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OleMash [197]
3 years ago
10

the average velocity of a bus is 15 m/s. If it starts off going 15m north and then 20m south, how long will the bus ride take?

Physics
1 answer:
Readme [11.4K]3 years ago
4 0
It’s going to be 34 miles
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(a) As you ride on a Ferris wheel, your apparent weight is different at the top and at the bottom. Explain. (b) Calculate your a
otez555 [7]

Answer:

a. The component of the net force which make up the apparent weight are added to each other at the bottom and subtracted (the centripetal force from the weight) at the top)

b. Apparent weight at the top is approximately 519.06 N

Apparent weight at the bottom is approximately 558.94 N

Explanation:

a. The apparent weight at the top is different from the apparent weight at the bottom of a moving Ferris wheel because of the opposite direction in which the centripetal force acts at the top and the bottom, which are upwards and downwards respectively, while the weight acts downwards constantly

b. The given parameters are

The radius of the Ferris wheel, r = 7.2 m

The period for one complete revolution, t = 28 seconds

The angle covered in one revolution, θ = 2·π radian

The mass of the person riding on the Ferris wheel, the passenger  = 55 kg

Therefore, we have;

The angular speed, ω = Δθ/Δt = 2·π/(28)

From which we have;

Centripetal force, F_c = m × ω² × r

Substituting the known values, we have F_c = 55 kg × (2·π/(28 s))² × 7.2 m ≈ 19.94 N

The centripetal force, F_c = 19.94 N always acting outward from the center

Weight = Mass × Acceleration due to gravity

The weight of the passenger = 55 kg × 9.8 m/s² = 539 N

The weight of the passenger = 539 N always acting downwards

At the top of the Ferris wheel the the centripetal force is acting upwards and the weight is acting downwards

Therefore;

The net force, which is the apparent weight of the passenger at the top F_{NET_{Top}} = 539 N - 19.94 N ≈ 519.06 N

Apparent weight at the top ≈ 519.06 N

At the bottom of the Ferris wheel the weight is acting downwards and the centripetal force is also acting downwards

Therefore;

The net force at the bottom, which is the apparent weight of the passenger at the bottom F_{NET_{bottom}} = 539 N + 19.94 N ≈ 558.94 N

Apparent weight at the bottom ≈ 558.94 N.

5 0
3 years ago
A box of mass 10 kg is pulled from the hold of a ship with an acceleration of 1 m/s^2 by a vertical rope attached to it .find th
wlad13 [49]
F=ma
Mass times acceleration
We have g (10ms^_2) and a (1 given)
So total would be
10 kg times (10+1) =
110 N
5 0
2 years ago
FCFrdfAQefgEwdfgedfsaewfdedasfewreagargferdfares
Ivan

Answer: AAAAAAAAGGGGGHHHHJJJGSSSUUUUUUUUYCCFVGBHNJM

Explanation: YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET

6 0
2 years ago
Read 2 more answers
Answer the following. (a) What is the surface temperature of Betelgeuse, a red giant star in the constellation of Orion, which r
bagirrra123 [75]

Answer:

(a) T = 2987.6 k

(b) T = 19986.2 k

Explanation:

The temperature of a star in terms of peak wavelength can be given by Wein's Displacement Law, which is as follows:

T = \frac{0.2898\ x\ 10^{-2}\ m.k}{\lambda_{max}}

where,

T = Radiated surface temperature

\lambda_{max} = peak wavelength

(a)

here,

\lambda_{max} = 970 nm = 9.7 x 10⁻⁷ m

Therefore,

T = \frac{0.2898\ x\ 10^{-2}\ m.k}{9.7\ x\ 10^{-7}\ m}

<u>T = 2987.6 k</u>

(b)

here,

\lambda_{max} = 145 nm = 1.45 x 10⁻⁷ m

Therefore,

T = \frac{0.2898\ x\ 10^{-2}\ m.k}{1.45\ x\ 10^{-7}\ m}

<u>T = 19986.2 k</u>

6 0
2 years ago
A skateboarder, starting from rest, rolls down a 10.9-m ramp. When she arrives at the bottom of the ramp her speed is 6.74 m/s.
JulijaS [17]

Answer:b

Explanation:b

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