later is ?5.00 cm, what is its acceleration? The object has moved to a particular coordinate in the positive x-direction with a certain velocity and constant acceleration; then it reverses its direction and moves in the negative x-direction to a particular x-coordinate in time t. We are given an initial velocity vi = 10.5 cm/s in the positive x-direction when the initial position is xi = 2.72 cm (t = 0). We are given that at t = 2.30 s, the final position is xf = ?5.00 cm. The acceleration is uniform so that we have the following equation in terms of the constant acceleration a. Xf-Xi=Vit-1/2at^2 Now we substitute the given values into this equation. (___cm)-(___cm)=(___cm/s)(__s)+1/2a(___s)
Answer:
Acceleration = 8.27 cm/s²
Explanation:
We are given;
initial velocity; v_i = 10.5 cm/s
Initial position; x_i = 2.72 cm
Time; t = 2.30 s
final position; x_f = 5.00 cm
To find the acceleration, we will make use of the formula;
x_f - x_i = (v_i * t) - (½at²)
Plugging in the relevant values, we have;
5 - 2.72 = (10.5 × 2.3) - (½ × a × 2.3²)
2.28 = 24.15 - 2.645a
24.15 - 2.28 = 2.645a
2.645a = 21.87
a = 21.87/2.645
a = 8.27 cm/s²
A 100V battery is connected to two oppositely charged plates that are 10cm apart.
i. What is the magnitude of the electric field?
ii. Calculate the Electric Force exerted on a +200 C point charge.
iii. What is the electric potential energy of the charge when it’s 8cm and 2cm from the negatively
charged plate?
iv. How much work was required to move the charge from 8cm to 2cm?
v. If the 10gram point charge was at rest at point A, what is the final speed at point B?
Answer:
A 100V battery is connected to two oppositely charged plates that are 10cm apart.
i. what is the magnitude of the electric field?
ii. Calculate the electric force exerted on a +200uC point charge.
Explanation:
5. 500 kg gold with a volume of 0.026 m
A. 0.193 kg/m3
B. 13 kg/m3
C. 19230 kg/m3
4
Answer:
19230 kg/m^3
Explanation:
Divide its mass by its volume to get its density:
500 kg/0.026 m = 19230 kg/m^3
Also, please provide the question next time since other's didn't know what they should solve for in this problem-- I only knew because I had this same problem.
Answer:
19230 kg/m3
Explanation:
I took the test, that is the answer
This interaction is not possible in nature because it violates a conservation law. Which quantity is not conserved?
p+ e-+ --> π+n+ve
Answer:
its D i just did it on the app
a.
13. A planet orbiting the sun
Newton's 1st Law
b. Newton's 2nd Law
c. Newton's 3rd Law
Newtons first law of motion.
For the following distance vs time graph
Answer:
3
5
4
Explanation:
x = (8=(9+9)
(9+9) = 4, 5, 3
A plane travels 1743 KM in 2 hours 30 minutes. How fast was the plane traveling?
Answer:
[tex]v=697.2km/h[/tex]
Explanation:
Hello.
In this case, since the velocity is computed via the division of the distance traveled by the elapsed time:
[tex]V=\frac{d}{t}[/tex]
The distance is clearly 1743 km and the time is:
[tex]t=2h+30min*\frac{1h}{60min} =2.5h[/tex]
Thus, the velocity turns out:
[tex]v=\frac{1743km}{2.5h}\\ \\v=697.2km/h[/tex]
Which is a typical velocity for a plane to allow it be stable when flying.
Best regards.
