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sineoko [7]
3 years ago
15

A frog falls from its rainforest tree. If we ignore wind resistance, (a) how much time does it take the frog to fall a distance

of 12.0 m? (b) how fast is the frog falling at this point?
Physics
1 answer:
gtnhenbr [62]3 years ago
5 0

Answer:

Explanation:

a) Using the equation of motion

S = ut + 1/2gt²

S is the distance of fall

g is the acceleration due to gravity

t is the time taken

Given S = 12.0m, g = 9.81m/s^2, un= 0m/s

12 = 0+1/2(9.81)t²

12 = 4.905t²²²

t² = 12/4.905

t² = 2.446

t = √2.446

t = 1.56secs

b) To determine how fast is the frog falling at this point, we need to calculate the speed of the frog. Using the equaton v = u+gt

v = 0+9.81(1.56)

v = 15.34m/s

Hence the frog is falling at the rate of 15.34m/s

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blagie [28]
Assuming Adam is on earth g= 9.8 m/s and m= weight/ gravity = 667/9.8 = 68 kg
8 0
3 years ago
The atmosphere is held together by
Vera_Pavlovna [14]

Answer:

D. gravity

Explanation:

Gravity keeps the atmosphere from escaping into space.

3 0
4 years ago
Consider a 20 cm thick granite wall with a thermal conductivity of 2.79 W/m·K. The temperature of the left surface is held const
kozerog [31]

Answer:

The right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

Explanation:

Thickness of the wall is  L=  20cm = 0.2m

Thermal conductivity of the wall is  K = 2.79 W/m·K

Temperature at the left side surface is T₁ =  50°C

Temperature of the air is T = 22°C

Convection heat transfer coefficient is  h = 15 W/m2·K

Heat conduction process through wall is equal to the heat convection process so

Q_{conduction} = Q_{convection}

Expression for the heat conduction process is

Q_{conduction} = \frac{K(T_1 - T)}{L}

Expression for the heat convection process is

Q_{convection} = h(T_2 - T)

Substitute the expressions of conduction and convection in equation above

Q_{conduction} = Q_{convection}

\frac{K(T_1 - T_2)}{L} = h(T_2 - T)

Substitute the values in above equation

\frac{2.79(50- T_2)}{0.2} = 15(T_2 - 22)\\\\T_2 = 35.5^\circC

Now heat flux through the wall can be calculated as

q_{flux} = Q_{conduction} \\\\q_{flux}  = \frac{K(T_1 - T_2)}{L}\\\\q_{flux}  = \frac{2.79(50 - 35.5)}{0.2}\\\\q_{flux} = 202.3W/m^2

Thus, the right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

6 0
3 years ago
A proton moving at 8.00 106 m/s through a magnetic field of magnitude 1.72 T experiences a magnetic force of magnitude 7.20 10-1
gladu [14]

Answer:

19.1 deg

Explanation:

v = speed of the proton = 8 x 10⁶ m/s

B = magnitude of the magnetic field = 1.72 T

q = magnitude of charge on the proton = 1.6 x 10⁻¹⁹ C

F = magnitude of magnetic force on the proton = 7.20 x 10⁻¹³ N

θ = Angle between proton's velocity and magnetic field

magnitude of magnetic force on the proton is given as

F = q v B Sinθ

7.20 x 10⁻¹³ = (1.6 x 10⁻¹⁹) (8 x 10⁶) (1.72) Sinθ

Sinθ = 0.327

θ = 19.1 deg

4 0
3 years ago
Why do metals make good electrical conductors?
alexdok [17]

Answer:

Explained

Explanation:

Metals are good conductors of electricity because they contain free electrons in their atoms. The outer shell of atom's of metal have free electrons. These free electrons are responsible of electrical conductivity of metals. These electron are not bounded by the attraction forces of the nucleous. They are free to wonder in lattice of positive ion and thus allow electrical conductivity.

3 0
3 years ago
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