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2 years ago

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A box contains 9 new light bulbs and 6 used light bulbs. Each light bulb is the same size and shape. Meredith will randomly sele

ct 2 light bulbs from the box without replacement. What is the probability Meredith will select a new light bulb and then a used light bulb

1 answer:

MAVERICK [17]2 years ago

8 0

Answer:

(9/35) = 0.257

Explanation:

Box contains 9 new light bulbs and 6 used light bulbs, total number of bulbs = 15.

Probability of selecting two bulbs; a new light bulb and then, a used light bulb in that order = [(probability of selecting a new bulb) × (probability of selecting a used bulb from the rest)] = [(9/15) × (6/14)] = (9/35) = 0.257

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A cyclist moving with a constant velocity of 6.0 m/s forward passes a car that is just starting. If the car has a constant accel

sergiy2304 [10]

After 6 seconds, the car will surpass the cyclist.

<h3><u>Explanation:</u></h3>

The speed of the cyclist = 6 m/s.

Let after time t sec, the car will overtake the cyclist.

So, distance covered by the cyclist in t sec = 6t m

Initial velocity of the car is 0 m/s, because the car is just starting.

Acceleration of the car = $2 m/s^2$ .

Final velocity of the car =6 m/s.

So to cover the distance 6t, the time required by the car = $\frac{1}{2} \times a \times t^2 = \frac{1}{2} \times 2 \times t^2$

$6t = \frac{1}{2} \times 2 \times t^2\\6t = t^2$

t =6 sec

So, after 6 seconds, the car will surpass the cycle.

6 0

2 years ago

Two objects are placed in thermal contact and are allowed to come to equilibrium in isolation. the heat capacity of object a is

Harman [31]

Given:
Ca = 3Cb (1)
where
Ca = heat capacity of object A
Cb = heat capacity f object B

Also,
Ta = 2Tb (2)
where
Ta = initial temperature of object A
Tb = initial temperature of object B.

Let
Tf = final equilibrium temperature of both objects,
Ma = mass of object A,
Mb = mass of object B.

Assuming that all heat exchange occurs exclusively between the two objects, then energy balance requires that
Ma*Ca*(Ta - Tf) = Mb*Cb*(Tf - Tb) (3)

Substitute (1) and (2) into (3).
Ma*(3Cb)*(2Tb - Tf) = Mb*Cb*(Tf - Tb)
3(Ma/Mb)*(2Tb - Tf) = Tf - Tb

Define k = Ma/Mb, the ratio f the masses.
Then
3k(2Tb - Tf) = Tf - Tb
Tf(1+3k) = Tb(1+6k)
Tf = [(1+6k)/(1+3k)]*Tb

Answer:
$T_{f} =( \frac{1+6k}{1+3k} )T_{b}= \frac{1}{2}( \frac{1+6k}{1+3k})T_{a}$
where
$k= \frac{M_{a}}{M_{b}}$

7 0

2 years ago

Lilli suggests that they explore the simulation starting with varying only a single parameter in order to understand the role of

mrs_skeptik [129]

Answer:

B.

Explanation:

One of the ways to address this issue is through the options given by the statement. The concepts related to the continuity equation and the Bernoulli equation.

Through these two equations it is possible to observe the behavior of the fluid, specifically the velocity at a constant height.

By definition the equation of continuity is,

$A_1V_1=A_2V_2$

In the problem $A_2$ is $2A_1$ , then

$A_1V_1=2A_1V_2$

$V_2 = \frac{V_1}{2}$

<em>Here we can conclude that by means of the continuity when increasing the Area, a decrease will be obtained - in the diminished times in the area - in the speed.</em>

For the particular case of Bernoulli we have to

$P_1 + \frac{1}{2}\rho V_1^2 = P_2 +\frac{1}{2}\rho V_2^2$

$P_2-P_1 = \frac{1}{2} \rho (V_1^2-V_2^2)$

For the previous definition we can now replace,

$P_2-P_1 = \frac{1}{2} \rho (V_1^2-(\frac{V_1}{2})^2)$

$\Delta P = \frac{3}{8} \rho V_1^2$

<em>Expressed from Bernoulli's equation we can identify that the greater the change that exists in pressure, fluid velocity will tend to decrease</em>

The correct answer is B: "If we increase A2 then by the continuity equation the speed of the fluid should decrease. Bernoulli's equation then shows that if the velocity of the fluid decreases (at constant height conditions) then the pressure of the fluid should increase"

4 0

2 years ago

A person is working on a steel structure while standing on the ground. An accident occurred where 5 A pass through the structure

matrenka [14]

Answer:

35mA

Explanation:

Hello!

