Which of these is the most effective way for Leanna to cool down after an intense bike ride
1 answer:
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The gravitational force between two masses m₁ and m₂ is
where
G = 6.67408 x 10⁻¹¹ m³/(kg-s²), the gravitational constant
d = distance between the masses.
Given:
F = 1.5 x 10⁻¹⁰ N
m₁ = 0.50 kg
m₂ = 0.1 kg
Therefore
1.5 x 10⁻¹⁰ N = (6.67408 x 10⁻¹¹ m³/(kg-s²))*[(0.5*0.1)/(d m)²]
d² = [(6.67408x10⁻¹¹)*(0.5*0.1)]/1.5x10⁻¹⁰
= 0.0222
d = 0.1492 m = 149.2 mm
Answer: 149.2 mm
where
G = 6.67408 x 10⁻¹¹ m³/(kg-s²), the gravitational constant
d = distance between the masses.
Given:
F = 1.5 x 10⁻¹⁰ N
m₁ = 0.50 kg
m₂ = 0.1 kg
Therefore
1.5 x 10⁻¹⁰ N = (6.67408 x 10⁻¹¹ m³/(kg-s²))*[(0.5*0.1)/(d m)²]
d² = [(6.67408x10⁻¹¹)*(0.5*0.1)]/1.5x10⁻¹⁰
= 0.0222
d = 0.1492 m = 149.2 mm
Answer: 149.2 mm
We are given information:
If we apply Newton's second law we can calculate acceleration:
F = m * a
a = F / m
a = 25000 / 10000
a = 2.5 m/s^2
Now we can use this information to calculate change of speed.
a = v / t
v = a * t
v = 2.5 * 120
v = 300 m/s
Force is being applied in direction that is opposite to a direction in which space craft is moving. This means that final speed will be reduced.
v = 1200 - 300
v = 900 m/s
Formula for momentum is:
p = m * v
Initial momentum:
p = 10000 * 1200
p = 12 000 000
p = 12 *10^6 kg*m/s
Final momentum:
p = 10000 * 900
p = 9 000 000
p = 9 *10^6 kg*m/s
If we apply Newton's second law we can calculate acceleration:
F = m * a
a = F / m
a = 25000 / 10000
a = 2.5 m/s^2
Now we can use this information to calculate change of speed.
a = v / t
v = a * t
v = 2.5 * 120
v = 300 m/s
Force is being applied in direction that is opposite to a direction in which space craft is moving. This means that final speed will be reduced.
v = 1200 - 300
v = 900 m/s
Formula for momentum is:
p = m * v
Initial momentum:
p = 10000 * 1200
p = 12 000 000
p = 12 *10^6 kg*m/s
Final momentum:
p = 10000 * 900
p = 9 000 000
p = 9 *10^6 kg*m/s
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.