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3 years ago

Riders in a carnival ride stand with their backs against the wall of a circular room of diameter

Physics

1 answer:

Veseljchak [2.6K]3 years ago

6 0

Answer:

option C

Explanation:

given,

diameter of circular room = 8 m

rotational velocity of the rider = 45 rev/min

= $45 \times \dfrac{2\pi}{60}$

=4.712 rad/s

here in this case normal force is equal to centripetal force

N = m r ω²

N = m x 4 x 4.712²

N = 88.83m

frictional force = μ N

= 88.83m x μ

now, for the body to not to slide

gravity force is equal to frictional force

m g = 88.83 m x μ

g = 88.83 x μ

9.8 = 88.83 x μ

μ = 0.11

hence, the correct answer is option C

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Janet wants to find the spring constant of a given spring, so she hangs the spring vertically and attaches a 0.50 kg mass to the

frez [133]

Given: Mass m = 0.50 Kg; Force = Weight = mg F = (0.50 Kg)(9.8 m/s²)

F = 4.9 N

Displacement x = 3.0 cm convert to meter   x = 0.03 m

Required: Spring constant k = "

Formula: F = kx

k = F/x

k = 4.9 N/0.03 m

k = 163.33 N/m

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3 years ago

Suppose a car travels 108 km at a speed of 30.0 m/s, and uses 1.90 gallons of gasoline. Only 30% of the gasoline goes into usefu

Aleksandr [31]

Answer:

686.11 N

1.7733 gallons

Explanation:

$\eta$ = Efficiency = 30%

V = Volume of gasoline

E = Energy content of gasoline = $1.3\times 10^8\ J/gal$

F = Force

s = Displacement = 108000 m

v = Velocity

Work done is given by

$W=F\times s\\\Rightarrow F=\frac{W}{s}\\\Rightarrow F=\frac{VE\eta}{s}\\\Rightarrow F=\frac{1.9\times 0.3\times 1.3\times 10^8}{108000}\\\Rightarrow F=686.11\ N$

The force required to keep the car moving at a constant speed is 686.11 N

Here the force is directly proportional to speed

$\\\Rightarrow F=v$

$\\\Rightarrow \frac{F_1}{v_1}=\frac{F_2}{v_2}\\\Rightarrow F_2=\frac{F_1\times v_2}{v_1}\\\Rightarrow F_2=\frac{686.11\times 28}{30}\\\Rightarrow F_2=640.36\ N$

$W=F\times s\\\Rightarrow 0.3\times 1.3\times 10^8\times V=640.36\times 108000\\\Rightarrow V=\frac{640.36\times 108000}{0.3\times 1.3\times 10^8}\\\Rightarrow V=1.7733\ gal$

The gallons that will be used is 1.7733 gallons

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3 years ago

Four 50-g point masses are at the corners of a square with 20-cm sides. What is the moment of inertia of this system about an ax

deff fn [24]

Answer:

moment of inertia I ≈ 4.0 x 10⁻³ kg.m²

Explanation:

given

point masses = 50g = 0.050kg

note: m₁=m₂=m₃=m₄=50g = 0.050kg

distance, r, from masses to eachother = 20cm = 0.20m

the distance, d, of each mass point from the centre of the mass, using pythagoras theorem is given by

= (20√2)/ 2 = 10√2 cm =14.12 x 10⁻² m

moment of inertia is a proportion of the opposition of a body to angular acceleration about a given pivot that is equivalent to the entirety of the products of every component of mass in the body and the square of the component's distance from the center

mathematically,

I = ∑m×d²

remember, a square will have 4 equal points

I = ∑m×d² = 4(m×d²)

I = 4 × 0.050 × (14.12 x 10⁻² m)²

I = 0.20 × 1.96 × 10⁻²

I = 3.92 x 10⁻³ kg.m²

I ≈ 4.0 x 10⁻³ kg.m²

attached is the diagram of the equation

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2 years ago

An object has a mass of 12 kg. On Planet A, the object weighs 117.6 N. The force of gravity on Planet A is less thangreater than

AveGali [126]

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3 years ago

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Butoxors [25]

A. The instantaneous velocity at any time is 3t^2 -12t +8.
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3 years ago