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cluponka [151]
2 years ago
7

What is a hypothesis?

Physics
1 answer:
Likurg_2 [28]2 years ago
4 0

Answer:

A

Explanation:

a statement that can be tested through the scientific method

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Which is less likely to be a reliable source of information, the webpage of a university or the webpage of a scientist who is tr
tatyana61 [14]

webpage of a scientist who is trying to sell a new invention

trying to dell = vested interest.

umiversity should be objective impartial

7 0
3 years ago
The motorcycle travels with a constant speed v0 along the path that, for a short distance, takes the form of a sine curve. Deter
Alexandra [31]

Let at any instant of time the speed is vo and the angle made by the bike with the horizontal is given

now we have

component of speed in x direction given as

v_x = v_0cos\theta

component of speed in y direction will be

v_y = v_0sin\theta

now from above two equations we can say that here

\theta = angle with the horizontal at any instant

and since here it is a sine curve so we know that

y = sin(x)

so we have slope of graph

tan\theta = \frac{dy}{dx} = cos(x)

6 0
3 years ago
the inertial mass of an object is measured by exerting a force on the object and measuring the object's BLANK using an inertial
JulijaS [17]

Answer:

acceleration

sorry i answered late

3 0
3 years ago
A vertical spring has a spring constant of 2900 N/m. The spring is compressed 80 cm and a 8 kg spider is placed on the spring. T
Serga [27]

Answer:

a)  k_{e} = 928 J , b)U = -62.7 J , c) K = 0 , d) Y = 11.0367 m,  e)  v = 15.23 m / s  

Explanation:

To solve this exercise we will use the concepts of mechanical energy.

a) The elastic potential energy is

      k_{e} = ½ k x²

      k_{e} = ½ 2900 0.80²

      k_{e} = 928 J

b) place the origin at the point of the uncompressed spring, the spider's potential energy

     U = m h and

     U = 8 9.8 (-0.80)

     U = -62.7 J

c) Before releasing the spring the spider is still, so its true speed and therefore the kinetic energy also

      K = ½ m v²

      K = 0

d) write the energy at two points, maximum compression and maximum height

     Em₀ = ke = ½ m x²

     E_{mf} = mg y

     Emo = E_{mf}

     ½ k x² = m g y

     y = ½ k x² / m g

     y = ½ 2900 0.8² / (8 9.8)

     y = 11.8367 m

As zero was placed for the spring without stretching the height from that reference is

     Y = y- 0.80

     Y = 11.8367 -0.80

     Y = 11.0367 m

Bonus

Energy for maximum compression and uncompressed spring

     Emo = ½ k x² = 928 J

     E_{mf}= ½ m v²

     Emo = E_{mf}

     Emo = ½ m v²

      v =√ 2Emo / m

     v = √ (2 928/8)

     v = 15.23 m / s

8 0
3 years ago
An object whose weight is 10kg is placed on smooth plane inclined at 30° to the horizontal. find the acceleration of the object
velikii [3]

Answer:

4.9 m/s²

Explanation:

Draw a free body diagram.  There are two forces on the object:

Weight force mg pulling straight down,

and normal force N pushing perpendicular to the plane.

Sum the forces in the parallel direction.

∑F = ma

mg sin θ = ma

a = g sin θ

a = (9.8 m/s²) (sin 30°)

a = 4.9 m/s²

8 0
3 years ago
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