A crate (100 kg) is in an elevator traveling upward and slowing down at 6 m/s2. Find the normal force exerted on the crate by the elevator. Assume g
Answer:
380 N
Explanation:
Forces: gravity mg (downward), Normal force N (upward)
Acceleration: Note velocity is up, but slowing down, so acceleration is opposite to velocity, or downward
Newton's 2nd law:
mg - N = ma
N = m(g - a) = 100(9.8 - 6) = 380 N
A teenager is standing at the rim of a large horizontal uniform wooden disk that can rotate freely about a vertical axis at its center. The mass of the disk (in kg) is M and its radius (in m) is R. The mass of the teenager (in kg) is m. The disk and teenager are initially at rest. The teenager then throws a large rock that has a mass (in kg) of m_rock. As it leaves the thrower's hands, the rock is traveling horizontally with speed v (in m/s) relative to the earth in a direction tangent to the rim of the disk. The teenager remains at rest relative to the disk and so rotates with it after throwing the rock. In terms of M, R, m, m_rock and v, what is the angular speed of the disk
Answer:
[tex]\omega = \frac{m_{rock} \cdot v}{\left(\frac{M}{2}+m\right) R}[/tex]
Explanation:
[tex](inertia.of.the.disk) Id =\frac{M R^{2}}{2}\\(inertia.of.the.solid) Is = m R^{2}\\\\Itotal = Id + Is = \left(\frac{M}{2}+m\right) R^{2}\\\\\\L=\vec{\gamma} \times \vec{p}={mvr} \quad \& \quad L=I \omega\\\\Mrock \timesv \times R=\left(\frac{M}{2}+m\right) R^{2} \times \omega\\\\\omega = \frac{m_{rock} \cdot v}{\left(\frac{M}{2}+m\right) R}\\[/tex]
The required angular speed of disk is [tex]\dfrac{m_{rock} \times v}{(\dfrac{MR}{2}+mR)}[/tex].
Given data:
The mass of disk is, M.
The radius of disk is, R.
The mass of teenager is, m.
The mass of rock is, [tex]m_{rock}[/tex].
The speed of rock is, v.
Here we need to apply the concept of angular momentum to obtain the value of angular speed of disk. The expression for the angular momentum is given as,
[tex]L = I_{total} \times \omega[/tex] .......................................(1)
Here, [tex]\omega[/tex] is the angular speed and [tex]I_{total}[/tex] is the total moment of inertia of system.
And its value is,
[tex]I_{total} = I_{disk}+I_{solid}\\\\I_{total} = \dfrac{MR^{2}}{2}+mR^{2}[/tex] ..........................................(2)
And the angular momentum is also expressed as,
[tex]L = p \times R\\L =( m_{rock}v) \times R[/tex] ....................................................(3)
Then, using the equation (1), (2) and (3) we have,
[tex]( m_{rock} \times v) \times R = (\dfrac{MR^{2}}{2}+mR^{2}) \times \omega\\\\m_{rock} \times v = (\dfrac{MR}{2}+mR) \times \omega\\\\\\\omega = \dfrac{m_{rock} \times v}{(\dfrac{MR}{2}+mR)}[/tex]
Thus, we can conclude that the required angular speed of disk is [tex]\dfrac{m_{rock} \times v}{(\dfrac{MR}{2}+mR)}[/tex].
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A skier of mass 58.0 kg starts from rest at the top of a ski slope of height 70.0 m . Part APart complete If frictional forces do −1.04×104 J of work on her as she descends, how fast is she going at the bottom of the slope? Take free fall acceleration to be g = 9.80 m/s2 . 31.8 m/s Previous Answers Correct Part BPart complete Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is μk = 0.250. If the patch is of width 68.0 m and the average force of air resistance on the skier is 150 N , how fast is she going after crossing the patch? 18.1 m/s Previous Answers Answer Requested Part C After crossing the patch of soft snow, the skier hits a snowdrift and penetrates a distance 2.30 m into it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her?
Answer:
A) v = 31.8 m/s
B) v_f = 18.1 m/s
C) F_avg = 4130.7 N
Explanation:
A) From law of conservation of energy, we know that;
mgh = W_f + ½mv²
v is the speed at which she is going at the bottom of the slope.
