Which line represents a stationary object?
A. Line A
B. Line B
C. Line C
D. Line D
PLZ HELP

Answer:

D. At a depth equal to 1/3 radius of the earth

Answer:

Uniform Speed: A body is said to be moving with uniform speed, if it covers equal distances in equal intervals of time. Variable Speed: A body is said to be moving with variable speed if it covers unequal distances in equal intervals of time.

Explanation:

I got it from google

Answer:

Speed is a measure of how fast an object is moving at a given instant of time. It is a scalar quantity. Its unit is m/s. For a body covering equal distances in equal time intervals, the speed is called uniform speed.

Explanation:

Answer:........... .. .....

Answer:

1) The velocity of the ball return to the thrower's hand is -15 meters per second.

2) The resulting velocity of the boat is [tex]\vec v_{B} = 6\,\hat{i}+18\,\hat{j}\,\left[\frac{m}{s} \right][/tex].

Explanation:

1) Let suppose that ball experiments a free fall, that is an uniform accelerated motion, in which effects from gravity and Earth's rotation can be neglected. The velocity of the ball is represented by the following equations of motion:

Position

[tex]v_{o}\cdot t -\frac{1}{2}\cdot g\cdot t^{2} = 0[/tex]

[tex]t\cdot \left(v_{o}-\frac{1}{2}\cdot g\cdot t \right) = 0[/tex] (1)

Velocity

[tex]v = v_{o}-g\cdot t[/tex] (2)

Where:

[tex]t[/tex] - Time, measured in seconds.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]v_{o}[/tex] - Initial velocity of the ball, measured in meters per second.

[tex]v[/tex] - Final velocity of the ball, measured in meters per second.

From (1), we get the time when the ball returns to the thrower's hand:

[tex]v_{o}-\frac{1}{2}\cdot g\cdot t = 0[/tex]

[tex]t = \frac{2\cdot v_{o}}{g}[/tex]

And then we apply this result in (2):

[tex]v = v_{o}-g\cdot \left(\frac{2\cdot v_{o}}{g} \right)[/tex]

[tex]v = -v_{o}[/tex] (3)

Then, the velocity of the ball return to the thrower's hand is -15 meters per second.

2) The resulting velocity of the boat ([tex]\vec v_{B}[/tex]) is represented by the vectorial sum of the velocity of the boat relative to the river ([tex]\vec v_{B/R}[/tex]) and the velocity of the river ([tex]\vec v_{R}[/tex]), both measured in meters per second, that is:

[tex]\vec v_{B} = \vec v_{R}+\vec {v}_{B/R}[/tex] (4)

If we know that [tex]\vec v_{R} = 6\,\hat{i}\,\left[\frac{m}{s} \right][/tex] and [tex]\vec v_{B/R} = 18\,\hat{j}\,\left[\frac{m}{s} \right][/tex], then the resulting velocity of the boat is:

[tex]\vec v_{B} = 6\,\hat{i}+18\,\hat{j}\,\left[\frac{m}{s} \right][/tex]

The resulting velocity of the boat is [tex]\vec v_{B} = 6\,\hat{i}+18\,\hat{j}\,\left[\frac{m}{s} \right][/tex].

Answer:

The resultant displacement of the airplane in 4 hours is 212.8 km.

Explanation:

The components of the airplane's velocity and wind's velocity are:

Airplane:

[tex] v_{a_{x}} = v_{a}cos(45) = 70 km/h*cos(45) = 49.50 km/h [/tex]

[tex] v_{a_{y}} = v_{a}sin(45) = 70 km/hsin(45) = 49.50 km/h [/tex]

Wind:

[tex] v_{w_{x}} = 0 [/tex]

[tex] v_{w_{y}} = v_{w} = -30 km/h [/tex]

Now, to know the new velocity of the airplane we to find the result vector:

[tex] v_{x} = v_{a_{x}} + v_{w_{x}} = 49.50 km/h + 0 = 49.50 km/h [/tex]

[tex]v_{y} = v_{a_{y}} + v_{w_{y}} = 49.50 km/h - 30 km/h = 19.50 km/h[/tex]  

Now, the magnitude of the new speed of the airplane is:

[tex] v_{a} = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{(49.50 km/h)^{2} + (19.50 km/h)^{2}} = 53.20 km/h [/tex]

Finally, after 4 hours the resultant displacement of the airplane is:

[tex] x = v*t = 53.20 km/h*4 h = 212.8 km [/tex]

Therefore, the resultant displacement of the airplane in 4 hours is 212.8 km.

I hope it helps you!                                        

Answer:

4,900

Explanation:

you find the weight by multiplying the ball in grams to the gravity

The weight or force due to gravity of a 500 g ball is 4.9 N.

A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

Given in the question  a 500 g ball,

Weight = mass.gravity

Weight = .5*9.8

Weight = 4.9 N

The weight or force due to gravity of a 500 g ball is 4.9 N.

