Mass of the car = 1200 kg
Mass of the truck = 2100 kg
Total mass of car and truck = 2100 + 1200 = 3300 kg
Since, the car pushes the truck. Hence, they will move together and will have same acceleration.
Let the acceleration be a.
According to Newton's second law:
F(net) = ma
F = 4500 N
4500 = 3300 × a
a = 1.36 m/s^2
Let the force applied by the car on truck be F.
F = F(net) on the truck
F = ma
F = 2100 × 1.36
F = 2856 N
Hence, the force applied by the car on the truck is 2856 N
Answer: A) In case of dark energy increasing significantly, the universe would accelerate its expansion at an ever-increasing rate which later on may result in "Big Rip".
B) In case of dark energy decreasing significantly, the universe would decelerate its expansion which later on may result in "Big Crunch".
Explanation:
In case of significant increase in dark energy, the universe would accelerate its expansion at more increasing rate. Ultimately the expansion would result in scale factor becoming infinite within a finite period of time. As a result due to the repulsive dominance of dark energy the whole universe would come apart resulting in a situation called "Big Rip"
If dark energy were to decrease significantly enough with time, the accelerating expansion of the universe that is observed now would start to decelerate as the mass once again dominates over dark energy. It will result in universe reversing its expansion resulting in its collapse, a situation called "Big Crunch".
Hope it helps!!!
To solve this problem we will apply the concepts related to the Doppler effect. The relationship given between the observed frequency and the actual frequency is given under the following mathematical function,
Here,
Rearrange the equation to obtain the frequency observed
Therefore the frequency of the emergency lights that you observe when it reaches you in your spaceship is
Answer:
The mass of the second weight is approximately 0.477 kg
Explanation:
The given parameters are;
The acceleration experienced by the two weights = 3.8 m/s²
The mass of the first weight = 1.08 kg
The formula for the acceleration, a, of weights attached to a friction pulley, is given as follows;
Where;
a = The common acceleration of the two weights
g = The acceleration due to gravity = 9.81 m/s²
M = The mass of the first weight = 1.08 kg
m = The mass of the second weight
Therefore, we have;
The mass of the second weight = m ≈ 0.477 kg
The mass of the second weight ≈ 0.477 kg.
Answer:
.
Explanation:
The box is sliding with a constant speed in a fixed direction (to the right.) In other words the velocity of this box is constant. Hence, this box would be in a translational equilibrium. The acceleration of this box would be zero.
By Newton's Second Law of motion, the net force on this box would be . In other words, forces on this box are balanced.
The question is asking for the size of the friction on the box. Assuming that the floor is horizontal. The friction on this box would also be horizontal,
The only other force that could balance that friction would be the push to the right. The direction of this push is horizontal (to the right.) Hence, the entirety of that would be in the horizontal direction.
Thus, forces on this box in the horizontal direction would be:
- The push to the right.
- Friction that opposes the rightward motion of the box (that is, to the left.)
Since these two forces must balance each other, the size of the friction would also be .