I believe the answer here is <span>C).seedless, seed (nonflowering), and seed (flowering).</span><span>
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Answer:
<em>600N.</em>
Explanation:
From the question, we are to calculate the net force acting on the car.
According to Newton's second law of motion:
F = ma
m is the mass of the car
a is the acceleration = change in velocity/Time
a = v-u/t
F = m(v-u)/t
v is the final velocity = 30m/s
u is the initial velocity = 20m/s
t is the time = 5secs
m = 300kg
Get the net force:
Recall that: F = m(v-u)/t
F = 300(30-20)/5
F = 60(30-20)
F = 60(10)
<em>F = 600N</em>
<em>Hence the net force acting on the car is 600N.</em>
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Explanation:
Given:
v₀ = 0 m/s
a = 3 m/s²
t = 4 s
Find: Δx and v
Δx = v₀ t + ½ at²
Δx = (0 m/s) (4 s) + ½ (3 m/s²) (4 s)²
Δx = 24 m
v = at + v₀
v = (3 m/s²) (4 s) + 0 m/s
v = 12 m/s
Answer:
the time interval that an earth observer measures is 4 seconds
Explanation:
Given the data in the question;
speed of the spacecraft as it moves past the is 0.6 times the speed of light
we know that speed of light c = 3 × 10⁸ m/s
so speed of spacecraft v = 0.6 × c = 0.6c
time interval between ticks of the spacecraft clock Δt₀ = 3.2 seconds
Now, from time dilation;
t = Δt₀ / √( 1 - ( v² / c² ) )
t = Δt₀ / √( 1 - ( v/c )² )
we substitute
t = 3.2 / √( 1 - ( 0.6c / c )² )
t = 3.2 / √( 1 - ( 0.6 )² )
t = 3.2 / √( 1 - 0.36 )
t = 3.2 / √0.64
t = 3.2 / 0.8
t = 4 seconds
Therefore, the time interval that an earth observer measures is 4 seconds
Answer:
20 m/s westward
Explanation:
Taking eastward as positive direction, we have:
is the velocity of Bill with respect to Amy (which is stationary)
is the velocity of Carlos with respect to Amy
Bill is moving 5 m/s eastward compared to Amy at rest, so the velocity of Bill's reference frame is
Therefore, Carlos velocity in Bill's reference frame will be
and the direction will be westward (negative sign).
