An airplane is traveling 835 km/h in a direction 41.5 ∘ west of north. Find the components of the velocity vector in the northerly and westerly directions. How far north and how far west has the plane traveled after 2.20 h ?

Answers

Answer 1
Answer: I assume the graph is looking like in the picture bellow.

North component:
cos(41.5) * 835 = 625.37 km/h

West component of speed:
sin(41.5) * 835 = 553.29 km/h

After 2.2 hours plane will fly:
2.2*625.37 = 1375.81 km north
2.2*553.29 = 1217.23 km  west
Answer 2
Answer:

Final answer:

To find the components of the velocity vector, you can use trigonometry. The north component is calculated using the sine function and the west component is calculated using the cosine function. After 2.20 hours, the distance traveled north and west can be found by multiplying the velocity components by the time.

Explanation:

To find the components of the velocity vector in the northerly and westerly directions, we can use trigonometry. The velocity vector is 835 km/h and is traveling in a direction 41.5° west of north. To find the north component, we can use the sine function: North component = velocity * sin(angle). To find the west component, we can use the cosine function: West component = velocity * cos(angle).

After 2.20 hours, we can find the distance traveled north and west by multiplying the velocity components by the time: Distance north = North component * time and Distance west = West component * time.

Let's calculate the values:

  1. North component = 835 km/h * sin(41.5°)
  2. West component = 835 km/h * cos(41.5°)
  3. Distance north = North component * 2.20 h
  4. Distance west = West component * 2.20 h

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You are traveling on an interstate highway at the posted speed limit of 70 mph when you see that the traffic in front of you has stopped due to an accident up ahead. You step on your brakes to slow down as quickly as possible. Assume that you to slow down to 30 mph in about 5 seconds. A) With this same average acceleration, how much longer would it take you to stop?B) What total distance would you travel from when you first apply the brakes until the car stops?

Answers

A.The time taken for the car to stop is 8.75 s

B.The distance travelled when the brakes were applied till the car stops is 136.89 m

A. Determination of the time taken for the car to stop.

  • We'llbegin bycalculatingthedecelerationof thecar

Initial velocity (u) = 70 mph = 0.447 × 70 = 31.29 m/s

Final velocity (v) = 30 mph = 0.447 × 30 = 13.41 m/s

Time (t) = 5 s

Deceleration (a) =?

a \:  =  (v \:  - u)/(t)  \n  \n a =  (13.41 - 31.29)/(5)  \n  \n a \:  =  ( - 17.88)/(5)  \n  \n

a = –3.576 m/s²

  • Finally,we shall determine the time taken for the car to stop.

Initial velocity (u) = 31.29 m/s

Final velocity (v) = 0 m/s

Deceleration (a) = –3.576 m/s²

Time (t) =?

v \:  = u \:  + at \n 0 \:  = 31.29 \:  +  \: ( - 3.576 * t) \n 0 \:  = 31.29 \:  - 3.576 * t \n collet \: like \: terms \n 0 - 31.29 \:  = - 3.576 * t  \n - 31.29 \:  = - 3.576 * t  \n divide \: both \: side \: by \:  - 3.576 \n t \:  =  (- 31.29)/(- 3.576)  \n

t = 8.75 s

Thus, the time taken for the car to stop is 8.75 s

B.Determination of the total distance travelled when the brakes were applied.

Initial velocity (u) = 31.29 m/s

Final velocity (v) = 0 m/s

Deceleration (a) = –3.576 m/s²

Distance (s) =?

{v}^(2)  =  {u}^(2)  + 2as \n {0}^(2)  =  {31.29}^(2)  + (2 *  - 3.576 * s) \n 0  = 979.0641   - 7.152 s \n collect \: like \: terms \n 0  -  979.0641  =  - 7.152 s \n -  979.0641  =  - 7.152 s \n divide \: both \: side \: by \: - 7.152 \n s  =  (-  979.0641)/(- 7.152)  \n  \n

s = 136.89 m

Therefore, the total distance travelled by the car when the brakes were applied is 136.89 m

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Answer:8.75 s,

136.89 m

Explanation:

Given

Initial velocity=70 mph\approx 31.29 m/s

velocity after 5 s is 30 mph\approx 13.41 m/s

Therefore acceleration during these 5 s

a=(v-u)/(t)

a=(13.41-31.29)/(5)=-3.576 m/s^2

therefore time required to stop

v=u+at

here v=final velocity =0 m/s

initial velocity =31.29 m/s

0=31.29-3.576* t

t=(31.29)/(3.576)=8.75 s

(b)total distance traveled before stoppage

v^2-u^2=2as

0^2-31.29^2=2* (-3.576)\cdot s

s=136.89 m

Two large parallel metal plates are 1.6 cm apart and have charges of equal magnitude but opposite signs on their facing surfaces. Take the potential of the negative plate to be zero. If the potential halfway between the plates is then +3.8 V, what is the electric field in the region between the plates?

Answers

Answer:

475 N/C

Explanation:

As we know that, the electric field in parallel plate capacitor is same (constant) throughout. And is potential gradient.

So, Electric field is given by

Electric field = potential gradient

Electric FIeld = (Change\: in\: Potential)/(Distance)

Here, the potential change is 3.8V and the distance from negative plate to positive plate is 1.6 cm. The potential from negative plate to the center is (1.6/2)cm i.e., 0.8 cm.

