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Bond [772]
3 years ago
12

Calculate the instantaneous speed of an apple 8 seconds after being dropped from rest.

Physics
1 answer:
Lynna [10]3 years ago
8 0
Vfinal = Vinitial + at
Vinitial = 0
a = 9.8 m/s^2
t = 8s
Vfinal = 0 + 9.8 * 8
ANSWER: 78.4 m/s
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If a proton and an electron are released when they are 7.00×10−10 m apart (typical atomic distances), find the initial accelerat
Ugo [173]

Answer:

The acceleration of the proton is 2.823 x 10¹⁷ m/s²

The acceleration of the electron is 5.175 x 10²⁰ m/s²

Explanation:

Given;

distance between the electron and proton, r = 7 x 10⁻¹⁰ m

mass of proton, m_p = 1.67 x 10⁻²⁷ kg

mass of electron, m_e = 9.11 x 10⁻³¹ kg

The attractive force between the two charges is given by Coulomb's law;

F = \frac{k(q_p)(q_e)}{r^2}

where;

k is Coulomb's constant = 9 x 10⁹ Nm²/c²

F = \frac{k(q_p)(q_e)}{r^2} \\\\F = \frac{(9*10^9)(1.602*10^{-19})(1.602*10^{-19})}{(7*10^{-10})^2} \\\\F = 4.714 *10^{-10} \ N

Acceleration of proton is given by;

F = ma

F = m_pa_p\\\\a_p = \frac{F}{m_p}\\\\a_p = \frac{4.714*10^{-10}}{1.67*10^{-27}}\\\\a_p = 2.823 *10^{17} \ m/s^2

Acceleration of the electron is given by;

F = m_ea_e\\\\a_e = \frac{F}{m_e}\\\\a_e = \frac{4.714*10^{-10}}{9.11*10^{-31}}\\\\a_e = 5.175 *10^{20} \ m/s^2

7 0
3 years ago
A 645 g block is released from rest at height h0 above a vertical spring with spring constant k = 530 N/m and negligible mass. T
Nina [5.8K]

Answer:

a)5.88J

b)-5.88J

c)0.78m

d)0.24m

Explanation:

a) W by the block on spring is given by

W= \frac{1}{2}kx² = \frac{1}{2}(530)(0.149)² =  5.88 J

b)  Workdone by the spring = - Workdone by the block = -5.88J

c) Taking x = 0 at the contact point we have U top = U bottom

So, mgh_o = \frac{1}{2}kx² - mgx

And, h_o= ( \frac{1}{2}kx² - mgx )/(mg) = [\frac{1}{2} (530)(0.149^2)-(0.645)(9.8)(0.149)]/(0.645x9.8)    

   h_o=   0.78m            

d) Now, if the initial initial height of block is 3h_o

h_o = 3 x 0.78 = 2.34m

then, \frac{1}{2}kx² - mgx - mgh_o =0

 

\frac{1}{2}(530)x²  - [(0.645)(9.8)x] - [(0.645)(9.8)(2.34) = 0

265x² - 6.321x - 14.8 = 0  

a=265

b=-6.321

c=-14.8

By using quadratic eq. formula, we'll have the roots

x= 0.24 or x=-0.225

Considering only positive root:

x= 0.24m (maximum  compression of the spring)

4 0
3 years ago
Ice has a density of 0.920 g/cm3. An ice cubes mass is 100 g. What is its volume?
Tasya [4]

Answer:

\boxed {\tt 108.695652 \ cm^3}

Explanation:

Volume can be found by dividing the mass by the density.

v=\frac{m}{d}

The mass is 100 grams and density is 0.920 grams per cubic centimeters.

Therefore,

m=100 \ g \\d= 0.920 \ g/cm^3

Substitute the values into the formula.

v=\frac{100 \ g}{0.920 \ g/cm^3}

Divide. Note that the grams, or "g" will cancel each other out.

v- 108.695652 \ cm^3

The volume of the ice cube is 108.697652 cubic centimeters.

3 0
3 years ago
What frequency (in Hz) is received by a person watching an oncoming ambulance moving at 116 km/h and emitting a steady 950 Hz so
Arte-miy333 [17]

To solve this problem we will apply the concepts related to the Doppler Effect, defined as the change in apparent frequency of a wave produced by the relative movement of the source with respect to its observer. Mathematically it can be written as

f_{obs} = f(\frac{v_w}{v_w-v_s})

Here,

f_s= Frequency of the source

v_w = Speed of the sound

v_s= Speed of source

Now the velocity we have that

v_s = 116km/h (\frac{1000m}{1km})(\frac{1h}{3600s})

v_s = 32.22m/s

Then replacing our values,

f_{obs} = (950Hz) (\frac{345m/s}{345m/s-32.22m/s})

f_{obs} = 1047.86Hz

Therefore the frequency of the observer is 1047.86Hz

8 0
3 years ago
The spiral spring of a spring balance is 25.0cm long when 5N hangs on it and 30cm long when the weight is 10N , what is the leng
Nutka1998 [239]

Answer:

stay away from my boyfriend.

Explanation:

3 0
4 years ago
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