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2 years ago

A 1150 kg car is on a 8.70° hill. using x-y axis tilted down the plane, what is the x-component of the weight?

1 answer:

6 0

I assume the x-y axis are tilted such that the x-axis is parallel to the surface of the hill while the y-axis is perpendicular to it.

In this case, the x-component of the weight is given by:
$W_x =mg \sin \theta$
where
m is the mass of the car
g is the acceleration of gravity
$\theta$ is the angle of the hill

Substituting numbers into the formula, we find
$W_x=(1150 kg)(9.81 m/s^2)(\sin 8.70^{\circ})=1706 N$

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The force shown in the attached figure is the net eastward force acting on a ball. The force starts rising at t = 0.012 s, falls

kifflom [539]

Impulse = Integral of F(t) dt from 0.012s to 0.062 s

Given that you do not know the function F(t) you have to make an approximation.

The integral is the area under the curve.

The problem suggest you to approximate the area to a triangle.

In this triangle the base is the time: 0.062 s - 0.012 s = 0.050 s

The height is the peak force: 35 N.

Then, the area is [1/2] (0.05s) (35N) = 0.875 N*s

Answer> 0.875 N*s

6 0

2 years ago

1. For each of the following scenarios, describe the force providing the centripetal force for the motion: a. a car making a tur

GenaCL600 [577]

Complete Question

For each of the following scenarios, describe the force providing the centripetal force for the motion:

a. a car making a turn

b. a child swinging around a pole

c. a person sitting on a bench facing the center of a carousel

d. a rock swinging on a string

e. the Earth orbiting the Sun.

Answer:

Considering a

The force providing the centripetal force is the frictional force on the tires \

i.e $\mu mg = \frac{mv^2}{r}$

where $\mu$ is the coefficient of static friction

Considering b

The force providing the centripetal force is the force experienced by the boys hand on the pole

Considering c

The force providing the centripetal force is the normal from the bench due to the boys weight

Considering d

The force providing the centripetal force is the tension on the string

Considering e

The force providing the centripetal force is the force of gravity between the earth and the sun

Explanation:

6 0

2 years ago

: Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at

ser-zykov [4K]

Complete Question

Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at sea level. One container is in Denver at an altitude of about 6,000 ft and the other is in New Orleans (at sea level). The surface area of the container lid is A=0.0155 m. The air pressure in Denver is PD = 79000 Pa. and in New Orleans is PNo = 100250 Pa. Assume the lid is weightless.

Part (a) Write an expression for the force FNo required to remove the container lid in New Orleans.

Part (b) Calculate the force FNo required to lift off the container lid in New Orleans, in newtons.

Part (c) Calculate the force Fp required to lift off the container lid in Denver, in newtons.

Part (d) is more force required to lift the lid in Denver (higher altitude, lower pressure) or New Orleans (lower altitude, higher pressure)?

Answer:

The expression is $F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

$F_{No}= 7771.125 \ N$

$F_p = 2.2*10^{6} N$

From the value obtained we can say the that the force required to open the lid is higher at Denver

Explanation:

The altitude of container in Denver is $d_D = 6000 \ ft = 6000 * 0.3048 = 1828.8m$

The surface area of the container lid is $A = 0.0155m^2$

The altitude of container in New Orleans is sea-level

The air pressure in Denver is $P_D = 79000 \ Pa$

The air pressure in new Orleans is $P_{ro} = 100250 \ Pa$

Generally force is mathematically represented as

$F_{No} = \Delta P A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

So the $\Delta P$ which is the difference in pressure within and outside the container is

$\Delta P = P_{No} - \frac{P_{sea}}{2}$

Therefore

$F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

Now substituting values

$F_{No} = 0.0155 [100250 - \frac{101000}{2}]$

$F_{No}= 7771.125 \ N$

The force to remove the lid in Denver is

$F_p = \Delta P_d A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

At sea level the air pressure in Denver is mathematically represented as

$P_D = \rho g h$

=> $g = \frac{P_D}{\rho h}$

Let height at sea level is h = 1

The air pressure at height $d_D$

$P_d__{D}} = \rho gd_D$

=> $g = \frac{P_d_D}{\rho d_D}$

Equating the both

$\frac{P_D}{\rho h} = \frac{P_d_D}{\rho d_D}$

$P_d_D = P_D * d_D$

Substituting value

$P_d__{D}} = 1828.2 * 79000$

$P_d__{D}} = 1.445*10^{8} Pa$

$\Delta P_d = P_{d} _D - \frac{P_{sea}}{2}$

=> $\Delta P_d = 1.445 *10^{8} - \frac{101000}{2}$

$\Delta P_d = 1.44*10^{8}Pa$

$F_p = \Delta P_d A$

$= 1.44*10^8 * 0.0155$

$F_p = 2.2*10^{6} N$

6 0

2 years ago

Una cuerda de violin vibra con una frecuencia fundamental de 435 Hz. Cual sera su frecuencia de vibracion si se le somete a una

