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Mazyrski [523]
3 years ago
14

Given a die, would it be more likely to get a single 6 in six rolls, at least two 6s in twelve rolls, or at least one-hundred 6s

in six-hundred rolls?
Physics
1 answer:
vichka [17]3 years ago
4 0

Answer:

Explanation:

In first case we are interested in one time 6 in six rolls

Thus probability = number of chances required/Total chances

= 1/6

Similarly in the second case probability = 2/12 = 1/6

In the same way in last case probability = 100/600 = 1/6

The probability is the same . Thus all the cases has equal chances  

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The Rank of a matrix can not be greater than the number of columns or rows
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I feel like my answer is wrong but I think it B) false or C) true
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4 years ago
In a particular lab, a cube of ice (Tice = -5.5˚C) is taken and dropped into a calorimeter cup (98g) partially filled with 326 g
Veseljchak [2.6K]

Answer:

The mass of the ice added = 16.71 g

Explanation:

The heat gained by the ice is equal to the heat lost by the calorimeter cup and the water in the cup.

But for this question, the cup is said to be perfectly insulated, hence, there is no loss of heat from the calorimeter cup.

Heat gained by the ice = Heat lost by the 326 g of water.

Let the mass of ice be m

The heat gained by the ice = (Heat gained by ice in temperature from -5.5°C to 0°C) + (Heat used by the ice to melt at 0°C) + (Heat required for the melted ice to rise in temperature from 0°C to 15°C)

Heat gained by ice in temperature from -5.5°C to 0°C = mCΔT

m = unknown mass of ice

C = Specific Heat capacity of ice = 2.108 J/g°C

ΔT = change in temperature = 0 - (-5.5) = 5.5°C

Heat gained by ice in temperature from -5.5°C to 0°C = m×2.108×5.5 = (11.594m) J

Heat used by the ice to melt at 0°C = mL

m = unknown mass of ice

L = Latent Heat of fusion of ice to water = 334 J/g

Heat used by the ice to melt at 0°C = m×334 = (334m) J

Heat required for the melted ice or water now, to rise in temperature from 0°C to 15°C = mCΔT

m = unknown mass of water (which was ice)

C = Specific Heat capacity of water = 4.186 J/g°C

ΔT = change in temperature = 15 - 0 = 15°C

Heat required for the melted ice or water now, to rise in temperature from 0°C to 15°C = m×4.186×15 = (62.79m) J

Total heat gained by the ice = 11.594m + 334m + 62.79m = (408.384m) J

Heat lost by the water in the calorimeter cup = MCΔT

M = mass of water in the calorimeter cup = 326 g

C = specific heat capacity of water = 4.186 J/g°C

ΔT = change in temperature = 20 - 15 = 5°C

Heat lost by the water in the calorimeter cup = 326×4.186×5 = 6,823.18 J

Heat gained by the ice = Heat lost by the 326 g of water.

408.384m = 6,823.18

m = (6,823.18/408.384)

m = 16.71 g

Hope this Helps!!!

7 0
3 years ago
To calculate the wavelength of a wave, take the wave speed and divide by the frequency of the wave. true or false
Serjik [45]

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4 0
3 years ago
a 2.0 kg block slides on the horizontal, frictionless surface until it counters a spring force constant with
icang [17]

Complete question:

a 2.0 kg block slides on the horizontal, frictionless surface until it counters a spring with force constant of  955 N/m. The block comes to rest after compressing the spring a distance of 4.6 cm. Find the initial speed (in m/s) of the block.

Answer:

The initial speed of the block is 1.422 m/s

Explanation:

Given;

mass of the block, m = 2.0 kg

force constant of the spring, K = 955 N/m

compression of the spring, x = 4.6 cm = 0.046 m

Apply Hook's law to determine applied force on the spring;

F = Kx

F = (955 N/m)(0.046 m)

F = 43.93 N

Apply Newton's 2nd law to determine the magnitude of deceleration of the block when it encounters the spring;

F = ma

a = F / m

a = 43.93 / 2

a = 21.965 m/s²

Apply kinematic equation to determine the initial speed of the block;

v² = u² + 2ax

where;

v is the final speed of the block = 0

u is the initial speed of the block

x is the distance traveled by the block = compression of the spring

a is the block deceleration = -21.965 m/s²

0 = u² + 2(-21.965 )(0.046)

0 = u²  - 2.021

u² =  2.021

u = √2.021

u = 1.422 m/s

Therefore, the initial speed of the block is 1.422 m/s

8 0
3 years ago
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