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damaskus
3 months ago
8

A Chevrolet Corvette convertible can brake to a stop from a speed of 60.0 mi/h in a distance of 123 ft on a level roadway. What

is its stopping distance on a roadway sloping downward at an angle of 26.0°?
Physics
1 answer:
kicyunya [3.2K]3 months ago
6 0

Response:

83.1946504051 m

Rationale:

u = Starting velocity = 60\ mph=\dfrac{60\times 1609.34}{3600}=26.82233\ m/s

s = Distance traveled = 123\ ft=\dfrac{123}{3.281}=37.4885705578\ m

\theta = Incline = 26^{\circ}

v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-26.82233^2}{2\times 37.4885705578}\\\Rightarrow a=-9.5954230306\ m/s^2

Friction coefficient

\mu=-\dfrac{a}{g}\\\Rightarrow \mu=\dfrac{9.5954230306}{9.81}\\\Rightarrow \mu=0.978126710561

mg sin\theta - u mg cos\theta = ma\\\Rightarrow a=9.81(sin26-0.978126710561cos26)\\\Rightarrow a=-4.32382\ m/s^2

v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-26.82233^2}{2\times -4.32382}\\\Rightarrow s=83.1946504051\ m

The calculated stopping distance is 83.1946504051 m

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Answer:

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This issue pertains to electrical circuits, specifically series RLC circuits, where the resistor, capacitor, and inductor are arranged in series.

In these types of circuits, impedance can be calculated as follows:

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2 months ago
A sled that has a mass of 8 kg is pulled at a 50 degree angle with a force of 20 N. The force of friction acting on the sled is
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3 months ago
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Answer:

Explanation:

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(2.5 \times 3.7 \times 10^{16} \times 10^6)MeV

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(2.5\times 3.7\times 10^{10}\times 60 \times 60)MeV

Gamma ray energy absorbed by meat = 30% (in 1 hour) of E is 0.3E

The dosage necessary to eliminate pathogens = 4000Gy=4000J/kg

The mass of meat that can be processed every hour is

\frac{0.3E/hr}{4000J/kg}\\\\E=(2.5\times 3.7\times 10^{10}\times 60 \times 60)MeV\\\\=333\times 10^{18}MeV\\\\eV=1.6\times 10^{-19}\\\\E=53.3J

The total meat produced =\frac{0.3\times 53.3}{4000}=3.996\times 10^{-3}

4 0
3 months ago
You are an investigator at the scene of a car crash. A Volvo of has hit an Alpha Romeo from the side and the two are inter-tangl
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Answer:

Let the Volvo's speed upon braking be v, and its mass be m1.

The speed of the Volvo just before the crash is represented as v1.

After applying the brakes, the skidding distance is noted as 30 m.

The equation governing this scenario is v1^2- v^2 = - 2 * a * s, where a= represents deceleration and s= denotes skidding distance.

Assuming the deceleration to be -0.3 g = - 2.94 m/s^2 (as provided by the Volvo website for the car's emergency system).

The equation can be modified to solve for v: v^2 = v1^2 + (2 * a * s )= v1^2 + 176.4.

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They then move 12.25 m and eventually come to a stop.

Using the equation (0)^2 -(V)^2 = - 2 * a' * s', where a' signifies the combined deceleration and s' denotes the distance traveled by the two cars.

Let's assume the combined system’s deceleration is notably higher at (- g).

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This results in V = 15.5 m/s.

Using momentum conservation in the X direction (East): m1 * v1 = (m1+ m2) * V * cos 15.

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Solving for v2 results in: v2 = (m1+m2) * V * sin 15/m2.

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2 months ago
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