EleoNora [17]

Answer:

a) f = 615.2 Hz b) f = 307.6 Hz

Explanation:

The speed in a wave on a string is

v = √ T / μ

also the speed a wave must meet the relationship

v = λ f

Let's use these expressions in our problem, for the initial conditions

v = √ T₀ /μ

√ (T₀/ μ) = λ₀ f₀

now it indicates that the tension is doubled

T = 2T₀

√ (T /μ) = λ f

√( 2To /μ) = λ f

√2 √ T₀ /μ = λ f

we substitute

√2 (λ₀ f₀) = λ f

if we suppose that in both cases the string is in the same fundamental harmonic, this means that the wavelength only depends on the length of the string, which does not change

λ₀ = λ

f = f₀ √2

f = 435 √ 2

f = 615.2 Hz

b) The tension is cut in half

T = T₀ / 2

√ (T₀ / 2muy) = f = λ f

√ (T₀ / μ) 1 /√2 = λ f

fo / √2 = f

f = 435 / √2

f = 307.6 Hz

Traslate

La velocidad en una onda en una cuerda es

v = √ T/μ

ademas la velocidad una onda debe cumplir la relación

v= λ f

Usemos estas expresión en nuestro problema, para las condiciones iniciales

v= √ To/μ

√ ( T₀/μ) = λ₀ f₀

ahora nos indica que la tensión se duplica

T = 2T₀

√ ( T/μ) = λf

√ ) 2T₀/μ = λ f

√ 2 √ T₀/μ = λ f

substituimos

√2 ( λ₀ f₀) = λ f

si suponemos que en los dos caso la cuerda este en el mismo armónico fundamental, esto es que la longitud de onda unicamente depende de la longitud de la cuerda, la cual no cambia

λ₀ = λ

f = f₀ √2

f = 435 √2

f = 615,2 Hz

b) La tension se reduce a la mitad

T = T₀/2

RA ( T₀/2μ) = λ f

Ra(T₀/μ) 1/ra 2 = λ f

fo /√ 2 = f

f = 435/√2

f = 307,6 Hz

5 0

2 years ago

A mass m slides down a frictionless ramp and approaches a frictionless loop with radius R. There is a section of the track with

Lana71 [14]

Answer:

h = 2 R (1 +μ)

Explanation:

This exercise must be solved in parts, first let us know how fast you must reach the curl to stay in the

let's use the mechanical energy conservation agreement

starting point. Lower, just at the curl

Em₀ = K = ½ m v₁²

final point. Highest point of the curl

$Em_{f}$ = U = m g y

Find the height y = 2R

Em₀ = Em_{f}

½ m v₁² = m g 2R

v₁ = √ 4 gR

Any speed greater than this the body remains in the loop.

In the second part we look for the speed that must have when arriving at the part with friction, we use Newton's second law

X axis

-fr = m a (1)

Y Axis

N - W = 0

N = mg

the friction force has the formula

fr = μ N

fr = μ m g

we substitute 1

- μ mg = m a

a = - μ g

having the acceleration, we can use the kinematic relations

v² = v₀² - 2 a x

v₀² = v² + 2 a x

the length of this zone is x = 2R

let's calculate

v₀ = √ (4 gR + 2 μ g 2R)

v₀ = √4gR( 1 + μ)

this is the speed so you must reach the area with fricticon

finally have the third part we use energy conservation

starting point. Highest on the ramp without rubbing

Em₀ = U = m g h

final point. Just before reaching the area with rubbing

$Em_{f}$ = K = ½ m v₀²

Em₀ = Em_{f}

mgh = ½ m 4gR(1 + μ)

h = ½ 4R (1+ μ)

h = 2 R (1 +μ)

7 0

2 years ago