In a moving car the outside looks to be moving. however if viewed from the outside, the car appears to be moving. so motion is relative to the person observing.
Answer:
xf = 5.68 × 10³ m
yf = 8.57 × 10³ m
Explanation:
given data
vi = 290 m/s
θ = 57.0°
t = 36.0 s
solution
firsa we get here origin (0,0) to where the shell is launched
xi = 0 yi = 0
xf = ? yf = ?
vxi = vicosθ vyi = visinθ
ax = 0 ay = −9.8 m/s
now we solve x motion: that is
xf = xi + vxi × t + 0.5 × ax × t² ............1
simplfy it we get
xf = 0 + vicosθ × t + 0
put here value and we get
xf = 0 + (290 m/s) cos(57) (36.0 s)
xf = 5.68 × 10³ m
and
now we solve for y motion: that is
yf = yi + vyi × t + 0.5 × ay × t
² ............2
put here value and we get
yf = 0 + (290 m/s) × sin(57) × (36.0 s) + 0.5 × (−9.8 m/s2) × (36.0 s) ²
yf = 8.57 × 10³ m
Answer:
The options are
A.on racetracks
B.in real-world conditions
C.in flooded environments
D.on closed courses
The answer is B. In real world conditions
The public is not yet able to purchase cars powered by hydrogen fuel cells because engineers have to determin
how the cars perform based on real world conditions.
This will ensure they encounter the real and first hand experiences about the challenges and also the advantages associated with using this type of fuel.
<h3>The answer is 0.89 L</h3>
Explanation:
The new volume can be found by using the formula for Boyle's law which is
Since we are finding the new volume
From the question we have
We have the final answer as
<h3>0.89 L</h3>
Hope this helps you
Answer:
B. 1700 Hz, 5100 Hz
Explanation:
Parameters given:
Length of ear canal = 5.2cm = 0.052 m
Speed of sound in warm air = 350 m/s
The ear canal is analogous to a tube that has one open end and one closed end. The frequency of standing wave modes in such a tube is given as:
f(m) = m * (v/4L)
Where m is an odd integer;
v = velocity
L = length of the tube
Hence, the two lowest frequencies at which a dog will have increased sensitivity are f(1) and f(3).
f(1) = 1 * [350/(4*0.052)]
f(1) = 1682.69 Hz
Approximately, f(1) = 1700 Hz
f(3) = 3 * [350/(4*0.052)]
f(3) = 5048 Hz
Approximately, f(3) = 5100 Hz
