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OleMash [197]
3 years ago
14

A car is traveling around a horizontal circular track with radius r = 210 m at a constant speed v = 23 m/s as shown. The angle θ

A = 23° above the x axis, and the angle θB = 53° below the x axis.
1) What is the magnitude of the car’s acceleration?
2) What is the x component of the car’s acceleration when it is at point A
3) What is the y component of the car’s acceleration when it is at point A
4) What is the x component of the car’s acceleration when it is at point B
5) What is the y component of the car’s acceleration when it is at point B
Physics
1 answer:
Fantom [35]3 years ago
6 0

Centripetal acceleration of car is given by formula

a_c = \frac{v^2}{R}

now plug in the values in this

a_c = \frac{23^2}{210}

a_c = 2.52 m/s^2

Part b)

At position A we have

x component of acceleration is given as

a_x = -a_c cos23

a_x = -2.32 m/s^2

Part c)

At position A we have

y component of acceleration is given as

a_y = -a_c sin23

a_y = -0.98 m/s^2

Part d)

At position B we have

x component of acceleration is given as

a_x = -a_c cos53

a_x = -1.52 m/s^2

Part e)

At position B we have

Y component of acceleration is given as

a_y = a_c sin53

a_y = 2.01 m/s^2

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MatroZZZ [7]

Answer:

See the answers below.

Explanation:

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i)

The center of gravity of the rod is concentrated in half the distance, that is, from the end of the bar to the center there is 40 [cm]. This can be seen in the attached free body diagram.

We have only two equilibrium equations, a summation of forces on the Y-axis equal to zero, and a summation of moments on any point equal to zero.

For the summation of forces we will take the forces upwards as positive and the negative forces downwards.

ΣF = 0

-15+T-W=0\\T-W=15

Now we perform a sum of moments equal to zero around the point of attachment of the string with the metal bar. Let's take as a positive the moment of the force that rotates the metal bar counterclockwise.

ii) In the free body diagram we can see that the force acts at 18 [cm] of the string.

ΣM = 0

(15*9) - (18*W) = 0\\135 = 18*W\\W = 7.5 [N]

7 0
2 years ago
Consider a monochromatic electromagnetic plane wave propagating in the x direction. At a particular point in space, the magnitud
Scrat [10]

Answer:

a)   S = 2.35 10³   J/m²2 ,  

b)and the tape recorder must be in the positive Z-axis direction.

the answer is 5

c) the direction of the positive x axis

Explanation:

a) The Poynting vector or intensity of an electromagnetic wave is

          S = 1 /μ₀ E x B

if we use that the fields are in phase

          B = E / c

we substitute

         S = E² /μ₀ c

let's calculate

        s = 941 2 / (4π 10⁻⁷  3 10⁸)

        S = 2.35 10³   J/m²2

 

b) the two fields are perpendicular to each other and in the direction of propagation of the radiation

In this case, the electro field is in the y direction and the wave propagates in the ax direction, so the magnetic cap must be in the y-axis direction, and the tape recorder must be in the positive Z-axis direction.

the answer is 5

C) The poynting electrode has the direction of the electric field, by which or which should be in the direction of the positive x axis

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Reil [10]

Answer: The sound will change due to changes in frequency and the wavelength of the airplane.

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The sound that the observer hears will change base on the illustration above.

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2. A solution that contains a small amount of salt and a large amount of water is said to be a _______ solution.
koban [17]
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Read 2 more answers
Batman (mass = 96.1 kg) jumps straight down from a bridge into a boat (mass = 458 kg) in which a criminal is fleeing. The veloci
MariettaO [177]

Answer:

The velocity of the boat after the batman lands in it is +9.26 m/s

Explanation:

Applying the law of conservation of momentum,

Total momentum before collision = Total momentum after collision.

Note: The collision between the Batman and the boat is an inelastic collision.

m'u'+mu = V(m+m').................... Equation 1

Where m' = mass of the Batman, u' = initial velcoity of the batman, m = mass of the boat, u = initial velocity of the boat, V = common velocity.

make V the subject of equation 1

V = (m'u'+mu)/(m+m')............... Equation 2

Given: m' = 96.1 kg, u' = 0 m/s, m = 458 kg, u = +11.3 m/s.

Substitute these values into equation 2

V = [(96.1×0)+(458×11.2)]/(96.1+458)

V = 5129.6/554.1

V = +9.26 m/s

8 0
3 years ago
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