A car travels along a straight line at a constant speed of 53.0 mi/h for a distance d and then another distance d in the same direction at another constant speed. the average velocity for the entire trip is 26.5 mi/h. (a) what is the constant speed with which the car moved during the second distance d?
Answers
A distance of d is covered with 53 mile/hr initially.Time taken to cover this distance t1 = d/53 hourNext distance of d is covered with x mile hours.Time taken to cover this distance t2 = d/x hours.We have average speed = 26.5 mile / hour
= Total distance traveled/ total time taken
=
Related Questions
A bridge is made with segments of concrete 91 m long (at the original temperature). If the linear expansion coefficient for concrete is 1.2 × 10−5 ( ◦C)−1 , how much spacing is needed to allow for expansion for an increase in temperature of 56◦F? Answer in units of cm.
If a single circular loop of wire carries a current of 61 A and produces a magnetic field at its center with a magnitude of 1.70 10-4 T, determine the radius of the loop.
A small segment of wire contains 10 nC of charge. The segment is shrunk to one-third of its original length. A proton is very far from the wire. What is the ratio Ff/Fi of the electric force on the proton after the segment is shrunk to the force before the segment was shrunk?
Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually leaks off of both spheres. Calculate the magnitude of the electrostatic force, when each of the spheres has lost half of its initial charge. (Your answer will be a function of F, since no values are giving)
Answers
thereforemass m1=4.8kg and the tension
in the horizontalspring T2=10N.
HOPEIT HELPS YOU
PLEASEMarkme as brainliest.
☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️✌️✌️✌️✌️✌️✌️✌️✌️❤️❤️❤️
To determine the tension in the string that connects M2 and M3, we can follow these steps:
Step 1: Identify the necessary variables. Given data (for example) could be:
- Mass of M2, which is 5 kg
- Mass of M3, which is 10 kg
- The acceleration due to gravity, which is approximately 9.8 m/s²
- The angle at which the string pulls on M2, which is 30 degrees
- Assume the system is in equilibrium, meaning there is no net acceleration, so the acceleration is 0 m/s²
Step 2: Calculate the weight of M3, which is its mass times the acceleration due to gravity. This is because weight is the force exerted by gravity on an object, which equals the object's mass times the acceleration due to gravity.
For M3, this calculation would be M3 * g = 10 kg * 9.8 m/s² = 98 N (Newtons).
Step 3: Determine the force exerted by M2 that acts along the line of the string. This won't be the full weight of M2, because the string pulls at an angle. This component of the force can be calculated using the sine of the angle, because sine gives us the ratio of the side opposite the angle (here, the force along the string) to the hypotenuse (here, the full weight of M2) in a right triangle.
The horizontal component of the force of M2 is then M2 * g * sin(30deg) = 5 kg * 9.8 m/s² * sin(30deg) = 24.5 N.
Step 4: The tension in the string is the force M3 exerts on it, which is its weight, minus the component of M2's weight that acts along the string. This is because M2 and M3 are pulling in opposite directions, so they subtract from each other.
The tension in the string is then the weight of M3, 98 N, minus the horizontal (along the string) component of M2's weight, 24.5 N.
So, the tension in the string is 98 N - 24.5 N = 73.5 N.
This is the force that the string needs to exert in order to keep M2 and M3 connected and in equilibrium.
Learn more about Tension in a string here:
#SPJ12
Answers
Answers
We are given:
The tuning fork vibrates at 15660 oscillations per minute
Period of one back-and forth movement:
the given data can be rewritten as:
1 minute / 15660 oscillations
60 seconds / 15660 oscillations (1 minute = 60 seconds)
dividing the values
0.0038 seconds / Oscillation
Therefore, one back and forth vibration takes 0.0038 seconds
Answers
Answer:
51 mph
Explanation:
Average speed = Total Distance/Total Time
From the given data, Total Distance = 100 + 30 + 75 miles
and Total Time = 2 + 1 + 1 hours
Average Speed = 205/4
Average Speed = 51.25 mph ( or 51mph to the nearest whole number)
Answers
Statements that are right as regards oscillation are:
A. The decrease in the amplitude of an oscillation caused by dissipative forces is called damping.
B. The increase in amplitude of an oscillation by a driving force is called forced oscillation.
C. In a mechanical system, the amplitude of an oscillation diminishes with time unless the lost mechanical energy is replaced.
D. An oscillation that is maintained by a driving force is called forced oscillation.
- Amplitude can be regarded as magnitude of change that is been experienced by oscillating variable with each oscillation.
- When there is a decrease in the amplitude of an oscillation as a result dissipative forces, then it is regarded as damping.
- When there is increase in amplitude of an oscillation as a result of driving force then it is termed forced oscillation.
Therefore, the options are correct.
Learn more at:
Answer:
right A, B, C, D
Explanation:
They ask which statements are true
A) Right. The decrease in amplitude is due to the dissipation of energy by friction and is called damping
B) Right. In resonant processes the amplitude of the oscillation increases, being a forced oscillation
C) Right. In a system with energy loss, the amplitude must decrease, therefore energy must be supplied to compensate for the loss.
D) Right. It is a resonant process the driving force keeps the oscillation of the system
Answers
Answer:
b
Explanation: