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givi [52]
3 years ago
7

A motorist is traveling at 20 m/s. He is 60 m from a stoplight when he sees it turn yellow. Is reaction time, before stepping on

the brake, is 0.50 s. What steady acceleration (slowing down) while braking will bring him to a stop right at the light?
Physics
1 answer:
Gnesinka [82]3 years ago
5 0
V₀ = V₁ + 2ax
V₀ = final velocity which is 0 m/s
V₁ = initial velocity which is 20 m/s
x = distance which is 60-(0.5 x 20) =50m
     this is because his reaction time is 0.5 sec so he traveled 10m before stepping on the break paddle.

a= (V₀-V₁) / 2x
  = (0-20) / 2*50
  = -0.2m/s
the negative is because it is a deceleration speed hence it is 0.2m/s
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When scientists want to determine the actual age of a fossil, they use_____________
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Answer:

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4 0
2 years ago
The chemical symbol for sulfuric acid is H2SO4. How many atoms are contained in each molecule of sulfuric acid?
cricket20 [7]
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5 0
3 years ago
You fill two balloons with gas, one with hydrogen and one with carbon dioxide. You hold a match to each balloon. The hydrogen ba
miv72 [106K]

Answer:E. Hydrogen was able to participate in an exergonic reaction and carbon dioxide couldn't

Explanation:

An exergonic reaction releases energy to the environment. The combustion of hydrogen contained in the balloon is a chemical reaction. The reaction can take place because hydrogen combines with oxygen in air, that is, the gas is combustible. CO2 does not support combustion, it does not combine with oxygen in air and it is also denser than air, hence does not participate in the exergonic reaction.

4 0
3 years ago
A car is travelling at a constant speed of 26.5 m/s. Its tires have a radius of 72 cm. If the car slows down at a constant rate
maksim [4K]

Answer:

Magnitude of angular acceleration = -3.95 rad/s²

Explanation:

Angular acceleration is the ratio of linear acceleration and radius.

That is

        \texttt{Angular acceleration}=\frac{\texttt{Linear acceleration}}{\texttt{Radius}}\\\\\alpha =\frac{a}{r}

Radius = 72 cm = 0.72 m

Linear acceleration is rate of change of velocity.

a=\frac{11.7-26.5}{5.2}=-2.85m/s^2

Angular acceleration

        \alpha =\frac{a}{r}=\frac{-2.85}{0.72}=-3.95rad/s^2

Angular acceleration = -3.95 rad/s²  

Magnitude =  3.95 rad/s²     

4 0
3 years ago
Consider two identical objects of mass m = 0.250 kg and charge q = 4.00 μC. The first charge is held in place at the origin of a
Gnom [1K]

Answer:

a = 640 m/s²

Explanation:

From work-kinetic energy principles,

The net force acting on the second object is the gravitational force and the electric force due to the first object.

So, the gravitational force on the mass is F₁ = Gm₁m₂/r² since m₁ = m₂ = m, U = -Gm²/r²

Also, the electric force on the charge is F₂ = kq₁q₂/r² since q₁ = q₂ = q, U = kq²/r²

The net Force F = ma

So, -F₁ + F₂ = F     (F₁ is negative since it is an attractive force in the negative x -direction and F₂ is positive since it is a repulsive force in the positive x- direction)

-Gm²/r² + kq²/r² = ma

ma = -Gm²/r² + kq²/r²

a = (-Gm²/r² + kq²/r²)/m

a = (-G + kq²/m²)m/r²

Since m = 0.250 kg, q = 4.00 μC = 4.00 × 10⁻⁶ C, r = 3.00 cm = 3.00 × 10⁻² m, G = 6.67 × 10⁻¹¹ Nm²/kg², k = 9 × 10⁹ Nm²/C² and a = acceleration of second mass.

Substituting the variables into the equation, we have

a = (m/r²)(-G + k(q/m)²)]

a = (0.250 kg/{3.00 × 10⁻² m}²)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(4.00 × 10⁻⁶ C/0.250 kg)²)

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(16 × 10⁻⁶ C/kg)²)]

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(256 × 10⁻¹² C²/kg²)]

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 2304 × 10⁻³ Nm²/kg²  ]

a = (0.250 kg/9.00 × 10⁻⁴ m)(2.304 Nm²/kg²)

a = 0.576 Nm²/kg /9.00 × 10⁻⁴ m²

a = 0.064 × 10⁴N/kg

a = 64 × 10 N/kg)

a = 640 m/s²

8 0
2 years ago
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