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VladimirAG [237]
3 years ago
6

A girl pushes on a table with a 5-newton force directed to the east. Her friend pushes on the same table with a 10- newton force

directed south. Which of the following diagrams best depicts the action of the two friends?

Physics
2 answers:
nalin [4]3 years ago
4 0

Think like this: 10 is twice as much as 5.

Answer A pushes 2 squares east and 4 squares sout, being doubled just as 10 is to 5. Happy to help :)

kifflom [539]3 years ago
4 0

Explanation:

It is given that, a girl pushes on a table with a 5-newton force directed to the east. Her friend pushes on the same table with a 10- newton force directed south.  

We know that force is a vector quantity. So, diagram that depicts the action of two friends is option (d). The magnitude of the resultant force is given by :

R=\sqrt{5^2+10^2}

R = 11.18 N

Hence, the correct option is (D).

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What is the net force​
nadya68 [22]

Answer:

20 N in West direction.

Explanation:

opposite forces cancel each other. so 20 N in north and 20N in south cancel each other. In west and east direction...

70N in west-50N in east= 20N in west

3 0
3 years ago
See the diagram above. Which of the following is the best prediction of what would happen if you increased the distance in secti
Inessa [10]

By the definition of wavelength, the answer is the letter D, the wavelength would decrease.

We can see in the diagram a wave motion.

A wave has some characteristics:

  • Has an amplitude, the distance from 0 to the crest (highest point in the y-direction, point (3) in the figure) it would see in the figure as (2)
  • Has wavelength, the distance between the crests.
  • Has a trough, the lowest point in the y-direction.

Now, if we increase the distance of the crests, by the definition shown above, we will increase the wavelength.

Therefore, the answer is letter D, the wavelength would increase.

You can learn more about wave motion here:  

brainly.com/question/22763521

7 0
3 years ago
A person wants to determine the spring constant of an exercise stretch cord. He pulls the cord with a force probe that exerts a
Burka [1]

Answer:

350 N/m

Explanation:

If we are assuming the stretch does not exceed the elastic range of the material, then by Hooke's law the spring constant of the cord is simply the ratio between the force 70N acting on the cord to stretch 20cm or 0.2m

k = 70 / 0.2 = 350 N/m

The spring constant is 350 N/m

4 0
3 years ago
Read 2 more answers
Non-metals can form cations when combined with metals and anions when combined with other non-metals.
exis [7]
Well i think it is false caus ebunch of products these day have metals incorparded in the product
4 0
3 years ago
Read 2 more answers
A 1.8-kg object is attached to a spring and placed on frictionless, horizontal surface. A force of 40 N stretches a spring 20 cm
Sergio [31]

Answer:

a) k = 200 N/m

b) E = 4 J

c) Δx = 6.3 cm

Explanation:

a)

  • In order to find force constant of the spring, k, we can use the the Hooke's Law, which reads as follows:

       F = - k * \Delta x (1)

  • where F = 40 N and Δx =- 0.2 m (since the force opposes to the displacement from the equilibrium position, we say that it's a restoring force).
  • Solving for k:

       k =- \frac{F}{\Delta x} =-\frac{40 N}{-0.2m} = 200 N/m (2)

b)

  • Assuming no friction present, total mechanical energy mus keep constant.
  • When the spring is stretched, all the energy is elastic potential, and can be expressed as follows:

        U = \frac{1}{2}* k* (\Delta x)^{2} (3)

  • Replacing k and Δx by their values, we get:

       U = \frac{1}{2}* k* (\Delta x)^{2} = \frac{1}{2}* 200 N/m* (0.2m)^{2} = 4 J (4)

c)

  • When the object is oscillating, at any time, its energy will be part elastic potential, and part kinetic energy.
  • We know that due to the conservation of energy, this sum will be equal to the total energy that we found in b).
  • So, we can write the following expression:

        \frac{1}{2}* k* \Delta x_{1} ^{2} + \frac{1}{2} * m* v^{2}  = \frac{1}{2}*k*\Delta x^{2}   (5)

  • Replacing the right side of (5) with (4), k, m, and v by the givens, and simplifying, we can solve for Δx₁, as follows:

        \frac{1}{2}* 200N/m* \Delta x_{1} ^{2} + \frac{1}{2} * 1.8kg* (-2.0m/s)^{2}  = 4J   (6)

⇒      \frac{1}{2}* 200N/m* \Delta x_{1} ^{2}   = 4J  - 3.6 J = 0.4 J (7)

⇒     \Delta x_{1}   = \sqrt{\frac{0.8J}{200N/m} } = 6.3 cm (8)

6 0
3 years ago
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