. When people don't clearly recognize the cause of their frustration, they

In 1991 at smith college, in massachusetts, ferdie adoboe ran 1.00 × 102 m backward in 13.6 s. suppose it takes adoboe 2.00 s to

Alik [6]

as it is given that it covers a total distance 1 * 10^2 m

total time taken by it = 13.6 s

now the average speed is given as ratio of total distance and total time

$v = \frac{d}{t}$

$v = \frac{1* 10^2 }{13.6}$

$v = 7.35 m/s$

so the average speed will be 7.35 m/s

now if it starts from rest and achieve the final speed as 7.35 m/s

now we can use kinematics

$v_f = v_i + at$

$7.35 = 0 + a* 2$

$a = 3.68 m/s^2$

so its acceleration will be 3.68 m/s^2

4 0

4 years ago

Water enters a cylindrical tank through two pipes at rates of 250 and 100 gal/min. If the level of the water in the tank remains

tester [92]

Answer:

The total amount of water that enters the tank is:

250 gal/min + 100 gal/min = 350gal/min.

Then, if the level of the water remains constant, this means that the water leaves the tank at a rate of 350gal/min.

We know that the diameter of the pipe is 8 inches, then the area of the pipe is:

A = pi*(d/2)^2 = 3.14*(4in)^2 = 50.24in^2

now, the flow can be calculated as:

Q = v*A = (velocity*area)

if we want to write our velocity in inches per minute, then we need to write the entering flow in cubic inches:

1 gallon = 231 in^3

then:

350gal/min = (350*231) in^3/min = 80,850 in^3/min.

Then the water that leaves the tank must be the same amoun, we have:

Q = 80,850 in^3/min. = v*A = v*50.24in^2

v = (80,850in^3/min)/50.24in^2 = 1609.3 in/min.

The velocity of the flow leaving the tank is 1609.3 in/min.

3 0

4 years ago

Two 10-cm-diameter charged rings face each other, 25 cm apart. The left ring is charged to ? 25 nC and the right ring is charged

MArishka [77]

Answer:

A) E = 0N/C

B) 0i + 0^^j

C) F = 0N

D) 0^i + 0^j

Explanation:

You assume that the rings are in the zy plane but in different positions.

Furthermore, you can consider that the origin of coordinates is at the midway between the rings.

A) In order to calculate the magnitude of the electric field at the middle of the rings, you take into account that the electric field produced by each ring at the origin is opposite to each other and parallel to the x axis.

You use the following formula for the electric field produced by a charge ring at a perpendicular distance of r:

$E=k\frac{rQ}{(r+R^2)^{3/2}}$ (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C

Q: charge of the ring

r: perpendicular distance to the center of the ring

R: radius of the ring

You use the equation (1) to calculate the net electric field at the midpoint between the rings:

$E=k\frac{rQ}{(r^2+R^2)^{3/2}}-k\frac{rQ}{(r^2+R^2)^{3/2}}=0\frac{N}{C}$

The electric field produced by each ring has the same magnitude but opposite direction, then, the net electric field is zero.

B) The direction of the electric field is 0^i + 0^j

C) The magnitude of the force on a proton at the midpoint between the rings is:

$F=qE=q(0N/C)=0N$

D) The direction of the force is 0^i + 0^j

6 0

3 years ago

M A hockey player is standing on his skates on a froze pond when an opposing player, moving with a uniforn speed of 12 m/s, skat

Anni [7]

<span>A car with a mass of 1500kg starts from rest and accelerates to a speed of 18m/s in 12s. Assume that the force of resistance remains contant at 400N during</span>

8 0

4 years ago

If an observer is hearing a rising pitch, that means that the frequency of the waves in their position is falling.

zloy xaker [14]

Is this a true or false question?

3 0

4 years ago

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