Answers
thereforemass m1=4.8kg and the tension
in the horizontalspring T2=10N.
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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.
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Related Questions
Imagine two billiard balls on a pool table. Ball A has a mass of 2 kilograms and ballB has a mass of 3 kilograms. The initial velocity of ball A is 9 meters per second to the right, and the initial velocity of the ball B is 6 meters per second to the left. The final velocity of ball A is 9 meters per second to the left, while the final velocity of ball B is 6 meters per second to the right. 1. Explain what happens to each ball after the collision. Why do you think this occurs? Which of Newton’s laws does this represent?
A solid 0.6950 kg ball rolls without slipping down a track toward a vertical loop of radius ????=0.8950 m . What minimum translational speed ????min must the ball have when it is a height H=1.377 m above the bottom of the loop in order to complete the loop without falling off the track? Assume that the radius of the ball itself is much smaller than the loop radius ???? . Use ????=9.810 m/s2 for the acceleration due to gravity.
How can scientific method solve real world problems examples
Circuit A in a house has a voltage of 218 V and is limited by a 45-A circuit breaker. Circuit B is at 120.0 V and has a 25-A circuit breaker.What is the ratio of the maximum power delivered by circuit A to that delivered by circuit B?
Answers
Answer:
6000 joules
Explanation:
I jus learned dis
Answer:6000j
Explanation:
Hope that helps
B. It won't move.
C. It will move at the same speed that the large object was moving initially.
D. It will move slower than the large object was moving initially.
Answers
Answer:
a ut will move faster than the large object was moving initially
Answer: It will move faster than the large object was moving initially.
Explanation:
Answers
Answer:
17.54N in -x direction.
Explanation:
Amplitude (A) = 3.54m
Force constant (k) = 5N/m
Mass (m) = 2.13kg
Angular frequency ω = √(k/m)
ω = √(5/2.13)
ω = 1.53 rad/s
The force acting on the object F(t) = ?
F(t) = -mAω²cos(ωt)
F(t) = -2.13 * 3.54 * (1.53)² * cos (1.53 * 3.50)
F(t) = -17.65 * cos (5.355)
F(t) = -17.57N
The force is 17.57 in -x direction
How far does the car move during the 2.52 s?
How long does it take the car to come to a complete stop?
Answers
Answer:
19.1 m/s
58.1 m
8.60 s
Explanation:
Take north to be positive and south to be negative.
Use Newton's second law to find the acceleration.
∑F = ma
-7850 N = (2500 kg) a
a = -3.14 m/s²
Given:
v₀ = 27.0 m/s
a = -3.14 m/s²
Find: v given t = 2.52 s
v = at + v₀
v = (-3.14 m/s²) (2.52 s) + 27.0 m/s
v = 19.1 m/s
Find: Δx given t = 2.52 s
Δx = v₀ t + ½ at²
Δx = (27.0 m/s) (2.52 s) + ½ (-3.14 m/s²) (2.52 s)²
Δx = 58.1 m
Find: t given v = 0 m/s
v = at + v₀
0 m/s = (-3.14 m/s²) t + 27.0 m/s
t = 8.60 s
track with a radius of 30 meters. What
is the car's rate of centripetal
acceleration?
Answers
The car's rate of centripetal acceleration in the circular path is 4.8 m/s².
The given parameters;
- mass of the car, m = 2,000 kg
- velocity of the car, v = 12 m/s
- radius of the circular path, r = 30 m
The centripetal acceleration of the car is calculated as follows;
where;
- v is the tangential speed of the car
- r is the radius of the circular path
Substitute the given parameters and solve for the centripetal acceleration;
Thus, the car's rate of centripetal acceleration is 4.8 m/s².
Learn more here:brainly.com/question/11700262
a = 12²/30 =4.8 m/s²
Answers
North component:
cos(41.5) * 835 = 625.37 km/h
West component of speed:
sin(41.5) * 835 = 553.29 km/h
After 2.2 hours plane will fly:
2.2*625.37 = 1375.81 km north
2.2*553.29 = 1217.23 km west
Final answer:
To find the components of the velocity vector, you can use trigonometry. The north component is calculated using the sine function and the west component is calculated using the cosine function. After 2.20 hours, the distance traveled north and west can be found by multiplying the velocity components by the time.
Explanation:
To find the components of the velocity vector in the northerly and westerly directions, we can use trigonometry. The velocity vector is 835 km/h and is traveling in a direction 41.5° west of north. To find the north component, we can use the sine function: North component = velocity * sin(angle). To find the west component, we can use the cosine function: West component = velocity * cos(angle).
After 2.20 hours, we can find the distance traveled north and west by multiplying the velocity components by the time: Distance north = North component * time and Distance west = West component * time.
Let's calculate the values:
- North component = 835 km/h * sin(41.5°)
- West component = 835 km/h * cos(41.5°)
- Distance north = North component * 2.20 h
- Distance west = West component * 2.20 h
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