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LiRa [457]
3 years ago
10

an attempt to estimate the height of a tree the Shadow of an upright metre rule was found to be 25 cm and the length of the Shad

ow of the tree was 7 m what is the height of the tree​
Physics
1 answer:
labwork [276]3 years ago
3 0

Answer:

The actual height of the tree is 28 m

Explanation:

The given information are;

The length of the shadow of an upright meter rule = 25 cm

The actual height of the meter rule = 100 cm

The length of the shadow of the tree = 7 m

The actual  height of the tree  = h

We have

\dfrac{The \ length \ of \ the \ shadow \ of  \ an \ upright \ metre \ rule}{The \  actual \ height \ of \  the \ metre \ rule} = \dfrac{The \ length \ of \ the \ shadow \ of  \ the \ tree}{The \  actual \ height \ of \  the \ tree}Which gives;

\dfrac{25 \ cm}{100 \ cm} = \dfrac{7 \ m}{The \  actual \ height \ of \  the \ tree}

Therefore;

The \  actual \ height \ of \  the \ tree = 7 \ m \times \dfrac{100 \ cm}{25\ cm} = 7 \ m \times 4 = 28 \ m

That is the actual height of the tree = 28 m.

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A student wearing a frictionless roller skates on a horizontal is pushed by a friend with a constant force of 55N. How far must
umka21 [38]

Answer:

6.58m

Explanation:

The kinetic energy = Workdone on the roller

Workdone = Force * distance

Given

KE = Workdone = 362J

Force = 55N

Required

Distance

Substitute into the formula;

Workdone = Force * distance

362 = 55d

d = 362/55

d = 6.58m

Hence the student must push at a distance of 6.58m

3 0
3 years ago
An object attached to one end of a spring makes 20 vibrations in 10 seconds. Its angular frequency is: 0. 79 rad/s 1. 57 rad/s 2
morpeh [17]

Angular frequency in radian per second for 20 vibrations in 10 seconds is 12.6 rad/s

<h3>What is Angular frequency?</h3>

Angular frequency is the number of vibrations in radian per second.

The total number of vibrations n is 20 and the time taken for these vibrations is 10 s

The frequency of the vibrations will be

f = 20 / 10 = 2 Hz

Angular frequency is related to the frequency as

ω = 2πf

ω=2π × 2

ω = 12.6 rad/s

Thus, the angular frequency is 12.6 rad/s.

Learn more about Angular frequency.

brainly.com/question/14244057

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5 0
3 years ago
The density of a block of wood is 694 kg/m3. Its mass is 689 g. We tie the block to the bottom of a swimming pool using a single
Serhud [2]

Answer:

<em>The tension in the string = 2.065 N</em>

Explanation:

From Archimedes principle,

R.d = density of the wood block/density of water = weight of the wood block/Upthrust of the wood block in water.

R.d = D₁/D₂ = W/U

W/U =D₁/D₂.................................. Equation 1

Where W = weight of the wood block, U = upthrust of the wood block in water, D₁ = Density of the wood block, D₂ = Density of water.

Making U the subject of the equation,

U = WD₂/D₁........................... Equation 2

Given: W =  mg = (689/1000)9.8 = 6.75 N,  D₁ = 694 kg/m³, D₂ = 1000 kg/m³.

Substituting these values into equation 2,

U = 6.75(694)/1000

U = 4684.5/1000

U = 4.685 N.

Note: Three forces act on the wood block in the pool. and they are

(i) The weight(W) acting downs

(ii) The upthrust (U) acting upwards,

(iii) The Tension (T) in the string, acting upwards.

Thus,

W = U+T

T = W - U ................................. Equation 3

Where W = 6.75 N, U = 4.685 N

T = 6.75 - 4.685

T = 2.065 N.

T = 2.065 N

<em>Thus the tension in the string = 2.065 N</em>

7 0
3 years ago
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