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Klio2033 [76]
3 years ago
5

PLEASE HELP! Daniel is 50.0 meters away from a building. He observes that his line-of-sight to the tip of the building makes an

angle of 63.0° with the
horizontal. What is the height of the building?
A. 174 m
B. 110 m
C. 98 m
D. 50 m

Physics
1 answer:
zavuch27 [327]3 years ago
5 0

Answer:

The height of building should be 98.13 m plus the height of Daniel. Since the 63° was measured from his eye level.

Explanation:

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A 1000 kg satellite and a 2000 kg satellite follow exactly the same orbit around the earth. What is the ratio F1/F2 of the gravi
Tamiku [17]

Answer:

the <em>ratio F1/F2 = 1/2</em>

the <em>ratio a1/a2 = 1</em>

Explanation:

The force that both satellites experience is:

F1 = G M_e m1 / r²       and

F2 = G M_e m2 / r²

where

  • m1 is the mass of satellite 1
  • m2 is the mass of satellite 2
  • r is the orbital radius
  • M_e is the mass of Earth

Therefore,

F1/F2 = [G M_e m1 / r²] / [G M_e m2 / r²]

F1/F2 = [G M_e m1 / r²] × [r² / G M_e m2]

F1/F2 = m1/m2

F1/F2 = 1000/2000

<em>F1/F2 = 1/2</em>

The other force that the two satellites experience is the centripetal force. Therefore,

F1c = m1 v² / r    and

F2c = m2 v² / r

where

  • m1 is the mass of satellite 1
  • m2 is the mass of satellite 2
  • v is the orbital velocity
  • r is the orbital velocity

Thus,

a1 = v² / r ⇒ v² = r a1    and

a2 = v² / r ⇒ v² = r a2

Therefore,

F1c = m1 a1 r / r = m1 a1

F2c = m2 a2 r / r = m2 a2

In order for the satellites to stay in orbit, the gravitational force must equal the centripetal force. Thus,

F1 = F1c

G M_e m1 / r² = m1 a1

a1 = G M_e / r²

also

a2 = G M_e / r²

Thus,

a1/a2 = [G M_e / r²] / [G M_e / r²]

<em>a1/a2 = 1</em>

4 0
3 years ago
A 5.50 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 3.20 m by a force of 25.
kiruha [24]

(a) 69.3 J

The work done by the applied force is given by:

W=Fd cos \theta

where:

F = 25.0 N is the magnitude of the applied force

d = 3.20 m is the displacement of the sled

\theta=30^{\circ} is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J

(b) 0

The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

\Delta K = W

where

\Delta K is the change in kinetic energy

W is the work done

Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

\Delta K=69.3 J

(d) 4.9 m/s

The change in kinetic energy of the sled can be rewritten as:

\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2 (1)

where

Kf is the final kinetic energy

Ki is the initial kinetic energy

m = 5.50 kg is the mass of the sled

u = 0 is the initial speed of the sled

v = ? is the final speed of the sled

We can calculate the variation of kinetic energy of the sled, \Delta K, after it has travelled for d=3 m. Using the work-energy theorem again, we find

\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J

And substituting into (1) and re-arrangin the equation, we find

v=\sqrt{\frac{2 \Delta K}{m}}=\sqrt{\frac{2(65.0 J)}{5.50 kg}}=4.9 m/s

6 0
3 years ago
Sorin hits a cricket ball straight up into the air. After 1.5\,\text{s}1.5s1, point, 5, start text, s, end text, the ball is fal
Illusion [34]

Answer:

8.2 m/s upward

Explanation:

The motion of the ball is a free fall motion, so we can use the suvat equation:

v=u+at

where:

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

Choosing upward as positive direction:

v=-6.5 m/s

a=g=-9.8 m/s^2 (acceleration of gravity)

t = 1.5 s

Solving for u, we find the initial velocity:

u=v-at=-6.5-(-9.8)(1.5)=8.2 m/s

And the direction is upward.

3 0
3 years ago
A boy on a bicycle approaches a brick wall as he sounds his horn at a frequency 400 hz. the sound he hears reflected back from t
Mashutka [201]
As the question is about changing in frequency of a wave for an observer who is moving relative to the wave source, the concept that should come to our minds is "Doppler's effect."

Now the general formula of the Doppler's effect is:
f = (\frac{g + v_{r}}{g + v_{s}})f_{o} -- (A)

Note: We do not need to worry about the signs, as everything is moving towards each other. If something/somebody were moving away, we would have the negative sign. However, in this problem it is not the issue.

Where,
g = Speed of sound = 340m/s.
v_{r} = Velocity of the receiver/observer relative to the medium = ?.
v_{s} = Velocity of the source with respect to medium = 0 m/s.
f_{o} =  Frequency emitted from source = 400 Hz.
f = Observed frequency = 408Hz.

Plug-in the above values in the equation (A), you would get:

408 = ( \frac{340 + v_{r}}{340 + 0})*400

\frac{408}{400} =  \frac{340 + v_{r}}{340}

Solving above would give you,
v_{r} = 6.8 m/s

The correct answer = 6.8m/s



7 0
3 years ago
Two forces and are applied to an object whose mass is 11.8 kg. The larger force is . When both forces point due east, the object
S_A_V [24]

Can you please fill in whatever goes in the blanks ?

Without them, the question makes no sense and has no answer.

Two forces (___) and (___) are applied to an object whose mass is 11.8 kg. The larger force is (___) . When both forces point due east, the object's acceleration has a magnitude of 0.408 m/s2. However, when (___) points due east and (___) points due west, the acceleration is 0.227 m/s2, due east. Find (a) the magnitude of (___) and (b) the magnitude of (___) .

6 0
3 years ago
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