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Leokris [45]
4 years ago
6

Which statement correctly describes a hypothesis?

Physics
1 answer:
tankabanditka [31]4 years ago
7 0

Answer:

D

Explanation:

A hypothesis is pretty much what you think something is, what you think the solution is, with the little information that you already have. this can go on to be tested

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A 1300-kg car moving on a horizontal surface has speed v = 60 km/h when it strikes a horizontal coiled spring and is brought to
slega [8]

The stiffness constant of the spring is 68,290.3 N/m

<h3> Stiffness constant of the spring</h3>

Apply the principle of conservation of energy;

U = K.E

¹/₂kx² = ¹/₂mv²

kx² = mv²

k = mv²/x²

where;

  • m is mass
  • v is speed = 60 km/h = 16.67 m/s
  • x is the distance

k = (1300 x 16.67²)/(2.3²)

k = 68,290.3 N/m

Thus, the stiffness constant of the spring is 68,290.3 N/m.

Learn more about stiffness constant here: brainly.com/question/1685393

#SPJ1

7 0
3 years ago
Fertilization occurs in the fallopian tubes.
Tamiku [17]
What’s the question? Is it true or false?
7 0
3 years ago
A skier skies down a slope with an average speed of 50 km/s. How far does the skier travel in 5 minutes?
aliya0001 [1]

Answer:

15000000 m

Explanation:

From the question,

Speed (S) = Distance(d)/Time(t)

S = d/t....................... Equation 1

Make d the subject of the equation

d = St..................... Equation 2

Given: S = 50 km/s = 50000 m/s, t = 5 minutes = 5*60 seconds = 300 seconds

Substitute these values into equation 2

d = 50000(300)

d = 15000000 m

Hence the skier travels 15000000 m

6 0
4 years ago
In transverse waves, the medium moves perpendicular to the direction of energy transport? True or false?
saul85 [17]

Answer:

This statement is true

8 0
4 years ago
To stretch a spring 8.00cm from its unstretched length, 16.0J of work must be done.A)What is the force constant of this spring?B
ad-work [718]

A) 5000 N/m

The force constant of the spring can be found by using the expression for the elastic potential energy stored in the spring (which is equal to the work done on it):

W=U=\frac{1}{2}kx^2

where

k is the spring constant

x is the stretching/compression of the spring

In this problem, we have

W = 16.0 J is the work done

x = 8.00 cm = 0.08 m is the stretching

Substituting into the formula and re-arranging it, we find

k=\frac{2W}{x^2}=\frac{2(16.0 J)}{(0.08 m)^2}=5000 N/m

B) 400 N

The magnitude of the force needed to stretch the spring by x = 8.00 cm = 0.08 m is given by Hook's law:

F=kx

where k=5000 N/m as we found previously. Substituting x=0.08 m, we find:

F=(5000 N/m)(0.08 m)=400 N

C) 4 J

The work done to compress the spring by x=4.00 cm=0.04 m is given by the same formula used for part A:

W=\frac{1}{2}kx^2

where in this case, k=5000 N/m and x=0.04 m. Substituting, we find

W=\frac{1}{2}(5000 N/m)(0.04 m)^2=4 J

D) 200 N

As we did in part B), the force needed to stretch this distance is given by Hook's law:

F=kx

where in this case, k=5000 N/m and x=0.04 m. Substituting, we find

F=(5000 N/m)(0.04 m)=200 N

8 0
3 years ago
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