All categories

3 years ago

The sun is 27 degree above the horizon. It makes a 60 m long shadow of a tall tree.

Physics

1 answer:

mestny [16]3 years ago

5 0

Answer:

h = 30.57 m

Explanation:

given,

angle made with horizon, θ= 27°

Length of the shadow,L = 60 m

now,

the height of the tree would be equal to

$tan \theta = \dfrac{height\ of tree}{shadow\ length}$

$tan 27^0 = \dfrac{h}{60}$

h = 60 x tan 27°

h = 30.57 m

hence, height of the tree is equal to 30.57 m

You might be interested in

What is the eccentricity of an ellipse with a foci distance of 50,000,000 km and

inysia [295]

Answer:

25,000,000 Km ;)

Explanation:

5 0

3 years ago

A man and his dog are “walking” on flat Street. He is pulling on his stubborn dog with a force of 70 N Directed at a 30° angle f

Delvig [45]

Answer:

X component of force is 60.62 N.

Y component of force is 35 N.

Force of gravity on the dog is 245 N.

Magnitude of normal force is 210 N.

Explanation:

Given:

Force of pull is, $F=70\ N$

Mass of the dog is, $m=25\ kg$

Angle of inclination is, $\theta =30$ °

Acceleration due to gravity is, $g=9.8\ m/s^2$

The free body diagram of the dog is shown below.

The X and Y components of force of pull is given as:

$F_X=F\cos \theta=(70)\cos (30)=60.62\ N\\F_Y=F\sin \theta=(70)\sin(30)=35\ N$

Therefore, the X and Y components of the force are 60.62 N and 35 N respectively.

Force of gravity on the dog is the product of its mass and acceleration due to gravity. Thus,

$F_g=mg=25\times 9.8=245\ N$

Therefore, the force of gravity on the dog is 245 N.

Now, consider the motion in the vertical direction of the dog. As there is no motion in the vertical direction, the net force along the Y direction is 0. In other words, the total Upward force is equal to the total downward force.

From the free body diagram,

$N+F_Y=mg\\N=mg-F_Y\\N=245-35=210\ N$

Therefore, the normal force acting on the dog is 210 N.

6 0

3 years ago

Why did Kenyatta want to distance himself from the Mau Mau? Check all that apply.

Savatey [412]

Answer:

B and C!

Explanation:

I just did it on edge. I hope this helps!

3 0

3 years ago

Near the equator, the Earth's magnetic field points almost horizontally to the north and has magnitude B=.5 x 10^-4T. What shoul

Nataly [62]

Base in your question about the magnetic field of the Earth near the equator where as its almost horizontally to the north and has magnitude of B=0.5x10^-4t, the answer is <span>Velocity of electron will be westwards.</span>

6 0

3 years ago

7. A block of copper of unknown mass has an initial temperature of 65.4oC. The copper is immersed in a beaker containing 95.7g o

dolphi86 [110]

Answer:

37.34372 kg

Explanation:

m = Mass

$\Delta T$ = Change in temperature

1 denotes water

2 denotes copper

c = Heat capacity

Heat is given by

$Q=mc\Delta T$

In this case the heat transfer will be equal

$m_1c_1\Delta T_1=m_2c_2\Delta T_2\\\Rightarrow m_2=\frac{m_1c_1\Delta T_1}{c_2\Delta T_2}\\\Rightarrow m_2=\frac{95.7\times 4.18(24.2-22.7)}{0.39(65.4-24.2)}\\\Rightarrow m_2=37.34372\ kg$

Mass of copper block is 37.34372 kg

5 0

3 years ago