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Angelina_Jolie [31]
2 years ago
14

A spring is 14cm long. Three masses are hung from it and then it is measured again. Now it is 19.5cm long. What force did the th

ree masses provide? The spring constant for the spring is 30N/m.
Physics
1 answer:
lukranit [14]2 years ago
3 0

Answer:

Gravity

Explanation:

The answer is gravity because when the 3 masses were hung from the spring, gravity pulled the spring towards the ground.

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Donna's doctor must have diagnosed her with muscle cramps. With the given symptoms above, it is likely that she is experiencing muscle cramps. Muscle cramps most likely happen when the individual experiencing it is lack of fluid intake, causing the muscles to tighten causing pain and for the muscle to contract involuntarily.
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The diagram below depicts all the forces acting on an object. Use both vector resolution and vector
maksim [4K]

Answer:

A

Explanation:

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2 years ago
In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co
labwork [276]

Answer:

The answer is given below

Explanation:

u is the initial velocity, v is the final velocity. Given that:

m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m

a)

The final velocity of cart 1 after collision is given as:

v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\  Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s

The final velocity of cart 2 after collision is given as:

v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\  Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s

b) Using the law of conservation of energy:

\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm

7 0
2 years ago
You traveled 150.meters south at a speed of 1.50m/s, and then traveled 600. Meters north at a rate of 2.00 m/s.
Jlenok [28]

Answer:

1.) 400s

2.) 1.875 m/s

3.) 1.125 m/s

Explanation:

Given that you traveled 150.meters south at a speed of 1.50m/s,

Time = distance/ speed

Substitute speed and distance into the formula

Time = 150/1.5

Time = 100 s

and then traveled 600. Meters north at a rate of 2.00 m/s.

Time = 600 / 2

Time = 300 s

1.) Total time = 100 + 300

Total time = 400 s

2.) The average speed will be total distance travelled over total time.

Total distance travelled = 150 + 600

Total distance travelled = 750 m

Substitute all the parameters into the formula

Average speed = 750/400

Average speed = 1.875 m/s

3.) Average velocity will be displacement over total time

Displacement = 600 - 150

Displacement = 450 m

Average velocity = 450/400

Average velocity = 1.125 m/s

6 0
2 years ago
How would the interference pattern change for this experiment if a. the grating was moved twice as far from the screen and b. th
Airida [17]

Answer:

See explanation

Explanation:

Solution:-

- Here we will assume that the grating has the line density ( N ) defined by the number of lines per mm.

- The angle that each fringe forms on the screen is defined by ( θ ).

- The order of bright/dark spot is defined by an integer ( n )

- The wavelength of the incident light is ( λ )

- Here we will use the relation given for diffraction grating by the Young's Experiment as follows:

                               n*lambda = \frac{sin(theta)}{N}

- The above given formulation is for constructive interference.

- We will inspect the effect of increasing the distance between the screen and the grating. Consider the length ( L ) from the center of the grating to the center of the screen. The distance ( yn ) will denote the distance between each fringe in vertical direction on the screen.

- For small angles ( θ ) we can make an approximation of sin ( θ ) ≈ tan ( θ ). Where,

                            sin ( θ ) ≈ tan ( θ ) = [ yn / L ]

- Substitute the above approximation in the given relation of diffraction gratings as follows:

                            y_n = n*lamda*L*N

- To double the distance between the screen and grating we will use the above relation with ( 2L ):

                            yn ∝ L

Result: The distance between each order of bright and dark fringe is doubled. The interference pattern would have twice the spread! This also means that less number of bright spots would be seen on the screen as the coverage area would require a larger screen to accommodate the entire interference pattern. The spread also reduces the intensity/contrast between the bright and dark fringes because the distance travelled by each ray of light has increased. The intensity is inversely proportional to the square of distance travelled.

- Similarly, the line density of the grating ( N ) was doubled. Then,

                            yn ∝ N

Result: The distance between each order of bright and dark fringe is doubled. The interference pattern would have twice the spread!This also means that less number of bright spots would be seen on the screen as the coverage area would require a larger screen to accommodate the entire interference pattern.

4 0
2 years ago
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