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Horizontal component = 75 · cosine(65°)  =  31.7 N  

Vertical component = 75 · sine(65°)  =  68.0 N

Answer:

60 km per hour

Explanation:

if you drive for 30 min and go 30km then if you drive for 60 min(1hr) then you would have driven 60km.

Answer:

60

Explanation:

hope this helps

Answer:

053.

Explanation:

Given that the radius of curvature of the path, R = 75 m.

Speed of the car on that path , [tex]v_1 = 35 m/s[/tex]

The centripetal force,[tex]F_c[/tex] acting on the body having mass, m, when it moves with the velocity v  on curved path having radiusR

Ris[tex]F_c = mv^2/R[/tex]

Gravitational force, [tex]F_g[/tex] = mg.

Let tha angle of superelevation is [tex]\theta.[/tex]

As the car does not skid even with zero friction, so

[tex]mg\sin\theta = (mv_1^2)/2 \cos\theta[/tex]

[tex]\Rightarrow \tan\theta = v_1^2/2g=\cdots(i)[/tex]

On sunny day, let the minimun static friction coefficient between the wheels and the pavement is [tex]\mu.[/tex]  

As [tex]v_2[/tex] = 118 m/s is greater than   v_so the car tends to skid in upper direction and the frictional

force,f, will acts is downward direction.

As there is no skidding, so

[tex]f+ mg\sin\theta= (m(v_2)^2)/R\cos\theta[/tex]

[tex]\Rightarrow f=(m(v_2)^2)/R\cos\theta - mg\sin\theta[/tex]

where [tex]f= \mu N.[/tex]

[tex]So, \mu = \frac {(m(v_2)^2)/R\cos\theta - mg\sin\theta}{N} \cdots(ii)[/tex]

Where N is the normal reaction can be determined by balancing the force in perpendicular direction of the plane.

[tex]N= mg\cos\theta+\frac {m(v_2)^2}{R}\sin\theta[/tex]

From equation (ii)

[tex]\mu = \frac {(m(v_2)^2)/R\cos\theta - mg\sin\theta}{mg\cos\theta+\frac {m(v_2)^2}{R}\sin\theta}[/tex]

[tex]=\frac{-g\tan\theta+v^2/R}{v_2^2\tan\theta+g}[/tex]

[tex]= \frac {(m(v_2)^2)/R\cos\theta - mg\sin\theta}{mg\cos\theta+\frac {m(v_2)^2}{R}\sin\theta}[/tex]

[tex]=\frac{-g(v_1^2/2g)+v^2/R}{v_2^2(v_1^2/2g)+g}[/tex]

[tex]= \frac{v_2^2/R-v_1^2/R}{g+v_2^2/R\times v_1^2/Rg}[/tex]

[tex]=\frac{118^2/75-35^2/75}{9.81+118^2/75\times 35^2/(75\times 9.81)}[/tex]

=0.53

Hence, the minimum coefficient of friction is 0.53.

Answer:

  a = 7.5 m / s²

Explanation:

For this exercise let's use Newton's second law, let's create a coordinate system with the x axis parallel to the plane and the y axis perpendicular to the plane

Y axis

       N - W cos θ = 0

       N = mg cos θ

X axis

       W sin θ = m a

       mg sin θ = m a

        a = g sin θ

let's calculate

        a = 9.8 cos 40

        a = 7.5 m / s²

Answer:

The one that is 100 cm from the center.

Explanation:

Given that merry-go-round means moving at uniform circular motion or constant angular speed. Hence the person who is closer to the center is moving at a slower speed, while the one far away from the center has a larger linear speed because he has moved a long distance in the same amount of time, despite being on these objects.

To illustrate

A with the 50 cm

B with the 100 cm

Hence, to get speed, we have

A = 2 * pi * 50 / T

B = 2 * pi * 100 / T

And by the explanation of angular speed

We have w = 2*pi/T where w is the angular speed in radians/s

Hence we have

A = 50w

B = 100w

Therefore, the one that is 100 cm from the center has a larger linear speed.

Answer:

trenchs and magma

Explanation:

Answer: Ridges form along cracks (divergent boundaries) in the oceanic crust (Molten rock (magma) rises through these cracks and pushes to both sides. When it cools, it forms new oceanic crust. The old crust is pushed away and new crust takes over. This is called Sea-Floor Spreading.

Answer: used concepts to simplify and order the world around us, categories: objects, events, ideas, or people

hierarchies: subdivide categories into smaller more detailed units, prototypes: best example

algorithm: time-consuming but thorough set of rules or procedures

heuristics: simpler way of thinking, solve problems but maybe incorrect solutions

insight: flash of inspiration that solves problem

representativeness heuristic: judge likelihood of things in terms of how they represent our prototype

availability heuristic: judge likelihood of things based on how vivid they are or how readily they come to mind.

Explanation: i hope that helped!

Answer:

Their function is to produce sound by allowing the free edges of the folds to vibrate against one another and also to act as the laryngeal sphincter when they are closed.

