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NISA [10]
3 years ago
15

What is the greatest velocity which a falling object can achieve while falling through the air?

Physics
1 answer:
Alik [6]3 years ago
6 0

Answer:

Terminal Velocity

Explanation:

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What is the magnitude of the gravitational force exerted by earth on a 9.0-kg brick when the brick is in free fall?
GarryVolchara [31]
The magnitude is 9.000kg
4 0
3 years ago
What's the closest to the thickness of a dime? A(millimeters)B(centimeters)C(decameters)D(meters)
djyliett [7]
B centimeters is the closest thickness to a dike because the dime's real thickness is 1.35
4 0
3 years ago
A 32.0 kg wheel, essentially a thin hoop with radius 1.20 m, is rotating at 280 rev/min. It must be brought to a stop in 15.0 s.
elena-14-01-66 [18.8K]

Answer:

A) Must be done 19806.62 joules of work.

B) The average power is 1320.44 Watts.

Explanation:

A) First, we're going to use the work-energy theorem that states total work (W) done on an object is equal to the change in its kinetic energy (\Delta K):

W=\Delta K = K_{f}-K_{i} (1)

So, all we must do is to find the change on kinetic energy. Because we're working with rotational body, we should use the equation K=\frac{I\omega^{2}}{2} for the kinetic energy so:

\Delta K=\frac{I(\omega_{f})^{2}}{2}-\frac{I(\omega_{i})^{2}}{2} (2)

with \omega_{i} the initial angular velocity, \omega_{f} the final angular velocity (is zero because the wheel stops) and I the moment of inertia that for a thin hoop is I=MR^{2}, using those on (2)

\Delta K=0-\frac{MR^{2}(\omega_{i})^{2}}{2} (3)

By (3) on (1):

W= \frac{MR^{2}(\omega_{i})^{2}}{2} = \frac{(32.0)(1.2)^{2}(29.32)^{2}}{2}

W=19806.62\,J

B) Average power is work done divided by the time interval:

P=\frac{W}{\Delta t}=\frac{19806.62}{15.0}

P=1320.44\,W

NOTE: We use the relation 1rpm*\frac{2\pi}{60s}=\frac{rad}{s} to convert 280 rev/min(rpm) to 29.32 rad/s

4 0
4 years ago
Please help me solve all of them ( a, b, c and d ) thankiew !! <br> I’m also kind of in a rush
Sergio039 [100]

Answer:

a-

V= IR

9V = I ×( 12+6)

I = 9/ 18 A = 0.5 A

b

V=IR

240 = 6 A ×( 20 + R)

40 = 20 + R

R = 20 ohm

c

resultant resistance of the 2 parallel resistances= Ro

1/Ro = 1/ 5 + 1/ 20

1/Ro =( 20+5)/100

= 1/Ro = 1/4

Ro= 4 ohm

V=IR

V = 2A × ( 1+ 4 OHM)

V = 10V

d

equivalent resistance = Ro

1/Ro = 1/(2+8) + 1/(5+5)

1/Ro = 1/10 +1/10

2/10 = 1/ Ro

Ro= 10/2 = 5 ohm

V = IR

12V = I × 5Ohm

I=2.4 A

6 0
2 years ago
A uniform electric field, with a magnitude of 370 N/C, is directed parallel to the positive x-axis. If the electric potential at
dem82 [27]

Answer:

\Delta U = 0.2072 J

Explanation:

Potential difference between two points in constant electric field is given by the formula

\Delta V = E.\Delta x

here we know that

E = 370 N/C

also we know that

\Delta x = 2.1 - 1.9 = 0.2 m

now we have

\Delta V = 370 (0.2) = 74 V

now change in potential energy is given as

\Delta U = Q\Delta V

\Delta U = (2.80 \times 10^{-3})(74)

\Delta U = 0.2072 J

3 0
4 years ago
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