But we have to take distance in SI units So, distance=0.8 * 10^(-2) m

So, Electric field is

Electric\: field=(3.8V)/(0.8 * 10^(-2)m )

Electric\: field=475 V/m

So, electric field is 475 Volts per meter.

Note : Also we can say 475 Newtons per coulomb

It's nighttime, and you've dropped your goggles into a 3.2-m-deep swimming pool. If you hold a laser pointer 0.90m above the edge of the pool, you can illuminate the goggles if the laser beam enters the water 2.2m from the edge.-How far are the goggles from the edge of the pool?

Answers

Answer:

the googles are 5.3 m from the edge

Explanation:

Given that

depth of pool , d = 3.2 m

Now, let i be the angle of incidence

a laser pointer 0.90 m above the edge of the pool and  laser beam enters the water 2.2 m from the edge

⇒tan i = 2.2/0.9

i=arctan(2.2/.90)

solving we get

i = 67.8°

Using snell's law ,

n1 ×sin(i) = n2 ×sin(r)

n1= refractive index of 1st medium= 1

n2=  refractive index of 2nd medium = 1.33

r= angle of reflection

therefore,

1* sin(67.8) = 1.33* sin(r)

r = 44.1°

Now,

distance of googles = 2.2 + d×tan(r)

distance of googles = 2.2 + 3.2×tan(44.1)

distance of googles = 5.3 m

the googles are 5.3 m from the edge

A 0.157kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.850m/s . It has a head-on collision with a 0.306kg glider that is moving to the left with a speed of 2.26m/s . Suppose the collision is elastic.Find the magnitude of the final velocity of the 0.157kg glider.Find the magnitude of the final velocity of the 0.306kg glider.

Answers

Answer:

V1 = -3.260 m/s,  V2 = 1.303 m/s

Explanation:

Let mass of the left glider m1 = 0.157 kg and velocity v1 = 0.850 m/s

mass of the right glider m2 = 0.306 Kg and v2 = -2.26 m/s (-ve sign mean it is opposite to direction of left glider)

To Find:  Final Velocity of Left Glider is V1=? m/s  and  Velocity of right Glider is V2 =? m/s  (After Collision)

from law of conservation of momentum and energy we deduce a formula:

V1 = (m1-m2) v1 /(m1+m2) + 2 m2 v2/(m1+m2)

V1 = (0.157 kg - 0.306 Kg) × 0.850 m/s / (0.157 kg + 0.306 Kg)  + 2 ×0.306 kg × -2.26 m/s / (0.157 kg + 0.306 Kg)

V1 = -0.273 -2.987

V1 = -3.260 m/s

and V2 Formula

V2 = (m2-m1) v2/(m1+m2) + 2 m1 v1/(m1+m2)

V2 = (0.157 kg - 0.306 Kg) × -2.26 m/s / (0.157 kg + 0.306 Kg)  + 2 ×0.157 kg × 0.850 m/s / (0.157 kg + 0.306 Kg)

V2 = 0.727 + 0.576

V2 = 1.303 m/s

-0.149,  0.463

An implanted pacemaker supplies the heart with 72 pulses per minute, each pulse providing 6.0 V for 0.65 ms. The resistance of the heart muscle between the pacemaker’s electrodes is 550 Ω. Find (a) the current that flows during a pulse, (b) the energy delivered in one pulse, and (c) the average power supplied by the pacemaker.

Answers

Answer:

a) Current = 11 mA

b) Energy = 66 mJ

c) Power = 101.54 W

Explanation:

a) Voltage, V = IR

   Voltage, V = 6 V, Resistance, R = 550 Ω

   Current, I =(6)/(550)=0.011A=11mA

b) Energy = Current x Voltage = 6 x 0.011 = 0.066 J = 66 mJ

c) \texttt{Power=}(Energy)/(Time)=(0.066)/(0.65* 10^(-3))=101.54W    

In a collision between two objects having unequal masses, how does magnitude of the impulse imparted to the lighter object by the heavier one compare with the magnitude of the impulse imparted to the heavier object by the lighter one?a. the lighter object receives a larger impulse

b. the heavier object receives a larger impulse

c. both objects receive the same impulse

d. the answer depends on the ratio of the masses

e. the answer depends on the ratio of the speeds

Answers

Answer:

Explanation:

The impulse has as an expression

      I = ∫ F dt

In addition, on impulse of isolated systems is

       I =ΔP = m vf - m v₀

Let's replace

      F dt = m (vf -vo)

The force in these shocks laqueunxcyerpoesjerce on the other, using the law of action and reaction state forces has the same magnitude and the time of the shock is equal for the two bodies, this implies that the impulse is equal for the two bodies

Let's check the answers

a) False as the forces are of action and reaction the impulse must be equal

b) False

c) True. Why do we have an action and reaction stop

d False

e) false

Final answer:

In collision between two objects, despite difference in their masses, both objects receive the same impulse, as per Newton's third law.

Explanation:

In a collision between two objects with unequal masses, according to Newton's third law of motion, 'For every action, there is an equal and opposite reaction', the magnitude of the impulse imparted to the lighter object by the heavy object is exactly the same as the magnitude of the impulse imparted to the heavier object by the lighter one.

This is regardless of the masses or speeds of the objects involved. The direction of these impulses will be opposite, but their magnitudes will be the equal. So, option c. 'both objects receive the same impulse' is correct.

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