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2 years ago

13

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]2 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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Explain how thermal energy (temperature) affects chemical changes.

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3 years ago

What is the force of gravity (from the Earth) on the 700kg satellite if it’s 10km above the Earths surface?

joja [24]

Answer:

The force of gravity on a 700 kg satellite if its 10 km above Earth's surface is given by

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Explanation:

The force of gravity on a 700 kg satellite if its 10 km above Earth's surface is given by

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2 years ago

A sealed tank containing seawater to a height of 12.8 m also contains air above the water at a gauge pressure of 2.90 atm . Wate

exis [7]

Answer:

The velocity of water at the bottom, $v_{b} = 28.63 m/s$

Given:

Height of water in the tank, h = 12.8 m

Gauge pressure of water, $P_{gauge} = 2.90 atm$

Solution:

Now,

Atmospheric pressue, $P_{atm} = 1 atm = 1.01\tiems 10^{5} Pa$

At the top, the absolute pressure, $P_{t} = P_{gauge} + P_{atm} = 2.90 + 1 = 3.90 atm = 3.94\times 10^{5} Pa$

Now, the pressure at the bottom will be equal to the atmopheric pressure, $P_{b} = 1 atm = 1.01\times 10^{5} Pa$

The velocity at the top, $v_{top} = 0 m/s$ , l;et the bottom velocity, be $v_{b}$ .

Now, by Bernoulli's eqn:

$P_{t} + \frac{1}{2}\rho v_{t}^{2} + \rho g h_{t} = P_{b} + \frac{1}{2}\rho v_{b}^{2} + \rho g h_{b}$

where

$h_{t} - h_{b} = 12.8 m$

Density of sea water, $\rho = 1030 kg/m^{3}$

$\sqrt{\frac{2(P_{t} - P_{b} + \rho g(h_{t} - h_{b}))}{\rho}} = v_{b}$

$\sqrt{\frac{2(3.94\times 10^{5} - 1.01\times 10^{5} + 1030\times 9.8\times 12.8}{1030}} = v_{b}$

$v_{b} = 28.63 m/s$

5 0

3 years ago

A spring with a spring constant of 120 J/m2 is fixed to a wall, free to oscillate. On the other end, a ball with a mass of 1500

Neporo4naja [7]

Answer:

A. 4.47 m/s

Explanation:

As the ball oscillates, it mechanical energy, aka the total kinetic and elastics energy stays the same. For the ball to be at maximum speed, its elastic energy i 0 and vice versa. When the ball is at rest, its kinetic energy is 0 and its elastic energy is at maximum at 50 cm, or 0.5 m

1500 g = 1.5 kg

$E_e = E_k$

$kx^2/2 = mv^2/2$

$120*0.5^2/2 = 1.5*v^2/2$

$15 = 0.75v^2$

$v^2 = 15 / 0.75 = 20$

$v = \sqrt{20} = 4.47 m/s$

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3 years ago