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natima [27]
3 years ago
11

1. Your big brother, despite warning, drops a penny off the observation

Physics
1 answer:
Tpy6a [65]3 years ago
4 0

Answer:

The marble will take 8.813 seconds to hit the ground.

Explanation:

The marble will fall according to this motion equation: y(final) = y(initial) + v(initial)×t + 1/2×a×t×t

y(final)=0 m, the marble's final position in relation to the ground

y(initial)=381 m, the marble's initial position in relation to the ground

v(initial)=0 m/s (meters per second), the marble's initial velocity (speed with direction) when fall first begins

t=? seconds, this is the amount of time that passes while falling, the variable in this problem

a=-9.81 m/s/s, the rate at which the velocity of the marble changes, acceleration. On Earth, neglecting friction, acceleration is a constant at negative 9.81 meters per second squared

Plug the values into the equation: 0= 381 + 0t + 1/2(-9.81)(t)(t)

Subtract 381 from both sides and disregard 0t because it has no value and no longer matters: -381= 1/2(-9.81)(t)(t)

Multiply out the equation: -381= -4.905(t)(t)

Divide both sides by -4.905 to isolate the variable: 77.676=(t)(t)

Find the square root of both sides: t=8.813 seconds

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QveST [7]
Plants in general I think. 
photosynthetic organisms.
7 0
4 years ago
A girl walks 5 blocks east and then two blocks north. Find her total displacement.
Vlad [161]
Her displacement would be -3 blocks

displacement = final position - initial position
4 0
4 years ago
Velocity of an object of mass 5 kg increases from 3 m/s to 7 m/s on applying a force continuously for 2 s. Find out the force ap
Alenkasestr [34]

Answer:

Force=10N

Velocity=13m/s

Explanation:

Given,

Mass = 5kg

initial velocity=3m/s

Final velocity=7m/s

Time=2s

Now acceleration= v-u/t

=7-3/2

=4/2

=2m/s²

So, acceleration=2m/s²

Now,

F=ma

=5*2

=10N

7 0
3 years ago
A single slit of width 0.3 mm is illuminated by a mercury light of wavelength 254 nm. Find the intensity at an 11° angle to the
MArishka [77]

Answer:

The  the intensity at an 11° angle to the axis in terms of the intensity of the central maximum is  

   I_c = \frac{I}{I_o}  =8.48 *10^{-8}

Explanation:

From the question we are told that

   The  width of the slit is  D  =  0.3 \ mm =  0.3 *10^{-3} \ m

    The  wavelength is  \lambda =  254 \ nm =  254 *10^{-9} \ m

     The angle is  \theta  =  11^o

The intensity of at 11^o to the axis in terms of the intensity of the central maximum. is mathematically represented as

        I_c = \frac{I}{I_o}  = [ \frac{sin \beta  }{\beta }] ^2

Where \beta is mathematically represented as

        \beta  =  \frac{D sin (\theta ) *  \pi}{\lambda }

substituting values

      \beta  =  \frac{0.3 *10^{-3} sin (11 ) * 3.142}{254 *10^{-9} }

     \beta  =  708.1 \ rad

So

  I_c = \frac{I}{I_o}  = [ \frac{sin (708.1)  }{(708.1)}] ^2

   I_c = \frac{I}{I_o}  =8.48 *10^{-8}

6 0
4 years ago
You are designing a thin transparent reflective coating for the front surface of a sheet of glass. The index of refraction of th
galben [10]

Answer:

The index of refraction of the coating  is 1.32.

Explanation:

Given that,

Index of refraction of the glass = 1.52

Thickness = 104 nm

Length = 550 nm

We need to calculate the index of refraction of the coating

Using formula of index

n = \dfrac{L}{4t}

Where, L = length

t = thickness

Put the value into the formula

n=\dfrac{550}{4\times104}

n=1.32

Hence, The index of refraction of the coating  is 1.32.

7 0
4 years ago
Read 2 more answers
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