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Debora [2.8K]
3 years ago
12

a car travels between the 100 meter and the 250 meter highway marker in 10 seconds calculate the velocity of the car during this

interval
Physics
1 answer:
kozerog [31]3 years ago
4 0

Given:\\\Delta s= 250m-100m=150m\\t=10s\\\\Find:\\v=?\\\\Roz.\\\\v=\frac{\Delta s}{t} \\\\v=\frac{150m}{10s} =15\frac{m}{s}

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A scale contains a spring with a spring constant of 291 N/m. Placing a mass on the scale causes the spring to be compressed by 2
WINSTONCH [101]
E = 1/2*k*x^2 = 0.5*291*0.0289^2 = 0.12 J
6 0
3 years ago
A light ray strikes a flat, smooth, reflecting surface at an angle of 80° to the normal. What angle does
charle [14.2K]

Answer:

80 angle of incidence=angle of reflection

Explanation:

7 0
3 years ago
Read 2 more answers
A new landowner has a triangular piece of flat land she wishes to fence. Starting at the first corner, she measures the first si
Lera25 [3.4K]

Answer:

Length of third side = 3.97m

Explanation:

First of all, let we draw the triangle from the given information assuming our first corner is A. second corner is B and third corner is C.

From A-B we have distance = 5.5m = Say it c

From B-C we have distance = 4.3m = Say it a

From A-C we have distance = ? = Say it b which we have to find out.

Using the Law of Cosines: The square of the unknown side equals to the sum of squares of other 2 sides and subtracting 2*(Product of other sides)*(Cos(Angle opposite to the unknown side)

For our case it is:

b² = a² + c² - 2acCos(B)         -  Say it equation 1

From the attached triangle you may see that, a & c are our known sides and B is the angle opposite to the side b.

There values are:

a = 4.3m;  c = 5.5m ; Angle B = 0.8 rad = 0.8 * 57.3 = 45.84 degrees where 1 rad = 57.3 degrees

Now by putting the respectve values in equation 1 we have:

b² = (4.3)² + (5.5)² - 2*4.3*5.5*Cos(45.84)

b² = 18.49 + 30.25 - 32.95

b² = 15.79

b  = √15.79

b = 3.97m

Thus the length of third side is 3.97m.

PS: The picture of triangle is being attached for yours understanding.

6 0
3 years ago
In one experiment the electric field is measured for points at distances r from a uniform line of charge that has charge per uni
Yuki888 [10]

Answer:

(a) A. Uniform line of charge and B. Uniformly charged sphere

(b) To three digits of precision:

λ = 1.50 * 10^-10 C/m

p = 2.81 * 10^-4 C/m^3

Explanation:

8 0
4 years ago
Two cars are moving towards each other and sound emitted by first car with real frequency of 3000 hertz is detected by a person
sertanlavr [38]

Answer:

 v ’= 21.44 m / s

Explanation:

This is a doppler effect exercise that changes the frequency of the sound due to the relative movement of the source and the observer, the expression that describes the phenomenon for body approaching s

           f ’= f (v + v₀) / (v-v_{s})

where it goes is the speed of sound 343 m / s, v_{s} the speed of the source v or the speed of the observer

in this exercise both the source and the observer are moving, we will assume that both have the same speed,

                v₀ = v_{s} = v ’

we substitute

               f ’= f (v + v’) / (v - v ’)

               f ’/ f (v-v’) = v + v ’

               v (f ’/ f -1) = v’ (1 + f ’/ f)

               v ’= (f’ / f-1) / (1 + f ’/ f) v

               v ’= (f’-f) / (f + f’) v

let's calculate

                v ’= (3400 -3000) / (3000 +3400) 343

                v ’= 400/6400 343

                v ’= 21.44 m / s

3 0
3 years ago
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