Answer:
<em>When the speed is doubled, K = 100 J, when the speed is tripled, K = 225 J</em>
Explanation:
<u>Kinetic Energy
</u>
Is the type of energy an object has due to its speed. It's proportional to the square of the speed.
The equation for the kinetic energy is:

Where:
m = mass of the object
v = speed at which the object moves
The kinetic energy is expressed in Joules (J)
The object has a kinetic energy of K=25 J when moving at v=5 m/s, thus the mass can be calculated by solving for m:


m = 2 Kg
If the speed is doubled, v=10 m/s, the new kinetic energy is:

K = 100 J
If the speed is tripled, v=15 m/s, the new kinetic energy is:

K = 225 J
When the speed is doubled, K = 100 J, when the speed is tripled, K = 225 J
To develop this problem it is necessary to apply the concepts related to Wavelength, The relationship between speed, voltage and linear density as well as frequency. By definition the speed as a function of the tension and the linear density is given by

Where,
T = Tension
Linear density
Our data are given by
Tension , T = 70 N
Linear density , 
Amplitude , A = 7 cm = 0.07 m
Period , t = 0.35 s
Replacing our values,



Speed can also be expressed as

Re-arrange to find \lambda

Where,
f = Frequency,
Which is also described in function of the Period as,



Therefore replacing to find 


Therefore the wavelength of the waves created in the string is 3.49m
Answer:
Fc=5253
N
Explanation:
Answer:
Fc=5253
N
Explanation:
sequel to the question given, this question would have taken precedence:
"The 86.0 kg pilot does not want the centripetal acceleration to exceed 6.23 times free-fall acceleration. a) Find the minimum radius of the plane’s path. Answer in units of m."
so we derive centripetal acceleration first
ac (centripetal acceleration) = v^2/r
make r the subject of the equation
r= v^2/ac
ac is 6.23*g which is 9.81
v is 101m/s
substituing the parameters into the equation, to get the radius
(101^2)/(6.23*9.81) = 167m
Now for part
( b) there are two forces namely, the centripetal and the weight of the pilot, but the seat is exerting the same force back due to newtons third law.
he net force that maintains circular motion exerted on the pilot by the seat belts, the friction against the seat, and so forth is the centripetal force.
Fc (Centripetal Force) = m*v^2/r
So (86kg* 101^2)/(167) =
Fc=5253
N
Answer:
Explanation:
Given that,
Mass of star M(star) = 1.99×10^30kg
Gravitational constant G
G = 6.67×10^−11 N⋅m²/kg²
Diameter d = 25km
d = 25,000m
R = d/2 = 25,000/2
R = 12,500m
Weight w = 690N
Then, the person mass which is constant can be determined using
W =mg
m = W/g
m = 690/9.81
m = 70.34kg
The acceleration due to gravity on the surface of the neutron star is can be determined using
g(star) = GM(star)/R²
g(star) = 6.67×10^-11 × 1.99×10^30 / 12500²
g (star) = 8.49 × 10¹¹ m/s²
Then, the person weight on neutron star is
W = mg
Mass is constant, m = 70.34kg
W = 70.34 × 8.49 × 10¹¹
W = 5.98 × 10¹³ N
The weight of the person on neutron star is 5.98 × 10¹³ N