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V125BC [204]
3 years ago
13

At a given time of​ day, the ratio of the height of an object to the length of its shadow is the same for all objects. If a 4​-f

t stick in the ground casts a shadow of 2.8 ​ft, find the height of a tree that casts a shadow that is 24.64 ft.
Physics
1 answer:
katrin2010 [14]3 years ago
8 0

Answer:35.2 ft

Explanation:

Given

height of stick =4 ft

shadow length =2.8 ft

Angle of elevation of sun is

tan\theta =\frac{4}{2.8}

let the height of tree be h

as \thetawill remain same thus

tan\theta =\frac{h}{24.64}

\frac{4}{2.8}=\frac{h}{24.64}

h=35.2 ft

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1. The Moon's mass is 7.34 x 1022 kg, and it is 3.8 x 105 km away from Earth. Earth's
sineoko [7]

Answer:

2.03 x 10²⁴N

Explanation:

Given parameters:

Mass of moon = 7.34 x 10²²kg

Mass of the earth  = 5.97 x 10²⁴kg

Distance  = 3.8 x 10⁵km

Unknown:

Gravitational force of attraction  = ?

Solution:

To find the gravitational force of attraction between the masses, we use the expression below;

   F = \frac{Gm_{1} m_{2}  }{r^{2} }

G is the universal gravitation constant

m is the mass

1 and 2 represents moon and earth

r is  the distance

  F = \frac{6.67 x 10^{-11}  x 7.34 x 10^{22} x 5.97 x 10^{24}  }{(3.8 x 10^{5})^{2}  }

 F = \frac{2.92 x 10^{35} }{1.44 x 10^{11} }  = 2.03 x 10²⁴N

8 0
3 years ago
A physics student throws a softball straight up into the air. The ball was in the air for a total of 3.56 s before it was caught
meriva

Answer:

The initial velocity of the softball is 14.711 meters per second.

Explanation:

This is a case of an object which experiments a free fall, that is, an uniform accelerated motion due to gravity and in which effects from air friction and Earth's rotation can be neglected.

From statement we must understand that the student threw the softball upwards and it is caught at original position 3.56 seconds later. Initial and final heights, time and gravitational acceleration are known and initial speed is unknown. The following equation of motion is used:

y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2} (Eq. 1)

Where:

y_{o} - Initial height of the softball, measured in meters.

y - Final height of the softball, measured in meters.

v_{o} - Initial velocity of the softball, measured in meters per second.

t - Time, measured in seconds.

g - Gravitational acceleration, measured in meters per square second.

If we know that y = y_{o}, t = 3.56\,s and g = -9.807\,\frac{m}{s^{2}}, the initial velocity of the softball is:

v_{o}\cdot (3\,s)+\frac{1}{2}\cdot (-9.807\,\frac{m}{s^{2}} )\cdot (3\,s)^{2} = 0

3\cdot v_{o} -44.132\,m= 0

v_{o} = 14.711\,\frac{m}{s}

The initial velocity of the softball is 14.711 meters per second.

8 0
3 years ago
Someone help me please
ivanzaharov [21]
C. cooked noodles and water
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6 0
3 years ago
Compute the work performed when 32 pounds is lifted 10 feet.
Murljashka [212]
W = force * displacement
W = 32 pounds * 10 feet
Now you need to convert it to newton and meters
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(I approximated the conversions- I hope it helps)
7 0
3 years ago
1. Which of the following is not included
Harrizon [31]

Answer:

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Explanation:

8 0
3 years ago
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