She can climb 0.92 m without losing weight.
<u>Explanation</u>:
Gravitational potential energy is the energy consisting of the product of mass, gravity and height.
1 cal = 4184 J
140 cal = 585760 J
Energy = 585760 J, m = 65.0 kg = 65000 g, Efficiency = 20 %
GPE = mgh
where m represents the mass
g represents the gravity,
h represents the height.
585760 = 65000 
9.8 h
h = 0.92 m.
At the entrance of most beaches, there is a bulletin board with notices about water conditions: maybe a faded sign warning about rip currents and a list of this week's tide tables. Most people pass them by without a second thought, but if you want to enter the ocean, it is important to know its movements, whether to avoid being caught in a riptide or to figure out when the waves will be at their best.
Hope this helps
Answer:
304.89m
Explanation:
Given
acceleration a = 2.52m/s²
final speed v = 39.2m/s
initial speed = 0m/s (car accelerates from rest)
Using the equation of motion below to get the distance of Doc brown from Marty;
v² = u²+2as
substitute the given parameters
39.2² = 0²+2(2.52)s
1536.64 = 0+5.04s
divide both sides by 5.04
1536.64/5.04 = 5.04s/5.04
rearrange the equation
5.04s/5.04 = 1536.64/5.04
s = 304.89m
Hence He and Marty must stand at 304.89m to allow the car to accelerate from rest to a speed of 39.2 m/s?
The banking angle of the curved part of the speedway is determined as 32⁰.
<h3>
Banking angle of the curved road</h3>
The banking angle of the curved part of the speedway is calculated as follows;
V(max) = √(rg tanθ)
where;
- r is radius of the path
- g is acceleration due to gravity
V² = rg tanθ
tanθ = V²/rg
tanθ = (34²)/(190 x 9.8)
tanθ = 0.62
θ = arc tan(0.62)
θ = 31.8
θ ≈ 32⁰
Learn more about banking angle here: brainly.com/question/8169892
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