Thus, making v the subject, we have;
v = √[2gh - (2W_f)/m)]
We are given;
h = 70 m
m = 58 kg
W_f = 1.04 × 10⁴ J = 10400 J
Thus;
v = √[(2 × 9.8 × 70) - (2 × 10400)/58)]
v = 31.8 m/s
B) Total force will be given by the formula;
F_t = F_k + F_r
Where;
F_k is force of kinetic friction = mg•μ_k
μ_k = 0.25
So, F_k = 58 × 9.8 × 0.25
F_k = 142.1 N
We are given force of air resistance(F_r) as 150 N
Thus;
F_t = 142.1 + 150
F_t = 292.1 N
Final velocity is gotten from the formula;
v_i² - v_f² = 2F_t•L/m
Thus;
v_f = √[v_i² - (2F_t•L/m)]
Where, v_i = 31.8 m/s, F_t = 292.1, m = 58 kg, L = 68 m
Thus;
v_f = √[31.8² - (2 × 292.1 × 68/58)]
v_f = 18.1 m/s
C) the average force exerted on her by the snowdrift as it stops her is given by the formula;
F_avg = m•v_f²/2l
F_avg = 58 × 18.1²/(2 × 2.3)
F_avg = 4130.7 N
How does the gravitational force that the Earth exerts on the Moon which has a much smaller mass compare to the gravitational force that the Moon exerts on the Earth?
a. The gravitational force the Moon exerts on the Earth is larger
b. The gravitational force the Earth exerts on the Moon is larger
C. Both the Earth and the Moon exerts equal gravitational force on each other.
I
A 57 kg person in a rollercoaster moving through the bottom of a curved track of radius 42.7 m feels a normal force of 995 N. How fast is the car moving?
Answer:
Use Fc centripetal force as positive and W the weight as negative
N = m v^2 / R + m g
v^2 = (N - m g) R / m
v^2 = (995 - 57 * 9.8) 42.7 / 57 = 327 m^2/s^2
v = 18.1 m/s
Note: N - m g is the net force producing the centripetal force
it's raining in 60 degree and man is walking in an inclined plain of base angle 30 degree find angle with vertical at which the man must hold his umbrella to be safe
A light spring obeys Hooke's law. The spring's unstretched length is 34.0 cm. One end of the spring is attached to the top of a doorframe and a weight with mass 7.00 kg is hung from the other end. The final length of the spring is 44.5 cm. (a) Find its spring constant (in N/m).
When the spring is extended by 44.5 cm - 34.0 cm = 10.5 cm = 0.105 m, it exerts a restoring force with magnitude R such that the net force on the mass is
∑ F = R - mg = 0
where mg = weight of the mass = (7.00 kg) g = 68.6 N.
It follows that R = 68.6 N, and by Hooke's law, the spring constant is k such that
k (0.105 m) = 68.6 N ⇒ k = (68.6 N) / (0.105 m) ≈ 653 N/m
“Relative” is an important word. Block L of mass mL = 1.90 kg and block R of mass mR = 0.460 kg are held in place with a compressed spring between them. When the blocks are released, the spring sends them sliding across a frictionless floor. (The spring has negligible mass and falls to the floor after the blocks leave it.) (a) If the spring gives block L a release speed of 1.10 m/s relative to the floor, how far does block R travel in the next 0.740 s? (b) If, instead, the spring gives block L a release speed of 1.10 m/s relative to the velocity that the spring gives block R, how far does block R travel in the next 0.740 s?
Answer:m
R
ν
R
+m
L
ν
L
=0 ⇒ (0.500kg)ν
R
+(1.00kg)(−1.20m/s)=0
which yields ν
R
=2.40m/s . Thus , Δx=ν
R
t=(2.40m/s)(0.800s)=1.92m .
(b) Now we have m
R
ν
R
+m
L
(ν
R
−1.20m/s)=0 which yields
ν
R
=
m
L
+m
R
(1.2m/s)m
L
=
1.00kg+0.500kg
(1.20m/s)(1.00kg)
=0.800m/s .
Consequently , Δx=ν
R
t=0.640m .
Explanation:
A scientist is building a 3-D model of a small and fast meat-eating dinosaur. How will the dinosaur look in the model?
A fast meat eating dinosaur must have strong claws for catching prey as well as strong teeth for eating meat.
What is a dinosaur?Dinosaurs are ancient reptiles that occurred in different forms and shapes and are currently considered a part of the evolutionary sequence.
