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olga2289 [7]
3 years ago
10

Susan, driving north at 53 mphmph , and Shawn, driving east at 63 mphmph , are approaching an intersection. Part A What is Shawn

's speed relative to Susan's reference frame?
Physics
1 answer:
mafiozo [28]3 years ago
5 0

Answer:

Shawn's speed relative to Susan's speed = 10 mph

Resultant velocity = 82.32 mph

Explanation:

The given data :-

i) Susan driving in north and speed of Susan is ( v₁ ) = 53 mph.

ii) Shawn driving in east and speed of Shawn is ( v₂ ) = 63 mph.

iii) The speed of both Susan and Shawn is relative to earth.

iv) The angle between Susan in north and Shawn in east is 90°.

We have to find Shawn's speed relative to Susan's speed.

v₂₁ = v₂ - v₁   = 63 - 53 = 10 mph

Resultant velocity,

v = \sqrt{v_{2} ^{2}+ v_{1} ^{2}  }  =\sqrt{63^{2} +53^{2} }

v = 82.32 mph

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tatiyna
Most likely, the light wave will be absorbed by the wall. Without any information as to the size and color of the wall, the location and size of the hole, or the location of the light wave, this is a generalized probability problem. For all of the places the light could be, it's more likely that it hits the wall than the hole (if the hole is less than 50% of the area of the wall).
5 0
3 years ago
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Martin has severe myopia, with a far point on only 17 cm. He wants to get glasses that he'll wear while using his computer whose
marusya05 [52]

Answer:

Explanation:

Far point = 17 cm . That means he can not see beyond this distance .

He wants to see at an object at 65 cm away . That means object placed at 65 has image at 17 cm by concave lens . Using lens formula

1 / v - 1 / u = 1 / f

1 / - 17 - 1 / - 65 = 1 / f

= 1 / 65 - 1 / 17

= -  .0434 = 1 / f

power = - 100 / f

= - 100 x .0434

= - 4.34 D .

6 0
3 years ago
The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.32 with the floor. If t
coldgirl [10]

Answer:

The shortest braking distance is 35.8 m

Explanation:

To solve this problem we must use Newton's second law applied to the boxes, on the vertical axis we have the norm up and the weight vertically down

On the horizontal axis we fear the force of friction (fr) that opposes the movement and acceleration of the train, write the equation for each axis

    Y axis

     N- W = 0

     N = W = mg

  X axis

     -Fr = m a

     -μ N = m a

     -μ mg = ma

     a = μ g

     a  = - 0.32 9.8

     a =  - 3.14 m/s²

We calculate the distance using the kinematics equations

    Vf² = Vo² + 2 a x

     x = (Vf² - Vo²) / 2 a

When the train stops the speed is zero (Vf = 0)

 Vo = 54 km/h (1000m/1km) (1 h/3600s)= 15 m/s

     x = ( 0 - 15²) / 2 (-3.14)

     x=  35.8 m

The shortest braking distance is  35.8 m

7 0
3 years ago
A flat coil having 160 turns, each with an area of 0.20 m 2, is placed with the plane of its area perpendicular to a magnetic fi
S_A_V [24]

Answer:

10.2 Watt

Explanation:

N  = number of turns in flat coil = 160

A  = area = 0.20 m²

B₀= initial magnetic field = 0.40 T

B  = final magnetic field = - 0.40 T

Change in magnetic field is given as

ΔB = B - B₀ = - 0.40 - 0.40 = - 0.80 T

t  = time taken for the magnetic field to change = 2.0 s

Induced emf is given as

E = \frac{- N A \Delta B}{t}

E = \frac{- (160) (0.20) (- 0.80)}{2}

E = 12.8 volts

R = Resistance of the coil = 16 Ω

Power is given as

P = \frac{E^{2}}{R}

P = \frac{(12.8)^{2}}{16}

P = 10.2 Watt

6 0
3 years ago
4:36 PM<br> How high can a human throw a ball if he can throw it with initial velocity 90 kph
choli [55]

Answer:

90 km/hr= 90000/3600=25 m/s initial velocity

-25/ 2 (-4.9)= 2.551 secs

h(2.551)= -4.9 (2.551)² + 25(2.551)=-31.8872449+63.775 = 31.8m

The distance would be 31.8 meters, also that person will have sore arm.

3 0
3 years ago
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