A plane maintains constant acceleration and its velocity increases
from 40m/s to 100m/s over a 10 second interval. What is the plane's
average Distance ?

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

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:

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.

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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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

The potential difference between points a and b is zero.

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

The 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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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

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)

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

The displacement of the airplane from the take off point after travelling the given distance is 15 km.

The 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;

R² = 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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A fast meat eating dinosaur  must have strong claws for catching prey as well as strong teeth for eating meat.

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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Answer:

3

5

4

Explanation:

x = (8=(9+9)

(9+9) = 4, 5, 3

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

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.

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.

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

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

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

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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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 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

c = 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

The 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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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.

Answer:

its D i just did it on the app

Answer:

the c is in the cycle lab where they were created and then it cycles then it reoves it to the right.

Explanation:

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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Newtons first law of motion.

Answer:17m/s

Explanation:

In this question we use first law of motion which is

vf=vi+at

By putting values

vf=32+10×-1.5

vf=17m/s

Answer:

yeah its 17m/s

Explanation: