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Lisa [10]
4 years ago
5

Determine the number of ways to throw two die and get the number 11, as well as the probability of getting 11.

Physics
1 answer:
Ivenika [448]4 years ago
8 0

Answer:

1/18

Explanation:

The number of ways in which two die can be thrown is the sample space of the experiment

(1,1), (1,2) (1,3), (1,4) (1,5) (1,6)

(2,1), (2,2) (2,3), (2,4) (2,5) (2,6)

(3,1), (3,2) (3,3), (3,4) (3,5) (3,6)

(4,1), (4,2) (4,3), (4,4) (4,5) (4,6)

(5,1), (5,2) (5,3), (5,4) (5,5) (5,6)

(6,1), (6,2) (6,3), (6,4) (6,5) (6,6)

From here it can be seen that there are only two cases where the sum on the two die is 11 i.e., (6,5) and (5,6)

\text{Probability of getting 11}=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\\\Rightarrow \text{Probability of getting 11}=\frac{2}{36}\\\Rightarrow \text{Probability of getting 11}=\frac{1}{18}=0.056

∴ Probability of getting 11 is<u> 1/18</u>

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If one cup falls down then there will be 59 cups left.

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4 years ago
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An unknown fluid flows at a speed of 31 m/s. Suppose the fluid has a mass of 47 kg runs at this speed. What is the fluid’s kinet
Leya [2.2K]

Answer:

22583.5J

Explanation:

KE=1/2 mv^2

=1/2*47Kg*(31m/s^2)

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7 0
2 years ago
Part complete How long must a simple pendulum be if it is to make exactly one swing per five seconds?
shtirl [24]

Answer:

L=6.21m

Explanation:

For the simple pendulum problem we need to remember that:

\frac{d^{2}\theta}{dt^{2}}+\frac{g}{L}sin(\theta)=0,

where \theta is the angular position, t is time, g is the gravity, and L is the length of the pendulum. We also need to remember that there is a relationship between the angular frequency and the length of the pendulum:

\omega^{2}=\frac{g}{L},

where \omega is the angular frequency.

There is also an equation that relates the oscillation period and the angular frequeny:

\omega=\frac{2\pi}{T},

where T is the oscillation period. Now, we can easily solve for L:

(\frac{2\pi}{T})^{2}=\frac{g}{L}\\\\L=g(\frac{T}{2\pi})^{2}\\\\L=9.8(\frac{5}{2\pi})^{2}\\\\L=6.21m

3 0
3 years ago
What does weight require?
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Answer:

kg

Explanation:

easy question to be honest

7 0
3 years ago
Mercury has a density of 13.56 g/mL. How many kilograms of mercury would you expect to fit in a cylindrical glass cup with a bot
vovangra [49]

Answer:

263.152kg

Explanation:

<em>The density of a substance is related to its mass and volume as follows;</em>

density = mass / volume      

mass = density x volume       -------------(i)

The substance in question here is <em>mercury </em>which has;

density = 13.56g/mL = 13.56g/cm³

Since the mercury is going to be put in the cylindrical glass, the volume of the cylindrical glass is going to be equal to the volume of the mercury that will be put.

And we know that the;

volume of a cylinder = πr²h

<em>Where;</em>

π = 3.142

r = bottom radius of the cylinder = 5.75inches

h = height of the cylinder = 0.950ft

<em>For uniformity, let's convert the radius and height of the cylinder to their corresponding values in cm</em>

r  = 5.75 inches = 5.75 x 2.54 cm = 14.605cm

h = 0.950 ft = 0.950 x 30.48 cm = 28.956cm

<em>Therefore, the volume of the cylinder;</em>

v = 3.142 x (14.605cm)² x 28.956cm = 19406.5cm³

v = 19406.5cm³ [This is also the volume of the mercury necessary to fit the cylinder]

<em>Now the following value has been found;</em>

volume = 19406.5cm³

<em>Substitute the values of density and volume into equation (i)  as follows;</em>

mass = 19406.5cm³ x 13.56g/cm³

mass = 263152.14g

<em>Convert the result to kg by dividing by 1000</em>

mass = 263.152kg

Therefore, 263.152kg kilograms of mercury would fit in the cylindrical glass.

3 0
4 years ago
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