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oee [108]
3 years ago
12

Determine the centripetal force on a vehicle rounding a circular curve with a radius of 80 m at a constant speed of 90 km/h if t

he vehicle's mass is 2,000 kg. (Hint: First express the velocity in meters per second)
Physics
1 answer:
quester [9]3 years ago
3 0
Mass of the vehicle = 2000 kg
Velocity of the vehicle = 90 km/hr
                                   = 25 m/s
Radius of the curve = 80 m
Then
Centripetal force = [2000 * (25)^2]/80
                            = (2000 * 625)/80
                            = 15625 kg m/s
                            = 15625 Newton second
I hope that this is the answer that you were looking for and the answer has come to your desired help. 
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Lesechka [4]

Answer:

The velocity of the hay bale is - 0.5 ft/s and the acceleration is 6.25\times 10^{- 3} ft/s^{2}

Solution:

As per the question:

Constant velocity of the horse in the horizontal, v_{x} = 1 ft/s

Distance of the horse on the horizontal axis, x = 10 ft

Vertical distance, y = 20 ft

Now,

Apply Pythagoras theorem to find the length:

20^{2} + 10^{2} = l^{2}

l^{2}= 500

Now,

x^{2} + y^{2} = 500                            (1)

Differentiating equation (1) w.r.t 't':

2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0

x\frac{dx}{dt} = - y\frac{dy}{dt}

where

\frac{dx}{dt} = Rate of change of displacement along the horizontal

\frac{dy}{dt} = Rate of change of displacement along the vertical

v_{x} = velocity along the x-axis.

v_{y} = velocity along the y-axis

xv_{x} = -yv_{y}

v_{y} = - 10\times \frac{1}{20} = - 0.5 ft/s

|v_{y}| = 0.5\ ft/s

Acceleration of the hay bale is given by the kinematic equation:

v_{y}^{2} = u_{y} + 2ay

(-0.5)^{2} =0 + 2ay

0.25 = 2ay

\frac{0.25}{2y} = a

a = \frac{0.25}{2\times 20} = 6.25\times 10^{- 3} ft/s^{2}

7 0
3 years ago
A frame hanging on a wall is held by two cables. The tension in each cable is 30 N, and the cables make an angle of 45° with the
Dimas [21]

Answer: option D) 42.4 N

The weight of the frame is balanced by the vertical component of tension.

W = T sin θ + T sin θ = 2 T sin θ

The tension in each cable is T = 30 N

Angle made by the cables with the horizontal, θ = 45°

⇒ W = 2×30 N × sin 45° = 2 × 30 N × 0.707 = 42.4 N

Hence, the weight of the frame is 42.4 N. Correct option is D.


6 0
3 years ago
A common carnival ride, called a gravitron, is a large r = 11 m cylinder in which people stand against the outer wall. the cylin
Elena-2011 [213]

Answer:

The minimum speed is 14.53 m/s.

Explanation:

Given that,

r = 11 m

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Suppose we need to find the minimum speed, that the cylinder must make a person move at to ensure they will stick to the wall

When frictional force becomes equal to or greater than the weight of person

Then, he sticks to the wall

We need to calculate the minimum speed

Using formula for speed

f_{s}=\mu N\geq mg

Where, N =\dfrac{mv^2}{r}

\mu\times\dfrac{mv^2}{r}\gep mg

v^2\geq\dfrac{gr}{\mu}

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v=\sqrt{\dfrac{9.8\times11}{0.51}}

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8 0
3 years ago
Two sound waves are traveling through the same medium. They have the same amplitude, wavelength, and direction of travel. If the
Sloan [31]

Answer:

Constructive interference

Explanation:

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Let the intensity of waves be I which is same for both

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I^{2}=I_{1}^{2}+I_{2}^{2}+2I_{1}I_{2}Cos\phi

where, Ф be the phase difference

So, I_{R}^{2}=I^{2}+I^{2}+2IICos6\pi

Here, IR is maximum so the interference is constructive in nature.

7 0
4 years ago
In the Hydrogen atom, the energy spacing between the is 4.07 x 101 J (Joules). When an is the frequency of the photons emitted?
agasfer [191]

Answer:

The frequency of the photon is 3.069\times10^{14}\ Hz.

Explanation:

Given that,

Energy E=4.07\times10^{-19}\ J

We need to calculate the energy

Using relation of energy

E_{4}-E_{2}=\Delta E

Where, \Delta E =  energy spacing

4h\nu-2h\nu=4.07\times10^{-19}

\nu=\dfrac{4.07\times10^{-19}}{2h}

Put the value of h into the formula

\nu=\dfrac{4.07\times10^{-19}}{2\times6.63\times10^{-34}}

\nu=3.069\times10^{14}\ Hz

Hence, The frequency of the photon is 3.069\times10^{14}\ Hz.

4 0
3 years ago
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