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3 years ago

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What is the displacement for the couple's hiking trip? Give your answer as an ordered pair, with magnitude first, followed by a

comma, then followed by the direction. Give the direction in terms of an angle measured CCW from the Easterly direction.

2 answers:

nekit [7.7K]3 years ago

7 0

Answer:

Explanation:

See the attached

Svetlanka [38]3 years ago

5 0

Answer:

displacement = ( 4.045 km , 72 degrees )

Explanation:

Given:

- The radius if the circular path r = 2.5 km

- Couple hikes 70 % of the circular track.

Find:

What is the displacement for the couple's hiking trip? Give the direction in terms of an angle measured CCW from the Easterly direction.

Solution:

- The start point of the journey is west and due south in a circular direction.

- First we will determine the 70 % of the circumference of the circular path traveled. Using the arc length S formula:

S = 0.70* ( 2*pi*r) = 7/5 pi*r

- The arc length calculated above is the distance traveled. However, we are interested in the "displacement" which is the shortest distance from start point to 70% of the track.

- The total amount of angle swept between the two points in radians is:

S = r*Q

Q = S / r = 7/5*pi

- The remaining angle angle to complete the full circle:

Q' = 2*p - Q = 2*p -7/5*pi

Q' = 0.6*pi.

- The displacement between two points can be calculated using sine rule as follows:

sin(Q') / displacement = sin (0.2*pi) / r

displacement = r*sin(Q') / sin (0.2*pi)

displacement = 2.5*sin(0.6*pi) / sin (0.2*pi)

displacement = 4.045 km

- The position with respect to the East direction is:

pi - Q' = pi - 0.6*pi = 0.4*pi

Convert into degrees: 0.4*pi * 180 /pi = 72 degrees

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A cart starts from rest and accelerates uniformly at 4.0 m/s2 for 5.0 s. It next maintains the velocity it has reached for 10 s.

wlad13 [49]

Answer:

12m/s

Explanation:

$v_f=v_o+at$

Let's call the velocity that the car maintains for 10 seconds $v_f_1$ , and the final velocity $v_f_2$ .

$v_f_1=0+(4)(5)=20m/s \\\\v_f_2=20+(-2)(4)=12m/s$

Hope this helps!

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3 years ago

What connects the upper motor neurons to lower motor neurons?

skad [1K]

Answer:

The upper motor neurons synapse in the spinal cord connect with anterior horn cells of lower motor neurons, usually via interneurons. The anterior horn cells are the cell bodies of the lower motor neurons and are located in the grey matter of the spinal cord.

Explanation:

Interneurons are the central nodes of neural circuits, enabling communication between the upper motor neurons, sensory or motor neurons located in the brain and spinal cord and they send signals to lower motor neurons or central nervous system (CNS) in the brain stem and spinal cord . When they get a signal from the upper motor neurons, they send another signal to your muscles to make them contract. They play vital roles in reflexes, neuronal oscillations, and neurogenesis in the adult mammalian brain.

Renshaw cells are among the very first identified interneurons. They are excited by the axon collaterals of the motor neurons. In addition, Renshaw cells make inhibitory connections to several groups of motor neurons.

7 0

3 years ago

Tsunamis are fast-moving waves often generated by underwater earthquakes. In the deep ocean their amplitude is barely noticable,

Andrei [34K]

Answer:

a) V = 195.70 m/s

b) f=3.02 × 10⁻⁴ Hz

c) T = 3311.25 seconds

Explanation:

Given:

Wavelength, λ = 646 Km = 646000 m

Distance traveled = 3410 Km = 3410000 m

Time = 4.84 h = 4.84 × 3600 s = 17424 seconds

a) The speed (V) of the wave is given as

V = distance / time

V = 3410000 m/ 17424 seconds

or

V = 195.70 m/s

b) The frequency (f) of the wave is given as:

f = V / λ

f= 195.70 / 646000

f=3.02 × 10⁻⁴ Hz

c) The time period (T) is given as:

T = 1/ f

T = 1/ (3.02 × 10⁻⁴) Hz

T = 3311.25 seconds

5 0

3 years ago

You and a friend work in buildings five equal-length blocks apart, and you plan to meet for lunch. Your friend strolls leisurely

Pavel [41]

Answer:

Your friend is 2.143 blocks from the restaurant.

You are 2.857 blocks from the restaurant.

Explanation:

Let t be the time both you and your friend take to walk to the restaurant.

The distance (m) from your building to the restaurant is your walking time t times your speed v1

$s_1 = v_1t = 1.6t$

Similarly the distance (m) from your friend building to the restaurant:

$s_2 = v_2t = 1.2t$

Let b be the length (in m) of a block, the total distance of 5 blocks is 5b

$s_1 + s_2 = 5b$

$1.6t + 1.2t = 5b$

$2.8t = 5b$

$t = 5b/2.8 = 25b/14$

$s_2 = 1.2t = 1.2(25b/14) = 2.143b$

So your friend are 2.143b meters from the restaurant, since each block is b meters long, 2.143b meters would equals to 2.143b/b = 2.143 blocks. And you are 5 - 2.143 = 2.857 blocks from the restaurant.

7 0

3 years ago

A loaded ore car has a mass of 950 kg. and rolls on rails ofnegligible friction. It starts from rest ans is pulled up a mineshaf

stiks02 [169]

(a) 10241 W

In this situation, the car is moving at constant speed: this means that its acceleration along the direction parallel to the slope is zero, and so the net force along this direction is also zero.

The equation of the forces along the parallel direction is:

$F - mg sin \theta = 0$

where

F is the force applied to pull the car

m = 950 kg is the mass of the car

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=30.0^{\circ}$ is the angle of the incline

Solving for F,

$F=mg sin \theta = (950)(9.8)(sin 30.0^{\circ})=4655 N$

Now we know that the car is moving at constant velocity of

v = 2.20 m/s

So we can find the power done by the motor during the constant speed phase as

$P=Fv = (4655)(2.20)=10241 W$

(b) 10624 W

The maximum power is provided during the phase of acceleration, because during this phase the force applied is maximum. The acceleration of the car can be found with the equation

$v=u+at$

where

v = 2.20 m/s is the final velocity

a is the acceleration

u = 0 is the initial velocity

t = 12.0 s is the time

Solving for a,

$a=\frac{v-u}{t}=\frac{2.20-0}{12.0}=0.183 m/s^2$

So now the equation of the forces along the direction parallel to the incline is

$F - mg sin \theta = ma$

And solving for F, we find the maximum force applied by the motor:

$F=ma+mgsin \theta =(950)(0.183)+(950)(9.8)(sin 30^{\circ})=4829 N$

The maximum power will be applied when the velocity is maximum, v = 2.20 m/s, and so it is:

$P=Fv=(4829)(2.20)=10624 W$

(c) $5.82\cdot 10^6 J$

Due to the law of conservation of energy, the total energy transferred out of the motor by work must be equal to the gravitational potential energy gained by the car.

The change in potential energy of the car is:

$\Delta U = mg \Delta h$

where

m = 950 kg is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

$\Delta h$ is the change in height, which is

$\Delta h = L sin 30^{\circ}$

where L = 1250 m is the total distance covered.

Substituting, we find the energy transferred:

$\Delta U = mg L sin \theta = (950)(9.8)(1250)(sin 30^{\circ})=5.82\cdot 10^6 J$

8 0

3 years ago