Which of the following are inertial reference frames? A. A car driving at steady speed on a straight and level road. B. A car dr
iving at steady speed up a 10∘ incline. C. A car speeding up after leaving a stop sign. D. A car driving at steady speed around a curve.
2 answers:
Answers:
A. A car driving at steady speed on a straight and level road.
B. A car driving at steady speed up a 10∘ incline.
Explanation:
An object is said to be in an inertial reference frame if the net force acting on the object is zero. According to Newton's second law, this also means that the acceleration of the object is also zero:
Since F=0, a=0 as well.
Let's now analyze each case.
A. A car driving at steady speed on a straight and level road. --> YES: this is an inertial reference frame, because the car is keeping a constant speed and a constant direction, so its velocity is not changing, and its acceleration is zero.
B. A car driving at steady speed up a 10∘ incline. --> YES: this is an inertial reference frame, because the car is keeping a constant speed and a constant direction, so its velocity is not changing, and its acceleration is zero.
C. A car speeding up after leaving a stop sign. --> NO: this is not an intertial reference frame, because the car is speeding up, so it is accelerating.
D. A car driving at steady speed around a curve. --> NO: this is not an inertial reference frame, because the car is changing direction, therefore its velocity is changing and so the car is accelerating.
So the only two choices which are correct are A and B.
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Answer:
The activation energy for this reaction, Ea = 159.98 kJ/mol
Explanation:
Using the Arrhenius equation as:
Where, Ea is the activation energy.
R is the gas constant having value 8.314 J/K.mol
K₂ and K₁ are the rate constants
T₂ and T₁ are the temperature values in kelvin.
Given:
K₂ = 8.66×10⁻⁷ s⁻¹ , T₂ = 425 K
K₁ = 3.61×10⁻¹⁵ s⁻¹ , T₁ = 298 K
Applying in the equation as:
Solving for Ea as:
Ea = 159982.23 J /mol
1 J/mol = 10⁻³ kJ/mol
Ea = 159.98 kJ/mol
Answer:
1)
2)
Explanation:
<u>Projectile Motion</u>
When an object is launched near the Earth's surface forming an angle with the horizontal plane, it describes a well-known path called a parabola. The only force acting (neglecting the effects of the wind) is the gravity, which acts on the vertical axis.
The heigh of an object can be computed as
Where is the initial height above the ground level,
is the vertical component of the initial velocity and t is the time
The y-component of the speed is
1) We'll find the vertical component of the initial speed since we have not enough data to compute the magnitude of
The object will reach the maximum height when . It allows us to compute the time to reach that point
Solving for
Thus, the maximum heigh is
We know this value is 8 meters
Solving for
Replacing the known values
2) We know at t=1.505 sec the ball is above Julie's head, we can compute