Jenny's acceleration was negative during the intervals of time when her speed was decreasing.
That's really all we can tell, because you really haven't given us any information except her name.
A 60.0 kg secretary running up a 4.0 m tall flight of stairs in 4.2 s has an average power of 560 W (Option b).
<h3>What is power?</h3>
Power is the work done over a period of time.
A secretary with a mass (m) of 60.0 kg runs up a 4.0 m (d) tall flight of stairs. Given gravity (g) is 9.81 m/s², the work (W) done is:
W = m × g × d = 60.0 kg × 9.81 m/s² × 4.0 m = 2.35 × 10³ J
They do 2.4 × 10³ J of work in 4.2 s (t). The average power (P) is:
P = W / t = 2.35 × 10³ J / 4.2 s = 560 W
A 60.0 kg secretary running up a 4.0 m tall flight of stairs in 4.2 s has an average power of 560 W (Option b).
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When hard stabilization structures such as groins are used to stabilize a shoreline, the change in the longshore current results <u>deposition of sediment. </u>
On the upcurrent side of the barrier, sediment is deposited as the longshore current slows.
What is Hard stabilization?
- Hard stabilization is the prevention of erosion through the use of artificial barriers.
- Other hard stabilization structures, such as breakwaters and seawalls, are built parallel to the beach to protect the coast from the force of waves.
- Hard stabilization structures, such as groins, are built at right angles to the shore to prevent the movement of sand down the coast and maintain the beach.
- These constructions are made to last for many years, but because they detract from the visual splendor of the beach, they are not always the ideal answer.
- Additionally, they affect the habitats and breeding sites of native shoreline species, interfering with the ecosystem's natural processes.
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Answer:
Explanation:
The question relates to motion on a circular path .
Let the radius of the circular path be R .
The centripetal force for circular motion is provided by frictional force
frictional force is equal to μmg , where μ is coefficient of friction and mg is weight
Equating cenrtipetal force and frictionl force in the case of car A
mv² / R = μmg
R = v² /μg
= 26.8 x 26.8 / .335 x 9.8
= 218.77 m
In case of moton of car B
mv² / R = μmg
v² = μRg
= .683 x 218.77x 9.8
= 1464.35
v = 38.26 m /s .