Answer:

Explanation:

Given

Height S = 8.4m

Initial velocity u = 24m/s

Required

Time it will take to reach the ground

Using the equation of motion

S = ut +1/2gt²

8.4 = 24t + 1/2(9.8)t²

8.4 = 24t+4.9t²

4.9t²+24t - 8.4 = 0

Multiply through by 10

49t²+240t-84 = 0

Factorize

t = -240±√240²-4(49(-84)/2(49)

t = -240±√57600+16464/98

t = -240±√74064/98

t = -240±272.15/98

t = 32.15/98

t = 0.328seconds

Hence it takes 0.328seconds before it lands on the ground below.

b) The horizontal distance is the range expressed as;

R = u²/g

R = 27²/9.8

R = 729/9.8.

R = 74.39m

a. The time it took before the motorcycle stunt driver lands on the ground below is 1.31 seconds.

b. The distance from the base of the cliff that the motorcycle stunt driver traveled while in  the air is 31.44 meters.

Given the following data:

We know that the acceleration due to gravity (g) of an object on planet Earth is equal to 9.8 [tex]m/s^2[/tex]

a. To determine how long (time) it took before the motorcycle stunt driver lands on the ground below:

At maximum height, time is given by the formula:

[tex]Time = \sqrt{\frac{2H}{g} }[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]Time = \sqrt{\frac{2\times 8.4}{9.8} }\\\\Time = \sqrt{\frac{16.8}{9.8} }\\\\Time =\sqrt{1.7413}[/tex]

Time = 1.31 seconds.

b. To determine how far (distance) away from the base of the cliff the motorcycle stunt driver traveled while in  the air:

[tex]Horizontal\;distance = horizontal\;speed \times time\\\\Horizontal\;distance = 24 \times 1.31[/tex]

Horizontal distance = 31.44 meters.

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Answer:

D. The United States

Explanation:

Answer:

graph b

Explanation:

graph a isnt constant

graph c is slower (less steep slope)

graph d is slowing down (starts high ends low)

Answer:

This question is incomplete, a possible question would be to calculate the work done by the ball.

Work done = 8J

Explanation:

Work done by an object is calculated my multiplying the force by the distance.

The weight of this ball is 2.0N, which also represents the force exerted by the ball. The ball travels through a distance of 4m

Hence, work done = F × d

Work done = 2N × 4m

Work done = 8Nm or 8J

Answer:

Electricity produced via friction is a development of electric charges on articles. Energizes construct when negative electrons are moved starting with one article then onto the next. The article that surrenders electrons turns out to be decidedly charged, and the item that acknowledges the electrons turns out to be contrarily charged. This can occur in a few different ways.

Explanation:

[tex]\mathrm {All, done!}[/tex]

Explanation:

N/B: Please see attached a rough sketch of the free body diagram for your reference.

From the diagram, the weight W is acting downward and the resultant R is in the positive direction

Given data

mass m=20kg

Forces

Jacob is pulling with 50 N,

Matthew with 40 N, and

Savannah is pulling with 60 N in a negative direction

Required

The net force

Net force= 50+40-(60)

               =90-60

               =30N in the possitive direction(in Jacob and Mathew's direction)

we know that F=ma

a=F/m

a= 30/20

a=1.5m/s^2

if Mathew decides to run for his life and stop pulling the net force will be

Net force=50-60=-10N (in Savannah's direction)

a=F/m

a=10/20

a=0.5m/s^2

Use the definition of average acceleration:

a = ∆v / t

If v is the starting speed, then ∆v = 0 - v, so solve for v :

-6.42 m/s² = (0 - v) / (2.85 s)

v = (6.42 m/s²) (2.85 s)

v ≈ 18.3 m/s

Explanation:

70N-40N=30N

Answer:

D

Explanation:

I took the test

The procedure that would best demonstrate that moving magnet causes an electric current to flow in a wire coil is connect a wire coil to an ammeter. Move a bar magnet into and out of the wire coil as you observe the ammeter. The correct option is D.

The flow of electricity in an electronic circuit, as well as the amount of electricity flowing through a circuit, are referred to as electric current. It is expressed in amperes (A).

The greater the value in amperes, the more electricity flows through the circuit.

When an electron is introduced into a wire, a free electron is attracted to a proton in order to be neutral.

Forcing electrons out of their orbits can result in an electron shortage. Electric current is the continuous movement of electrons in a wire.

Connecting a wire coil to an ammeter is the best way to demonstrate that a moving magnet causes an electric current to flow in a wire coil. As you watch the ammeter, move a bar magnet in and out of the wire coil.

Thus, the correct option is D.

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Answer:

0.5m/s2

..............

Answer:

Explanation:

Initial velocity , u = 27 m/s

displacement before stop,  s = 41.5 m .

final velocity,  v = 0

acceleration a = ?

v² = u² + 2 a s

0 = 27² + 2 x a x 41.5

a = - 27² / 2 x 41.5

= - 8.78 m /s