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2 years ago

A current of 2.0 A flows through a flashlight bulb when it is connected

Physics

1 answer:

tatiyna2 years ago

8 0

V=5V
I=2A
t=1h

Using Joules law

$\\ \sf\Rrightarrow H=VIt$

$\\ \sf\Rrightarrow H=5(2)(1)$

$\\ \sf\Rrightarrow H=10Kwh$

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We dont see objects. We see the light ____ off objects.

Romashka [77]

The answer is BBBBBBBBB

4 0

3 years ago

A 22.0 kg bucket of concrete is connected over a very light frictionless pulley to a 375 N box on the roof of a building as show

Veronika [31]

Answer:

vf = 3.27[m/s]

Explanation:

In order to solve this problem we must analyze each body individually and find the respective equations. The free body diagram of each body (box and bucket) should be made, in the attached image we can see the free body diagrams and the respective equations.

With the first free body diagram, we determine that the tension T should be equal to the product of the mass of the box by the acceleration of this.

With the second free body diagram we determine another equation that relates the tension to the acceleration of the bucket and the mass of the bucket.

Then we equalize the two stress equations and we can clear the acceleration.

a = 3.58 [m/s^2]

As we know that the bucket descends 1.5 [m], this same distance is traveled by the box, as they are connected by the same rope.

$x = \frac{1}{2} *a*t^{2}\\1.5 = \frac{1}{2}*(3.58) *t^{2} \\t = 0.91 [s]$

And the speed can be calculated as follows:

$v_{f}=v_{o}+a*t\\v_{f}=0+(3.58*0.915)\\v_{f}= 3.27[m/s]$

7 0

3 years ago

Part 1 :

dybincka [34]

Answer:

1. 218.55 N

2. $30.96^{o}$

3. $2.1 m/s^{2}$

Explanation:

Part 1;

Net force $F=mg sin \theta$ where m is mass, g is gravitational force and $\theta$ is the angle of inclination

$F= 46*9.8*sin 29^{o}= 218.55N$

Frictional force, $F_{r}$ is given by

$F_{r} = \mu_{s}mg cos \theta$ where $\mu_{s}$ is the coefficient of static friction

$F_{r} = 0.6*46*9.8 cos 29$

$F_{r}=236.57N$

Since $F_{r}>F$ , therefore, the block doesn’t slip and the frictional force acting is mgh=218.55N

Part 2.

Using the relationship that

Frictional force $F_{s} = \mu_{s} mg cos \theta$

$mg sin \theta =\mu_{s} mg cos \theta$

$\mu_{s}= \frac {sin \theta}{cos \theta}$

$\mu_{s}= tan \theta$

The maximum angle of inclination $\theta = tan^{-1} \mu_{s}$

$\theta = tan^{-1} (0.6)$

$\theta= 30.96^{o}$

Part 3:

Net force on the object is given by

$ma = mg sin 38 - \mu_{k} mg cos 38$ where $\mu_{k}$ is the coefficient of kinetic friction

$a = g ( sin 38 - \mu_{k} cos 38 )$

= 9.8 ( sin 38 - (0.51) cos 38 )

= $2.1m/s^{2}$

7 0

3 years ago

15. Missy Diwater, the former platform diver or the Ringling Brother's Circus had a speed of 25m/s when she hit the

timurjin [86]

K.E.= 1/2 x MV^2 = 1/2 x 40(kg) x (25x25) =12500J

5 0

2 years ago

You have a flag pole at an angle of 45 degrees to a wall it is designed to support a max torque of 40Nm. You bought a new one wi

In-s [12.5K]

Answer:

0.915 rad or 52.44 degrees

Explanation:

Let g = 10 m/s2. We can calculate the gravity force acting on center of mass of the new 8.75 kg pole:

$F = mg = 8.75 * 10 = 87.5 kg$

The moment is the dot product of force and moment arm

$M = \vec{F} \cdot \vec{r} = Frcos\theta = 87.5*0.75cos\theta = 65.625cos\theta$

where θ is the angle between the gravity force and the pole, or between the pole and the wall. As M can be at its max, which is 40 Nm, we can solve for θ:

$62.625cos\theta = 40$

$cos\theta = 40 / 62.625 = 0.61$

$\theta = cos^{-1}0.61 = 0.915 rad = 0.915 * 180 /\pi = 52.44^o$

7 0

3 years ago