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3 years ago

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Given the height of rod,length of shadow of tree and length of shadow of the rod, estimate the height of the tree? Given:height

of rod=150cm,Length of shadow of the rod=120cm and length of shadow of the tree =800cm

1 answer:

diamong [38]3 years ago

5 0

Answer:

1000 cm.

Explanation:

To obtain the estimated tree height :

(Height of rod / length of rod shadow) = (height of tree / length of tree shadow)

Substituting values into the formula :

(150cm / 120 cm) = (height of tree / 800 cm)

Using cross multiplication :

Height of tree * 120 = 150 * 800

Height of tree = (150 * 800) / 120

Height of tree = 120,000 / 120

Height of tree = 1000

Hence, estimate height of tree = 1000 cm

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A 2.7-kg block is released from rest and allowed to slide down a frictionless surface and into a spring. The far end of the spri

exis [7]

a) The speed of the block at a height of 0.25 m is 2.38 m/s

b) The compression of the spring is 0.25 m

c) The final height of the block is 0.54 m

Explanation:

a)

We can solve the problem by using the law of conservation of energy. In fact, the total mechanical energy (sum of kinetic+gravitational potential energy) must be conserved in absence of friction. So we can write:

$U_i +K_i = U_f + K_f$

where

$U_i$ is the initial potential energy, at the top

$K_i$ is the initial kinetic energy, at the top

$U_f$ is the final potential energy, at halfway

$K_f$ is the final kinetic energy, at halfway

The equation can be rewritten as

$mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2$

where:

m = 2.7 kg is the mass of the block

$g=9.8 m/s^2$ is the acceleration of gravity

$h_i = 0.54$ is the initial height

u = 0 is the initial speed

$h_f = 0.25 m$ is the final height of the block

v is the final speed when the block is at a height of 0.25 m

Solving for v,

$v=\sqrt{u^2+2g(h_i-h_f)}=\sqrt{0+2(9.8)(0.54-0.25)}=2.38 m/s$

b)

The total mechanical energy of the block can be calculated from the initial conditions, and it is

$E=K_i + U_i = 0 + mgh_i = (2.7)(9.8)(0.54)=14.3 J$

At the bottom of the ramp, the gravitational potential energy has become zero (because the final heigth is zero), and all the energy has been converted into kinetic energy. However, then the block compresses the spring, and the maximum compression of the spring occurs when the block stops: at that moment, all the energy of the block has been converted into elastic potential energy of the spring. So we can write

$E=E_e = \frac{1}{2}kx^2$

where

k = 453 N/m is the spring constant

x is the compression of the spring

And solving for x, we find

$x=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(14.3)}{453}}=0.25 m$

c)

If there is no friction acting on the block, we can apply again the law of conservation of energy. This time, the initial energy is the elastic potential energy stored in the spring:

$E=E_e = 14.3 J$

while the final energy is the energy at the point of maximum height, where all the energy has been converted into gravitational potetial energy:

$E=U_f = mg h_f$

where $h_f$ is the maximum height reached. Solving for this quantity, we find

$h_f = \frac{E}{mg}=\frac{14.3}{(2.7)(9.8)}=0.54 m$

which is the initial height: this is correct, because the total mechanical energy is conserved, so the block must return to its initial position.

Learn more about kinetic and potential energy:

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5 0

3 years ago

Acceleration = change of velocity divided by time interval = Δv/Δt.

MariettaO [177]

Answer:

a=2.378 m/s^2

Explanation:

a=Δv/Δt------eq(1)

Δv=Vf-Vi=120 km/h-0 km/h=120 km/h

or Δv=33.3 m/sec

or time=t=14s

putting values in eq(1)

a=33.3/14

a=2.378 m/s^2

6 0

3 years ago

Explain how the basic unit are combined to give the derived units of force, velocity, pressure and work

LuckyWell [14K]

Velocity:

Velocity is change in displacement with respect to time:

$\frac{\Delta x}{\Delta t}$

Analysing the units, meters (displacement) and seconds (time) are basic units:

$\frac{m}{s}$

Therefore the unit of velocity is m/s

Force:

Newton's second law of motion:

$F = ma$

Kilogram (mass) is a basic unit, and accelerations unit can be found using the equation:

$a=\frac{\Delta v}{\Delta t}$

Analysing the units:

$\frac{\frac{m}{s}}{s}=\frac{m}{s^2}$

Therefore, the unit of force is:

$kg\frac{m}{s^2}$

Pressure:

Pressure is given by the equation:

$P=\frac{F}{S}$ where S is area of effect, F is force

Area for a basic rectangle (geometric shape is arbitrary for dimensional analysis) is found by multiplying two lengths:

$[l^2]=m^2$ , the unit of area

Dividing the aforementioned unit of force by the unit of area:

$\frac{kg\frac{m}{s^2}}{m^2}=\frac{kg}{ms^2}$ , the unit of pressure

Work:

Work is given by the equation:

$W=\vec{F}\cdot \vec{x}$ , (dot product may be assumed as normal multiplication for the purposes of unit analysis)

Knowing displacement's (x) unit is m:

$[W]=\frac{kgm}{s^2}m=\frac{kgm^2}{s^2}$ , the unit of work.

3 0

3 years ago

The arrows in the chart below represent phase transitions.

Vikentia [17]

Answer:

1, 2, and 3.

Explanation:

Hello.

In this process, since the phase transitions that require energy are those that pass from a state with less energy or more molecular order to a state with more energy or less molecular order, say, from solid to liquid (melting), from liquid to gas (boiling) and from solid to gas (sublimation), we can conclude that the arrows representing heat energy gained are 1, 2, and 3 since 1 represents boiling, 2 melting and 3 sublimation.

Best regards.

6 0

3 years ago

Suppose you have two solid bars, both with square cross-sections of 1 cm2. They are both 24.6 cm long, but one is made of copper

vodka [1.7K]

Explanation:

Expression to calculate thermal resistance for iron ( $R_{I}$ ) is as follows.

$R_{I} = \frac{L_{I}}{k_{I} \times A_{I}}$

where, $L_{I}$ = length of the iron bar

$k_{I}$ = thermal conductivity of iron

$A_{I}$ = Area of cross-section for the iron bar

Thermal resistance for copper ( $R_{c}$ ) = \frac{L_{c}}{k_{c} \times A_{c}}[/tex]

where, $L_{c}$ = length of copper bar

$k_{c}$ = thermal conductivity of copper

$A_{c}$ = Area of cross-section for the copper bar

Now, expression for the transfer of heat per unit cell is as follows.

Q = $\frac{(100^{o} - 0^{o}}{\frac{L_{I}}{k_{I}.A_{I}} + \frac{L_{c}}{k_{c}.A_{c}}}$

Putting the given values into the above formula as follows.

Q = $\frac{(100^{o} - 0^{o})}{\frac{L_{I}}{k_{I}.A_{I}} + \frac{L_{c}}{k_{c}.A_{c}}}$

= $\frac{(100^{o} - 0^{o})}{21 \times 10^{-2} m[\frac{1}{73 \times 10^{-4}m^{2}} + \frac{1}{386 \times 10^{-4}m^{2}}}$

= 2.92 Joule

It is known that heat transfer per unit time is equal to the power conducted through the rod. Hence,

P = $\frac{Q}{T}$

Here, T is 1 second so, power conducted is equal to heat transferred.

So, P = 2.92 watt

Thus, we can conclude that 2.92 watt power will be conducted through the rod when it reaches steady state.

7 0

3 years ago