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katen-ka-za [31]
3 years ago
7

An acorn falls from a tree and hits the ground in 0.8 s. How far did the acorn fall . Use g = 9.8 m/s^2. Round your final result

to the nearest tenth.
Physics
1 answer:
mylen [45]3 years ago
8 0

The distance covered by the acorn is 3.136 m.

<u>Explanation:</u>

The time taken for the acorn to hit the ground is 0.8 s. As it is a free fall, the acorn will be completely under the influence of gravity. So the acceleration will be acceleration due to gravity.

Then using the second law of equation,

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

Since the initial velocity and time is zero, then the time taken to reach the ground is stated as 0.8 s, so

   s=0+\left(\frac{1}{2} \times 9.8 \times 0.8 \times 0.8\right)=\frac{6.272}{2}=3.136 \mathrm{m}

So the distance covered by the acorn is 3.136 m.

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What is the reason eclipses do not happen every full moon and new moon
ioda
The answer is D, hope this helped!
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3 years ago
A guitar string with a linear density of 2.0 g/m is stretched between supports that are 60 cm apart. The string is observed to f
Butoxors [25]

<h2>The frequency of driver is 700 Hz</h2>

Explanation:

The frequency of wave in a string is given by the relation

n = \frac{p}{2l} \sqrt{\frac{T}{m} }

here n is the frequency

p is the number of antinodes and l is the length of string .

T is the tension in string and m is the mass per unit length

Thus 420 =\frac{3}{120} \sqrt{\frac{T}{2} }            I

Now if there is 5 antinodes , the value of p = 5

Thus n = \frac{5}{120} \sqrt{\frac{T}{2} }          II

Dividing II by I , we have

n/420 = 5/3

or n = 5/3 x 420 = 700 Hz

3 0
3 years ago
A circular grill of diameter 0.25 m has an emissivity of 0.8. If the surface temperature is maintained at 150°C, determine the r
Schach [20]

Answer:

the required electrical power when the room air and surroundings are at 30°C.= 52.51822 Watt

Explanation:

Power required to maintain the surface temperature at 150°C from 20°C

P= εσA(T^4-t^4)

P= power in watt

ε= emissivity

A=  area of surface

T= 150°C= 423 K

t= 20°C= 303K

/sigma= 5.67×10^{-8} watt/m^2K^4

putting vales we get

= 0.8\times5.67\times10^{-8} \pi\frac{0.25^4}{4}(423^4-303^4)

P=52.51822 Watt

the required electrical power when the room air and surroundings are at 30°C.= 52.51822 Watt

4 0
3 years ago
As an object slides across a rough horizontal
tiny-mole [99]

When the object slides across the rough surface some of its potential energy will be lost to friction.

<h3>Conservation of mechanical energy</h3>

The law of conservation of mechanical energy states that the total mechanical energy of an isolated system is always constant.

M.A = P.E + K.E

When the object slides across the rough surface, some of the potential energy of the object will be converted into kinetic energy while the remaining potential energy will be converted into thermal energy due to frictional force of the rough surface.

P.E = K.E + thermal energy

Learn more about conservation of energy here: brainly.com/question/166559

6 0
2 years ago
An object is placed 100 cm in front of a diverging lens of focal length -25cm. A converging lens of focal length 33 1/3 cm is pl
bagirrra123 [75]
We use 1/o + 1/i = 1/f  where o is the distance of the object, i as distance of the image and f is the focal length.
Substituting, <span>1/ 100 + 1 / i = - 1 /25 </span>
<span>i = - 20 cm </span>

<span>For the case of the problem,</span>

<span>o = (20 + 30)  = 50 cm </span>

<span>f = 33.33. </span>Using 1<span> / i + 1 / o = 1/f , </span><span> </span><span>i = 100 cm </span>

<span>M = magnification = - i / o </span>

<span>m1 = -(-20)/100 = 20/100 = 0.2 </span>

<span>m2 = -100/50 = -2 </span>

<span>M = m1*m2 = -2 x 0.2 = -0.4.</span>
8 0
3 years ago
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