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3 years ago

Why is it necessary to be able to replicate the results of a scientific experiment

1 answer:

5 0

To ensure the results are not a mistake and to add credibility and validity to the results. A good theory can always be tested.

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A pulley system lifts a 500-lb. block 2.0 feet with an effort of only 50 lb. If the 50 lb. moves 30 feet, calculate the efficien

tatiyna

The answer to this is 67%

4 0

3 years ago

Read 2 more answers

A 20.0 kg crate sits at rest at the bottom of a 15.0-m-long ramp that is inclined at 34.0° above horizontal. A constant horizont

ankoles [38]

Answer:

987 joules, 3.01s

Explanation:

(A)

from the attached diagram

net force, Fnet, pulling the crate up the ramp is given by

Fnet = FcosФ - WsinФ - Fr

where FcosФ is the component of horizontal force 290N resolved parallel to the plane

WsinФ = mgsinФ = component of the weight of the crate resolved parallel to the plane

Fr = constant opposing frictional force

Fnet = 290cos34⁰ - 20 × 9.8 × sin34° - 65

Fnet = 240.421 - 109.602 - 65

Fnet = 65.82N

Work done on the crate up the ramp, W, is given by

W = Fnet × d (distance up the plane)

W = 65.819 × 15

W = 987.285 joules

W = 987 joules (to 3 significant Figures)

(B)

to calculate the time of travel up the ramp

we use the equation of motion

$s = ut + \frac{1}{2}at^{2}$

where s = distance up the plane, 15m

u = Initial velocity of the crate, which is 0 for a body that is initially at rest

a = acceleration up the plane, given by

$a = \frac{Fnet}{m}$

where m = mass of the crate, 20 kg

now, $a = \frac{65.819}{20} \\a = 3.291\frac{m^{2} }{s}$

from, $s = ut + \frac{1}{2}at^{2}$

$15 = 0*t + \frac{1}{2}* 3.291 * t^{2}$

15 = 0 + 1.645 $t^{2}$

15 = 1.645 $t^{2}$

$t = \sqrt{\frac{15}{1.645} }$

$t = 3.019$

t = 3.01s (to 3 sig fig)

7 0

3 years ago

A ball hits a wall horizontally at 6m/s and rebounces at 4.4m/s the ball is in contact with wall for 0.04 sec. what is the accel

Svetlanka [38]

Answer:

Acceleration (a) = 40 m/s²

Explanation:

Given:

Initial velocity (u) = 6 m/s

Final velocity (v) = 4.4 m/s

Time taken (t) = 0.04sec

Find:

Acceleration (a) = ?

Computation:

We know that,

⇒ v = u + at

⇒ a = (v - u) / t

⇒ Acceleration (a) = (4.4 - 6) / 0.04

⇒ Acceleration (a) = (-1.6) / 0.04

Acceleration (a) = 40 m/s²

8 0

3 years ago

What is the unit for the volt?<br> W<br> V =<br> 9<br> What is the unit for the volt

Likurg_2 [28]

the unit for volt is v

The volt (symbol: V) is the derived unit for electric potential, electric potential difference (voltage), and electromotive force.

4 0

3 years ago

An electron that has a velocity with x component 2.6 × 106 m/s and y component 3.2 × 106 m/s moves through a uniform magnetic fi

elena55 [62]

Answer:

a) $\vec F_{B} = 8.766\times 10^{-14}\,T\,k$ , b) $\vec F_{B} = -8.766\times 10^{-14}\,T\,k$

Explanation:

a) The magnetic force experimented by a particle has the following vectorial form:

$\vec F_{B} = q\cdot \vec v \times \vec B$

The charge of the electron is equal to $-1.602\times 10^{-19}\,C$ . Then, cross product can be solved by using determinants:

$\vec F_{B} = \begin{vmatrix}i&j&k\\-4.165\times 10^{-13}\, C\cdot \frac{m}{s} &-5.126\times 10^{-13}\,C\cdot \frac{m}{s} &0\,C\cdot \frac{m}{s} \\0.041\,T&-0.16\,T&0\,T\end{vmatrix}$

The magnetic force is:

$\vec F_{B} = 8.766\times 10^{-14}\,T\,k$

b) The charge of the proton is equal to $1.602\times 10^{-19}\,C$ . Then, cross product has the following determinant:

$\vec F_{B} = \begin{vmatrix}i&j&k\\4.165\times 10^{-13}\, C\cdot \frac{m}{s} &5.126\times 10^{-13}\,C\cdot \frac{m}{s} &0\,C\cdot \frac{m}{s} \\0.041\,T&-0.16\,T&0\,T\end{vmatrix}$

The magnetic force is:

$\vec F_{B} = -8.766\times 10^{-14}\,T\,k$

8 0

3 years ago