50.0 meters away from a building. Tip of the building makes an angle of 63.0° with the horizontal. What is the height of the bui
1 answer:
Answer:
98.13m
Explanation:
Complete question
Daniel is 50.0 meters away from a building. Tip of the building makes an angle of 63.0° with the horizontal. What is the height of the building
CHECK THE ATTACHMENT
From the figure, using trigonometry
Tan(θ ) = opposite/adjacent
Where Angle (θ )= 63°
Opposite= X = height of the building
Adjacent= 50 m
Then substitute the values we have
Tan(63)= X/50
1.9626= X/50
X= 1.9626 × 50
X= 98.13m
Hence, the height of the building is 98.13m
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Answer:
a) L=0. b) L = 262 k ^ Kg m²/s and c) L = 1020.7 k^ kg m²/s
Explanation:
It is angular momentum given by
L = r x p
Bold are vectors; where L is the angular momentum, r the position of the particle and p its linear momentum
One of the easiest ways to make this vector product is with the use of determinants
Let's apply this relationship to our case
Let's start by breaking down the speed
v₀ₓ = v₀ cosn 45
voy =v₀ sin 45
v₀ₓ = 9 cos 45
voy = 9 without 45
v₀ₓ = 6.36 m / s
voy = 6.36 m / s
a) at launch point r = 0 whereby L = 0
. b) let's find the position for maximum height, we can use kinematics, at this point the vertical speed is zero
vfy² = voy²- 2 g y
y = voy² / 2g
y = (6.36)²/2 9.8
y = 2.06 m
Let's calculate the angular momentum
L=
L = -px y k ^
L = - (m vox) (2.06) k ^
L = - 20 6.36 2.06 k ^
L = 262 k ^ Kg m² / s
The angular momentum is on the z axis
c) At the point of impact, at this point the height is zero and the position on the x-axis is the range
R = vo² sin 2θ / g
R = 9² sin (2 45) /9.8
R = 8.26 m
L =
L = - x py k ^
L = - x m voy
L = - 8.26 20 6.36 k ^
L = 1020.7 k^ kg m² /s
Answer:
A) 1568.60 Hz
B) 1437.15 Hz
Explanation:
This change is frequency happens due to doppler effect
The Doppler effect is the change in frequency of a wave in relation to an observer who is moving relative to the wave source
where
C = the propagation speed of waves in the medium;
Vr= is the speed of the receiver relative to the medium,(added to C, if the receiver is moving towards the source, subtracted if the receiver is moving away from the source;
Vs= the speed of the source relative to the medium, added to C, if the source is moving away from the receiver, subtracted if the source is moving towards the receiver.
A) Here the Source is moving towards the receiver(C-Vs)
and the receiver is standing still (Vr=0) therefore the observed frequency should get higher
B)Here the Source is moving away the receiver(C+Vs)
and the receiver is still not moving (Vr=0) therefore the observed frequency should be lesser
Answer:
S = 122.5m
Explanation:
Given the following data;
Acceleration due to gravity = 9.8m/s²
Time, t = 5 seconds
Since it's a free fall, initial velocity, u = 0
To find the displacement, we would use the second equation of motion given by the formula;
Where;
- S represents the displacement or height measured in meters.
- u represents the initial velocity measured in meters per seconds.
- t represents the time measured in seconds.
- a represents acceleration measured in meters per seconds square.
Substituting into the equation, we have;
S = 122.5m.