A carnivorous dinosaur must have strong claws for catching prey as well as strong teeth for eating meat.
The image of a carnivorous dinosaur is shown in the image attached to this answer.
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After an airplane takes off, it travels 10.4 km west, 8.7 km north, and 2.1 km up. How far is it from the take off point?
The displacement of the airplane from the take off point after travelling the given distance is 15 km.
Displacement of the airplaneThe displacement of the airplane is the shortest distance between the initial position of the airplane and the final position of the airplane. The displacement of the of the airplane is calculated as follows;
total upward distance = 8.7 km + 2.1 km = 10.8 kmtotal horizontal distance = 10.4 kmR² = Rx² + Ry²
R² = 10.4² + 10.8²
R² = 224.8
R = √224.8
R = 15 km
Thus, the displacement of the airplane from the take off point is 15 km.
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Describe cytochrome c and how it can be used to provide evidence of evolutionary relationships. Then, explain how a scientist might determine which two species in a group of five species are more closely related using evidence from cytochrome c. You may use the evidence in the graph to support your answer.
Write 4-6 sentences:
Answer:
the c is in the cycle lab where they were created and then it cycles then it reoves it to the right.
Explanation:
A car with a mass of 950 kg is tied to a crane by a strong cable and being lowered at a rate of 1.2 m/s2. Draw a free-body diagram showing all the forces acting on the car. Include the values of each force. Show all calculations.
Answer:
Data that we know:
Mass of the car = 950kg.
Gravitational acceleration = -9.8m/s^2.
Acceleration of the car = -1.2 m/s^2.
(the negative signs are because that accelerations point downwards)
Now, the total acceleration of the car is:
-1.2m/s^2 = -9.8m/s^2 + Ac.
Where Ac is the acceleration caused by the cable, that is opposite to the acceleration of the car:
Ac = 9.8m/s^2 - 1.2m/s^2 = 8.6m/s^2.
Now we have the accelerations, remember that by second's newton law we have:
F = m*a
Then the force of the cable is:
Fc = 8.6m/s^2*950kg = 8,170N.
The gravitational force is:
Fg = -9.8m/s^2*950kg = -9,310N
The net force is:
Ft = 8,170N - 9,310N = -1,140N.
Below is the free body-diagram.
If a→=3i^-2j^-k^ and b→=i^+4j^+k^, find a unit vector n^ normal to the plane containing a→ and b→ such that a→, b→ and n^, in this order, form a right-handed system.
Since vector a = 3i - 2j - k and vector b = i + 4j + k, the unit vector, n, normal to the plane is (7i - j + 7k)/3√11 where a, b and n, in this order, form a right-handed system.
The normal vector to the planeThe normal vector to the plane is c = a × b
Since a = 3i - 2j - k and b = i + 4j + k,
c = a × b
c = (3i - 2j - k) × (i + 4j + k)
= 3i × i + (- 2j) × i + (-k) × i + 3i × 4j + (- 2j) × 4j + (-k) × 4j + 3i × k + (- 2j) × k + (-k) × k
= 3i × i - 2j × i - k × i + 12i × j - 8j × j - 4k × 4j + 3i × k - 2j × k - k × k
Given that
i × j = k, k × i = j, j × k = i, j × i = -k, i × k = -jk × j = -i, i × i = 0, j × j = 0 and k × k = 0. we havec = 3(0) - 2(-k) - (-j) + 12k - 8(0) - 16(-i) + 3(-j) - 2i - 0
c = 0 + 2k + j + 12k - 0 + 16i - 3j - 2i
c = 16i - 2i + j - 3j + 2k + 12k
c = 14i -2j + 14k
Unit vector normal to the plane nThe unit vector normal to the plane n = c/|c| where |c| is the magnitude of c = √(14² + (-2)² + 14²) = √(196 + 4 + 196)
= √396
= √(36 × 11)
= 6√11
So, n = c/|c|
= (14i -2j + 14k)/6√11
= 2(7i - j + 7k)/6√11
= (7i - j + 7k)/3√11
So, the unit vector normal to the plane is (7i - j + 7k)/3√11
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Help as soon as possible
PLS
Answer:
B-A-C
Explanation:
The idea is to evaluate the slopes of the three graphs. Notice that the three graphs are a representation of position (x) as a function of time (t).
