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ra1l [238]
3 years ago
6

What happens when the data in an experiment does not support the hypothesis?

Physics
1 answer:
liq [111]3 years ago
6 0
Say in your conclusion that your Hypothesis was incorrect 
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Two objects, M = 15.3 ks and m = 8.29 kg are connected with an ideal string and suspended by a pulley (which rotates with no fri
Scilla [17]

Answer:

(a) 7 m/s

(b) 931 rad/s

(c) 0.716 s

Explanation:

Gravity would be exerting on the 2 masses

G_1 = Mg = 15.3*9.81 = 150.093N

G_2 = mg = 8.29*9.81= 81.32N

Since heavier, mass 1 (M) would be the one pulling down, while mass 2 is being pulled up.

So the net force on mass 1 is

F = G_1 - G_2 = 68.77N

This force would generate torque on the solid pulley

T = FR = 68.77 * 0.0075 = 0.5158 Nm

We can also calculate the pulley moments of inertia, with it being solid

I = 0.5MpR^2 = 0.5*14.1*0.0075^2 = 0.000396563kgm^2

From there we can calculate the angular acceleration of the pulley, which generates the entire system motion

\alpha = \frac{T}{I} = \frac{0.5158}{0.000396563} = 1300.58 rad/s^2

Since the system is moved by a distance of d = 2.5m, the pulley would have turn an angle of

\theta = \frac{d}{R} = \frac{2.5}{0.0075} = 333 rad

(c)The time it takes to get to this distance is

\theta = \frac{\alpha t^2}{2}

t^2 = \frac{2\theta}{\alpha} = \frac{2*333}{1300.58} =0.513

t = \sqrt{0.513} = 0.716s

(b)The final angular speed of the disk is

\omega = \alpha t = 1300.58*0.716 = 931 rad/s

(a) And so the perimeter speed of the pulley, which is also speed of mass 1 when it comes to d = 2.5 m is

v = \omega R = 931*0.0075 \approx 7m/s

5 0
3 years ago
"On a movie set, an alien spacecraft is to be lifted to a height of 32.0 m for use in a scene. The 260.0-kg spacecraft is attach
Lisa [10]

Answer:

<em>The time interval required to lift the spacecraft to this specified height is 123.94 seconds</em>

Explanation:

Height through which the spacecraft is to be lifted = 32.0 m

Mass of the spacecraft = 260.0 kg

Four crew member each pull with a power of 135 W

18.0% of the mechanical energy is lost to friction.

work done in this situation is proportional to the mechanical energy used to move the spacecraft up

work done = (weight of spacecraft) x (the height through which it is lifted)

but the weight of spacecraft = mg

where m is the mass,

and g is acceleration due to gravity 9.81 m/s

weight of spacecraft = 260 x 9.81 = 2550.6 N

work done on the space craft = weight x height

==> work = 2550.6 x 32 = 81619.2 J

this is equal to the mechanical energy delivered to the system

18.0% of this mechanical energy delivered to the pulley is lost to friction.

this means that

0.18 x 81619.2  = <em>14691.456 J  </em> is lost to friction.

Total useful mechanical energy =  81619.2 J - 14691.456 J = 66927.74<em> J</em>

Total power delivered by the crew to do this work = 135 x 4 = 540 W

But we know tat power is the rate at which work is done i.e

P = \frac{w}{t}

where p is the power

where w is the useful work done

t is the time taken to do this work

imputing values, we'll have

540 = 66927.74/t

t = 66927.74/540

time taken t = <em>123.94 seconds</em>

8 0
3 years ago
Jenny and Fred both play a long note on their trombones. A stationary observer hears the sound increasing and decreasing repeate
lilavasa [31]
E=the sound waves from the trombones have slightly different velocity
7 0
3 years ago
Read 2 more answers
A small sphere has a harge of 9uC and other small sphere has a charge of 4uC.
Helga [31]

Answer:

Electrical force, F = 90 N

Explanation:

It is given that,

Charge on sphere 1, q_1=9\ \mu C=9\times 10^{-6}\ C

Charge on sphere 2, q_1=4\ \mu C=4\times 10^{-6}\ C

Distance between two spheres, d = 6 cm = 0.06 m

Let F is the electrical force between them. It is given by the formula of electric force which is directly proportional to the product of charges and inversely proportional to the square of distance between them such that,

F=k\dfrac{q_1q_2}{d^2}

F=9\times 10^9\times \dfrac{9\times 10^{-6}\times 4\times 10^{-6}}{(0.06)^2}

F = 90 N

So, the electrical force between them is 90 N. Hence, this is the required solution.

7 0
3 years ago
Martha was leaning out of the window on the second floor of her house and speaking to Steve. Suddenly, her glasses slipped from
DiKsa [7]

Answer:

s = 23.72 m

v = 21.56 m/s²  

Explanation:

given

time to reach the ground (t) = 2.2 second

we know that

a) s = u t + 0.5 g t²

   u = 0 m/s

   g = 9.8 m/s²

   s = 0 + 0.5 × 9.8 × 2.2²

  s = 23.72 m

b) impact velocity

      v = √(2gh)

      v = √(2× 9.8 × 23.72)    

      v =  √464.912

      v = 21.56 m/s²  

6 0
3 years ago
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