To solve this problem we must use the following steps

1. Find the electrical resistance of the metal rod using the following equation

$R=\alpha \frac{l}{a}$

WHERE

α=

metal rod resistivity=2x10^-4 Ωm

l=leght=2m

A= Cross-sectional area

$A=\frac{\pi }{4} d^2=\frac{\pi }{4} (0.06)^2=0.00283$

solving

$R=(2x10^-4)\frac{2}{0.00283} =0.14$

2. Now we model the system as a circuit with parallel resistors, where we will call 1 the metal rod and 2 the man(see attached image)

3.we know that the sum of the currents in 1 and 2 must be equal to 5A, by the law of conservation of energy

I1+I2=5

4.as the voltage on both nodes is the same we can use ohm's law in resitance 1 and 2 (V=IR)

V1=V2

(0.14I1)=2000(i2)

solving for i1

I1=14285.7i2

5.Now we use the equation found in step 3

14285.7i2+i2=5

$i2=\frac{5}{14285.7+1} =3.5x10^-4A=35mA$

6 0

2 years ago

Determine the force P required to maintain the 200-kg engine in the position for which θ = 30°. The diameter of the pulley at B

gregori [183]

Answer:

The force P required is 1759.22 N

Explanation:

The missing diagram is seen in the first image below.

From the second image, we can see the schematic diagram of the engine hanging over the pulley.

To start with determining the value of the angle ∝;

$tan \ \alpha = \dfrac{CD}{BD}$

where;

BD = AB-AD

Then;

$tan \ \alpha = \dfrac{CD}{AB-AD}$

$\alpha = tan^{-1} \bigg(\dfrac{CD}{AB-AD} \bigg )$

replacing their respective values, where;

CD = 2 sin 30° m, AB = 2m and AD = 2 cos 30° m

$\alpha = tan^{-1} \bigg(\dfrac{2 \ sin \ 30^0}{2-2 \ cos \ 30^0} \bigg )$

$\alpha = tan^{-1} \bigg(\dfrac{1}{2-1.732} \bigg )$

$\alpha = tan^{-1} \bigg(\dfrac{1}{0.268} \bigg )$

$\alpha = tan^{-1} \bigg(3.73\bigg )$

$\alpha \simeq 75^0$

From the third diagram attached below:

The tension occurring in the thread BC is equal to force P

$T_{BC} = P$

Using the force equilibrium expression along the horizontal direction.

$\sum F_x = 0\\\\ -T_{AC} \ cos \ 30^0 + Pcos \alpha = 0$

replacing the value of $\alpha \simeq 75^0$

$-T_{AC} \ cos 30^0 + P cos 75^0 = 0$

$P \ cos \ 75^0 = T_{AC} \ cos \ 30^0$

$P =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0} \ \ \ - - - (1)$

Along the vertical direction, the force equilibrium equation can be expressed as:

$\sum F_y =0$

$-W + P \ sin \alpha + T_{AC} \ sin \ 30^0 = 0$

$W = P \ sin \ \alpha + T_{AC} \ sin \ 30^0$

replacing $\alpha \simeq 75^0$ and $P =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0}$

$W =\dfrac{T_{AC} \ cos \ 30^0}{cos \ 75^0}\times sin \ 75^0 + T_{AC} \ sin \ 30^0$

Also, replacing W for (200 × 9.81) N

$200 \times 9.81 =\dfrac{T_{AC} \ cos \ 30^0}{cos \ 75^0}\times sin \ 75^0 + T_{AC} \ sin \ 30^0$

$200 \times 9.81 = T_{AC} \ cos \ 30^0 \ tan \ 75^0 + T_{AC} \ sin \ 30^0$

$1962= T_{AC} \ ( cos \ 30^0 \ tan \ 75^0 + \ sin \ 30^0)$

$1962= T_{AC} \ (0.8660\times 3.732 + 0.5)$

$1962= T_{AC} \ (3.231912 + 0.5)$

$1962= T_{AC} \ (3.731912)$

$T_{AC} = \dfrac{1962}{ \ (3.731912)}$

$T_{AC} = 525.736 \ N$

From $P =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0}$

$P =\dfrac{ 525.736 \ cos \ 30^0}{\ cos \ 75^0}$

$P =\dfrac{ 525.736 \times0.866}{0.2588}$

P = 1759.22 N

Thus, the force P required is 1759.22 N

6 0

1 year ago