A 1600 watt blow dryer is plugged into a source of 120 volts. What will the current through the blow dryer

If two different theories describe experimental observations equally well, can one be said to be more valid than the other (assu

Andreas93 [3]

Answer:

No

Explanation:

In such situations we cannot determine which one is more valid as both serves the purpose well.

Two theories are carried out in different environment and circumstance keeping different parameters and one can opt for any number of ways to carry out that experiment but what matter at the end is the accuracy they bring.

Each of the theory is a new discovery and follows all the possible logical rules hence it is not possible to decide which one is more valid.

3 0

3 years ago

A long, straight, vertical wire carries a current upward. due east of this wire, in what direction does the magnetic field point

Solnce55 [7]

(I assume that the 4 directions north-south-east-west are meant with respect to the wire seen from the top.)

We can use the right-hand rule to understand the direction of the magnetic field generated by the wire. The thumb follows the direction of the current in the wire (upward), while the other fingers give the direction of the field in every point around the wire. Seen from the top, the field has an anti-clockwise direction. Therefore, if we take a point at east with respect to the wire, in this point the field has direction south.

8 0

3 years ago

Josh is refilling his pool with clean water. How many gallons of water will it take if the pool is 50m by 25m by 1.5m?

horsena [70]

Answer:

V = 493421.05 [gal]

Explanation:

This is a problem that consists of handling units, we can calculate by first-hand the volume, then convert units from cubic meters to gallons.

V = 50 * 25 * 1.5

V = 1875 [m^3]

Now we need to convert units, using the proper conversion factor.

$1875[m^3]*\frac{1000lt}{1m^3} *\frac{1gal}{3.8lt} \\493421.05[gal]$

7 0

3 years ago

Water, which we can treat as ideal and incompressible, flows at 12 m/s in a horizontal pipe with a pressure of 3.0 x 10^4 Pa. If

frez [133]

Answer:

p2 = 9.8×10^4 Pa

Explanation:

Total pressure is constant and PT = P = 1/2×ρ×v^2

So p1 + 1/2×ρ×(v1)^2 = p2 + 1/2×ρ×(v2)^2

from continuity we have ρ×A1×v1 = ρ×A2×v2

v2 = v1×A1/A2

and

r2 = 2×r1

then:

A2 = 4×A1

so,

v2 = (v1)/4

then:

p2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v2)^2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v1/4)^2

p2 = 3.0×10^4 Pa + 1/2×(1000 kg/m^3)×(12m/s)^2 - 1/2×(1000kg/m^3)×(12^2/16)

= 9.75×10^4 Pa

= 9.8×10^4 Pa

Therefore, the pressure in the wider section is 9.8×10^4 Pa

5 0

3 years ago

A ball of mass m = 0.1kg is connected to a rope of length L = 1.2 m. The ball is swung around in a vertical circle and ball is m

bonufazy [111]

Answer:

The speed of the ball is approximately 5.94 m/s

The Tension of the string at the bottom is 3.92 N

Explanation:

We need to find the speed of the ball, which is constant due to the fact that we are in a uniform circular motion. Notice as well that the speed of the ball is the magnitude of the tangential velocity " $v_t$ " (vector that changes direction with the position of the ball but doesn't change magnitude in this case).

We analyze first the top position of the circular motion, for which information on the tension of the string is given (see first free body diagram in the attached picture). We are told that the tension at the top of the movement equals twice the force of gravity on the ball's mass: T - 2*m*g = 1.96 N. And we know that there are two forces acting on the ball in that position (illustrated with the green arrows pointing down): one is the ball's weight due to gravity, and the other is the string's tension. So we can write Newton's second law for this situation:

$F_{net}= T_{top}+W\\F_{net}=2\,W+W\\F_{net}=3\,W\\F_{net}=2.94 N\\$

Newton's second law tells us that the net force should equal the mass of the ball times its acceleration (F = m * a), and in this motion, the acceleration is the centripetal acceleration. Therefore weuse this equation to solve for the centripetal acceleration of the ball:

$m\,a_c=2.94\,N\\a_c=\frac{2.94\,N}{0.1\,kg} \\a_c=29.4\,\frac{m}{s^2}$

The centripetal acceleration is defined as the square of the tangential velocity divided the radius of the circular motion. Then we use it to derive the magnitude of the tangential velocity (speed of the ball):

$a_c=\frac{v^2}{R} \\29.4\,\frac{m}{s^2} =\frac{v_t^2}{R} \\v_t^2=29.4\,(1.2)\,\frac{m^2}{s^2} \\v_t=5.94\,\frac{m}{s}$

So we have found the speed of the ball.

Now we focus our attention to the bottom of the motion, and again use Newton's second law to solve for the string tension (see second free body diagram in the attached picture).

We notice here that the tension and the weight are acting in opposite directions, so we have such into account when finding the net force on the ball, and then solve for the tension knowing the value of the centripetal acceleration (recall that the magnitude of the tangential velocity is the same because of the uniform circular motion).

$F_{net}= T_{bot}-W\\m\,a_c=T_{bot}-0.98\,N\\2.94\,N=T_{bot}-0.98\,N\\T_{bot}=(2.94+0.98)N\\T_{bot}=3.92\,N$

4 0

3 years ago