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seropon [69]
3 years ago
14

How do you find the weight of an object on an incline in physics?​

Physics
1 answer:
Nezavi [6.7K]3 years ago
8 0

Refers to the attachment for the answer.

Let us assume that the object of mass m is kept on the Inclined plane.

Now, there will act one force called as Component of the Weight along the Incline which is given by the Relation,

mgsinθ,

where θ is the angle which the incline makes with the surface or Angle of the Incline.

Now, If there will be no friction and the object is moving along the incline

Force = mgsinθ

⇒ ma = mgsinθ

∴ a = gsinθ

This case is valid when the angle of the Incline is greater than the angle of repose, which means the object is moving with no cause or acing of the force.

But sometimes when the object does not move without the action of force, I mean that the angle of repose is greater than the angle of the incline, then we need to apply the force so that the object can move then,

Force applied = mgsinθ

∴ a = gsinθ

It will change the cases when friction is involved.

Now, For velocity, It can be found by using the equation of Motions. Time, Distance or initial velocity, etc must be given if the question will be asked related to the velocity. So by using them, you can find that.

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3 years ago
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I was driving along at 20 m/s , trying to change a CD and not watching where I was going. When I looked up, I found myself 46 m
vampirchik [111]

Answer:

6.46393559312 m/s²

Explanation:

Time taken to cover 56 m

t=\dfrac{56}{29}\ s

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From equation of linear motion

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

8.4+20(\dfrac{56}{29}-0.42)+\dfrac{1}{2}a(\dfrac{56}{29}-0.42)^2=46\\\Rightarrow a=(46-8.4-20(\dfrac{56}{29}-0.42))\times\dfrac{2}{(\dfrac{56}{29}-0.42)^2}\\\Rightarrow a=6.46393559312\ m/s^2

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5 0
3 years ago
A certain electromagnetic wave traveling in seawater was observed to have an amplitude of 98.02 (V/m) at a depth of 10 m, and an
maksim [4K]

Answer:

The  value is   \alpha =  0.002 Np/m

Explanation:

From the question we are told that

  The first amplitude of the wave is  E_{max}1 =  98.02 \  V/m

  The first  depth  is  D_1 =  10 \  m

   The second amplitude is  E_{max}2 =  81.87 \  (V/m)

   The second depth is D_2 = 100 \ m

Generally from the spatial wave equation we have

   v(x) =  Ae^{-\alpha d}cos(\beta x  + \phi_o)

=>       \frac{v(x)}{v(x)} =\frac{  Ae^{-\alpha d}cos(\beta x  + \phi_o)}{ Ae^{-\alpha d}cos(\beta x  + \phi_o)}

So considering the ratio of the equation for the  two depth

\frac{A}{A_S}  =  \frac{e^{-D_1 \alpha }}{e^{-D_2 \alpha }}

=>   \frac{98.02}{81.87}  =  \frac{e^{-10 \alpha }}{e^{-100 \alpha }}

=>   \alpha  =  \frac{0.18}{90}

=>    \alpha =  0.002 Np/m

       

4 0
3 years ago
how long does it take sound to travel the distance between the two microphones? Given.:wave 1 of microphone 1 has T=2 sec and f=
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Answer:

0.00583 seconds

Explanation:

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In an experiment, you find the mass of a cart to be 250 grams. What is the mass of the cart in kilograms?
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