Answer:
34 m/s
Explanation:
Potential energy at top = kinetic energy at bottom + work done by friction
PE = KE + W
mgh = ½ mv² + Fd
mg (d sin θ) = ½ mv² + Fd
Solving for v:
½ mv² = mg (d sin θ) − Fd
mv² = 2mg (d sin θ) − 2Fd
v² = 2g (d sin θ) − 2Fd/m
v = √(2g (d sin θ) − 2Fd/m)
Given g = 9.8 m/s², d = 150 m, θ = 28°, F = 50 N, and m = 65 kg:
v = √(2 (9.8 m/s²) (150 m sin 28°) − 2 (50 N) (150 m) / (65 kg))
v = 33.9 m/s
Rounded to two significant figures, her velocity at the bottom of the hill is 34 m/s.
C because when the part gets out of the probe it would no longer stay contacted
Answer: Radiation
Explanation: Radiation is the energy that comes from a source in form of electromagnetic waves, subatomic particles, light, or heat which travels through space.
Examples of radiation include the light, heat, and particles emitted from the Sun.
Using a foil barrier to prevent heat transfer is possible because foil has a silver color, and silver reflects light and heat instead of absorbing them. This is the opposite of black surfaces that absorb heat.
So in homes where these foil reflective barriers are used, the transfer of heat through Radiation is highly reduced.
Answer:
a) 19.4 m/s
b) 19 m/s
Explanation:
a) In the given question,
the potential energy at the initial point = Ui = 0
the potential energy at the final point = Uf = mgh
the kinetic energy at the initial point = Ki = 1/2 mv₀².
the kinetic energy at the final point = Kf = 0
work done by air= Ea= fh = 0.262 N
Now, using the law of conservation of energy
initial energy= final energy
Ki +Ui = Kf + Uf +Ea
1/2 mv₀² + 0 = 0 + mgh + fh
1/2 mv₀² = mgh + fh
h = v₀²/ 2g (1 +f/w)
calculate m
m= w/g = 5.29 /9.8
= 0.54 kg
h = 20 ²/ (2 x9.80) x (1 0.265/5.29)
h = 19.4 m.
b) 1/2 mv² + 2fh = 1/2 mv₀²
Vg = 19 m/s
Answer:
Explanation:
Given data
Time t=2.5 minutes=150 seconds
Distance A=1600 ft=487.68 m........east
Distance B=2500 ft=762m ........north
To find
Average velocity
Solution
First we need to find the resultant distance magnitude.To find that we apply Pythagorean theorem to find hypotenuse
So