Then, the slope of those lines are giving information of the change in position over the change in time (rise over the run). For the graph with positive slope the velocity is positive, for those with negative slope (line going down) the velocity is negative. But we are asked to compare speeds, which are the magnitude of each velocity (slope) so it is important to determine the graph with the largest slope (graph C), and that with the smallest (graph B)
Then the order is: B-A-C (third option answer option)
Which of the following experiments demonstrates Newton's Third Law? Select two answers
a rocket goes forward by burning fuel behind it
A student jumps 2 feet by using his muscles to exert a force on the ground
Matter cannot be created or destroyed
A person claps because they enjoy Physical Science
Which trait is found in early embryos of both birds and humans but then disappears from each develpment
A boat and a trailer are being pulled over an undulating road at a velocity v. The contour of the road is such that it can be approximated by a sine wave having a wavelength of l = 10 ft and an amplitude of Y = 0.5 in. The total static deflection of the springs and tires of the trailer due to the weight of the boat and trailer has been measured as 1.5 in. Assume that the damping inherent in the system is viscous and the damping ratio is 0.05. Determine: (a) The speed v at which the amplitude of the boat and the trailer will be a maximum; (b) The value of this maximum amplitude referred to in part (a); (c) The amplitude when the boat and trailer are traveling at the speed of 55 mph
Answer:
a) v = 25.54 ft/s
b) Xmax = 0.4180
c) Xmax = 0.0048 ft
Explanation:
a) determine the speed v at which the amplitude of the boat and the trailer will be a maximum;
to calculate the distance travelled
S = vt
given an expression of wavelength
l = vt
l = 2πv / ω
ω = 2πv / l
equation for counter of the road
y = Y sin ωt
= Y sin ( (2πv / l)t)
next we calculate the angular frequency
ω = √ ( k/m)
= √ ( g / Sst )
= √( 32.2 / ( 1.5/12))
= 16.05 rad/s
Now we calculate the speed v at which amplitude of the boat will be maximum
ω = 2πv / l
16.05 = (2 × π × v ) / 10
160.5 = 2πv
v = 160.5 / 2π
v = 25.54 ft/s
b)
we calculate the maximum amplitude using the following expression;
X/Y = [ √( 1 + ( 2Sr)²)] / [ √(( 1 - r²)² + ( 2Sr)²))]
Xmax / (0.5/12) = [ √( 1 + ( 2×0.05×1)²)] / [ √(( 1 - 1²)² + ( 2×0.05×r)²))]
Xmax / 0.0416 = 1.004987 / 0.1
Xmax 0.1 = 0.0418
Xmax = 0.0418 / 0.1
Xmax = 0.4180
c)
we convert speed of the boat from mph velocity m/s
v = 55 mph × 1.46667ft/s / 1mph
v = 80.66 ft/s
next we calculate angular velocity
ω = 2πv / l
= 2π × 80.66 / 10
= 50.68 rad/s
next is the frequency ratio
r = 50.68 / 16.05
= 3.16
finally we find the amplitude of the boat
X/Y = [ √( 1 + ( 2Sr)²)] / [ √(( 1 - r²)² + ( 2Sr)²))]
Xmax / (0.5/12) = [ √( 1 + ( 2×0.05×3.16)²)] / [ √(( 1 - 3.16²)² + ( 2×0.05×3.16)²))]
Xmax / 0.0416 = √1.0998 / √ ( 80.74 + 0.0998)
Xmax / 0.0416 = 1.0487 / 8.9910
Xmax 8.9910 = 0.0436
Xmax = 0.0436 / 8.9910
Xmax = 0.0048 ft
What is the standard unit of measure for pressure?
Answer:
Pascal
Explanation:
The standard SI unit for pressure measurement is the Pascal (Pa) which is equivalent to one Newton per square meter (N/m2) or the KiloPascal (kPa) where 1 kPa = 1000 Pa.