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3 years ago

A 2500-kg car is being pushed up a hill at an angle of 35 degrees. Determine the gravitational

Physics

1 answer:

velikii [3]3 years ago

3 0

Answer:

The force that pulls the car down is Wₓ = 14052.6 N and the one that pushes the car up is F = 26552.6 N

Explanation:

For this exercise we will use Newton's second law, let's set a reference system where the x axis is parallel to the plane, in the adjoint we can see the forces in the system.

sin 35 = Wₓ / W

cos 35 = W_y / W

Wₓ = W sin 35

W_y = W cos 35

Wₓ = 2500 9.8 sin 35

Wₓ = 14052.6 N

let's write the equations for each axis

and

Y axis

N-W_y = 0

N = W_y

X axis

F -Wₓ = m a

F = Wₓ + m a = mg sin 35 + m a

F = m (a + g sin 35)

let's calculate

F = 2500 (5 + 9.8 sin 35)

F = 26552.6 N

The force that pulls the car down is Wₓ = 14052.6 N and the one that pushes the car up is F = 26552.6 N

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Which substance is uniquely a product of incomplete combustion of a fossil fuel?

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I can think of two of them:

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3 years ago

You illuminate the grating in a spectrometer at normal incidence θi=0° with a beam of light that has a wavelength of 6562.8 Å. T

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Answer:

a) θ₁ = 23.14 ° , b) θ₂ = 51.81 °

Explanation:

An address network is described by the expression

d sin θ = m λ

Where is the distance between lines, λ is the wavelength and m is the order of the spectrum

The distance between one lines, we can find used a rule of proportions

d = 1/600

d = 1.67 10⁻³ mm

d = 1-67 10⁻³ m

Let's calculate the angle

sin θ = m λ / d

θ = sin⁻¹ (m λ / d)

First order

θ₁ = sin⁻¹ (1 6.5628 10⁻⁷ / 1.67 10⁻⁶)

θ₁ = sin⁻¹ (3.93 10⁻¹)

θ₁ = 23.14 °

Second order

θ₂ = sin⁻¹ (2 6.5628 10⁻⁷ / 1.67 10⁻⁶)

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θ₂ = 51.81 °

3 0

3 years ago

g A 1.5-kg mass attached to spring with a force constant of 20.0 N/m oscillates on a horizontal, frictionless track. At t = 0, t

jok3333 [9.3K]

Answer:

(a) f = 0.58Hz

(b) vmax = 0.364m/s

(d) E = 0.1J

(e) x(t)=0.1m*cos(2π(0.58s^{-1})t)

Explanation:

(a) The frequency of the oscillation, in a spring-mass system, is calculated by using the following formula:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$ (1)

k: spring constant = 20.0N/m

m: mass = 1.5kg

you replace the values of m and k for getting f:

$f=\frac{1}{2\pi}\sqrt{\frac{20.0N/m}{1.5kg}}=0.58s^{-1}=0.58Hz$

The frequency of the oscillation is 0.58Hz

(b) The maximum speed is given by the following relation:

$v_{max}=\omega A=2\pi f A$ (2)

A: amplitude of the oscillations = 10.0cm = 0.10m

$v_{max}=2\pi (0.58s^{-1})(0.10m)=0.364\frac{m}{s}$

The maximum speed of the mass is 0.364 m/s.

The maximum speed occurs when the mass passes trough the equilibrium point of the oscillation.

(c) The maximum acceleration is given by the following formula:

$a_{max}=\omega^2A=(2\pi f)^2 A$

$a_{max}=(2\pi (0.58s^{-1}))(0.10m)=1.32\frac{m}{s^2}$

The maximum acceleration is 1.32 m/s^2

The maximum acceleration occurs where the elastic force is a maximum, that is, where the mass is at the maximum distance from the equilibrium point, that is, the acceleration.

(d) The total energy of the system is calculated with the maximum potential elastic energy:

$E=\frac{1}{2}kA^2=\frac{1}{2}(20.0N/m)(0.10m)^2=0.1J$

The total energy is 0.1J

(e) The displacement as a function of time is:

$x(t)=Acos(\omega t)=Acos(2\pi ft)\\\\x(t)=0.1m\ cos(2\pi(0.58s^{-1})t)$

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2 years ago

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Answer:

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3 years ago

A rubber-wheeled 50 kg cart rolls down a 35 degree concrete incline. Coefficient of rolling friction between rubber and concrete

lapo4ka [179]

Answer:

(a) 5.62 m/s²

(b) 5.46 m/s²

Explanation:

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mass,

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