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natima [27]
3 years ago
10

An object is placed 8.5 cm in front of a convex spherical mirror of focal length −14.0 cm. What is the image distance? Show all

work and include units of measure.
Physics
1 answer:
Pani-rosa [81]3 years ago
5 0

Given:

u = 8.5 cm

f = -14.0 cm

v = image distance

Using the mirror formula 1/u + 1/v = 1/f

1/8.5 + 1/v = 1/-14.0

Rewrite to solve for v:

v = (8.5 * -14.0) / (8.5 - (-14.0))

v = -119 / 22.5

v = -5.29 cm  ( round answer as needed.)

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3 A picture is supported by two vertical strings; if the weighi
andrezito [222]

Answer:

<em>The force exerted by each string is 25 N</em>

Explanation:

<u>Net Force</u>

The net force is the vector sum of forces acting on a body. The net force is a single force that represents the effect of the original forces on the body's motion. It gives the particle the same acceleration as all those actual forces together as described by Newton's second law of motion.

The picture described in the problem is hanging at rest supported by two vertical strings. This means that the net force acting on it is zero.

Assume the magnitude of each of these equal forces is F, and the picture has a weight of W=50 N, thus the net force is:

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The positive signs indicate an upwards direction and the negative sign means a downwards direction. Since the net force is zero:

F + F - W = 0

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7 0
3 years ago
A 3.80-m-long, 500 kg steel beam extends horizontally from the point where it has been bolted to the framework of a new building
Virty [35]

Answer:

Explanation:

The question is ;

3.80 m-long, 500. kg steel beam extends horizontally from the point

where it has been bolted to the framework of a new building under

construction. A 67.0 kg construction worker stands at the far end of

the beam. What is the magnitude of the torque about the point where the beam is bolted into place?

Solution:

Here two torques are in action

a) torque T1 due to weight of worker at the edge of the beam - at center of the beam

b) torque T2 due to weight of the uniform beam - at the point where the beam is bolted

So we first calculate the torque produced due to weight of the worker;

We can see that;

Distance of worker from the center of the beam = 1.9m

Mass of the worker = 67 kg

Value of g= 9.8 m/sec2

T1=force × distance from point of rotation

Here force is weight of the worker which is = mass × g=67×9.8=656.6 N

So the torque is

T1= 656.6×1.9=12225.892 Nm

or

T1 = 12225.892 Nm

Now torque by the beam itself ;

Length of the beam from its center point to the bolt point = 1.9 m

Weight force of beam acting at center point= mass× g= 500×9.8=4900 N

Torque T2 at bolt point by the beam weight = weight of beam × length of beam from its center to the bolt point = 4900×1.9=9310 Nm

T2=9310Nm

So total torque = T1+T2= 12225.892+9310=21565.892 Nm

8 0
3 years ago
A silver wire has a cross sectional area a = 2.0 mm2. a total of 9.4 × 1018 electrons pass through the wire in 3.0 s. the conduc
marta [7]
This problem uses the relationships among current I, current density J, and drift speed vd. We are given the total of electrons that pass through the wire in t = 3s and the area A, so we use the following equation to to find vd, from J and the known electron density n, so: 

v_{d} =  \frac{J}{n\left | q \right |}

<span>The current I is any motion of charge from one region to another, so this is given by:

</span>I = \frac{\Delta Q}{\Delta t} = \frac{9.4x1018electrons}{3s} = 3189.73(A)

The magnitude of the current density is:

J = \frac{I}{A} = \frac{3189.73}{2x10^{-6}} = 1594.86(A/m^{2})

Being:

A=2mm^{2} = 2x10^{-6}m^{2}
<span>
Finally, for the drift velocity magnitude vd, we find:

</span>v_{d} = \frac{1594.86}{5.8x1028\left |1.60x10^{-19}|\right } = 1.67x10^{18}(m/s)

Notice: The current I is very high for this wire. The given values of the variables are a little bit odd
6 0
3 years ago
Uma pilha AA (pilha de controle remoto) fornece ao circuito 1,5V, já uma bateria de carro fornece 12V. Qual a relação entre a en
katrin [286]

Answer:

The ratio of energy of AA battery to the car battery is 0.0156.

Explanation:

An AA battery (remote control battery) supplies the circuit with 1.5V, while a car battery supplies 12V. What is the relationship between the energy supplied between these circuit elements?

Supply by AA battery = 1.5 V

Supply by car battery = 12 V

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8 0
3 years ago
A student weighs 400 Newtons climbs a 3 meter ladder in 4 seconds how much work is the student do
Y_Kistochka [10]
400 * 3 = 1200 Joules
7 0
4 years ago
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