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rodikova [14]
3 years ago
9

ENERG

Physics
2 answers:
Romashka [77]3 years ago
6 0
Answer: B) Uranium

Fossil fuels are natural fuels that were formed from the buried remains of living organisms that lived millions of years ago. Uranium is not a fossil fuel, but it is considered as a nonrenewable fuel.
ladessa [460]3 years ago
5 0

Answer:

Uranium

Explanation:

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A golfer is on the edge of a 12.5 m bluff overlooking the 18th hole which is located 60 m from the base of the bluff. She launch
Lina20 [59]

Answer:

The ball impact velocity i.e(velocity right before landing) is 6.359 m/s

Explanation:

This problem is related to parabolic motion and can be solved by the following equations:

x=V_{o}cos \theta t----------------------(1)

y=y_{o}+V_{o} sin \theta t - \frac{1}{2}gt^{2}---------(2)

V=V_{o}-gt ----------------------- (3)

Where:

x = m is the horizontal distance travelled by the golf ball

V_{o} is the golf ball's initial velocity

\theta=0\° is the angle (it was  a horizontal shot)

t is the time

y is the final height of the ball

y_{o} is the initial height of the ball

g is the acceleration due gravity

V is the final velocity of the ball

Step 1: finding t

Let use the equation(2)

t=\sqrt{\frac{2 y_{o}}{g}}

t=\sqrt{\frac{2 (12.5 m)}{9.8 m/s^{2}}}

t=1.597s

Substituting (6) in (1):

67.1 =V_{o} cos(0\°) 1.597-------------------(4)

Step 2:  Finding V_{o}:

From equation(4)

67.1 =V_{o}(1) 1.597

V_0 = \frac{6.71}{1.597}

V_{o}=42.01 m/s (8)  

Substituting V_{o} in (3):

V=42.01 -(9.8)(1.597)

v =42 .01 - 15.3566  

V=26.359 m/s

5 0
3 years ago
The velocity profile in fully developed laminar flow in a circular pipe of inner radius R 5 2 cm, in m/s, is given by u(r) 5 4(1
xxMikexx [17]

The question is not clear and the complete clear question is;

The velocity profile in fully developed laminar flow In a circular pipe of inner radius R = 2 cm, in m/s, is given By u(r) = 4(1 - r²/R²). Determine the average and maximum Velocities in the pipe and the volume flow rate.

Answer:

A) V_max = 4 m/s

B) V_avg = 2 m/s

C) Flow rate = 0.00251 m³/s

Explanation:

A) We are given that;

u(r) = 4(1 - (r²/R²))

To obtain the maximum velocity, let's apply the maximum condition for a single-variable continual real valued problem to obtain;

(d/dr)(u(r)) = 0

Thus,

(d/dr)•4(1 - (r²/R²)) = 0

4(d/dr)(1 - (r²/R²)) = 0

If we differentiate, we have;

4(0 - (2r/R²)) = 0

-8r/R² = 0

Thus, r = 0 and with that, the maximum velocity is at the centre of the pipe.

Thus, for maximum velocity, let's put 0 for r in the U(r) function.

Thus,

V_max = 4(1 - 0²/R²) = 4 - 0 = 4 m/s

B) Average velocity is given by;

V_avg = V_max/2

V_avg = 4/2 = 2 m/s

C) the flow can be calculated from;

Flow rate ΔV = A•V_avg

A is area = πr²

From question, r = 2cm = 0.02m

A = π x 0.02²

Hence,

ΔV = π x 0.02² x 2 = 0.00251 m³/s

8 0
3 years ago
The magnetic field inside a superconducting solenoid is 4.00 T. The solenoid has an inner diameter of 6.20 cm and a length of 26
Delvig [45]

Answer:

(a) The magnetic energy density in the field is 6.366 J/m³

(b) The energy stored in the magnetic field within the solenoid is 5 kJ

Explanation:

magnitude of magnetic field inside solenoid, B = 4 T

inner diameter of solenoid, d = 6.2 cm

inner radius of the solenoid, r = 3.1 cm = 0.031 m

length of solenoid, L = 26 cm = 0.26 m

(a) The magnetic energy density in the field is given by;

u _B = \frac{B^2}{2\mu_o} \\\\u _B = \frac{(4)^2}{2(4\pi*10^{-7})}\\\\u_B = 6.366*10^6 \ J/m^3

(b) The energy stored in the magnetic field within the solenoid

U_B = u_B V\\\\U_B = u_B AL

U_B = u_B(A)(L)\\\\U_B = 6.366*10^6(\pi * 0.031^2)(0.26) \\\\U_B = 4997.69 J\\\\U_B = 5 \ KJ\\

6 0
3 years ago
How does light travel?
Ivenika [448]

Answer:

C

Explanation:

light can travel in a vacuum Anne the sped varies

7 0
3 years ago
Read 2 more answers
A flat, wide cloud floats horizontally a few kilometers above the surface of Earth. Its lower surface carries a uniform surface
valentinak56 [21]

Answer:

\frac{kQ}{r^2} r^

Explanation:

Electric field strength= Force/unit charge

E= (kQq/r²)/q ₓ r

where r is the unit vector in the direction of unit charge

E= \frac{kQ}{r^2} r^

4 0
3 years ago
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