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qaws [65]
3 years ago
12

If a dice is thrown at random find the probability of getting 4​

Physics
2 answers:
Brrunno [24]3 years ago
8 0

Answer:

☆<《<em><u>HOPE IT WILL HELP YOU</u></em><em><u> </u></em>》>☆

Explanation:

The probability of getting 4 is 1

Serhud [2]3 years ago
5 0

Answer:

1/6

Explanation:

There are 6 faces, and one of them is 4. Therefore the answer is 1/6.

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A roller coaster starts from rest at point a and continues to point b on a frictionless track. what best describes the changes i
saw5 [17]

<span>At first when the car is at the datum point where is no elevation, the kinetic energy increases while the potential energy is zero. As it travel the path and goes upward the kinetic enrgy decreases while the potential energy increases. When it goes down again the kinetic enrgy increases again while the potential energy decreases </span>

3 0
3 years ago
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Carbon dioxide being released from a fire extinguisher a good example of how volume __________.
kirill [66]
"<span>increases when pressure decreases". Pressure and volume of gasses are related from Boyle's law, which states that Pressure is proportional to 1/V, so as pressure decreases, volume increases. </span>
6 0
3 years ago
HELP ASAP TIMED TEST
balu736 [363]

Answer:

<em>Correct choice: b 4H</em>

Explanation:

<u>Conservation of the mechanical energy</u>

The mechanical energy is the sum of the gravitational potential energy GPE (U) and the kinetic energy KE (K):

E = U + K

The GPE is calculated as:

U = mgh

And the kinetic energy is:

\displaystyle K=\frac{1}{2}mv^2

Where:

m = mass of the object

g = gravitational acceleration

h = height of the object

v = speed at which the object moves

When the snowball is dropped from a height H, it has zero speed and therefore zero kinetic energy, thus the mechanical energy is:

U_1 = mgH

When the snowball reaches the ground, the height is zero and the GPE is also zero, thus the mechanical energy is:

\displaystyle U_2=\frac{1}{2}mv^2

Since the energy is conserved, U1=U2

\displaystyle mgH=\frac{1}{2}mv^2    \qquad\qquad [1]

For the speed to be double, we need to drop the snowball from a height H', and:

\displaystyle mgH'=\frac{1}{2}m(2v)^2

Operating:

\displaystyle mgH'=4\frac{1}{2}m(v)^2 \qquad\qquad [2]

Dividing [2] by [1]

\displaystyle \frac{mgH'}{mgH}=\frac{4\frac{1}{2}m(v)^2}{\frac{1}{2}m(v)^2}

Simplifying:

\displaystyle \frac{H'}{H}=4

Thus:

H' = 4H

Correct choice: b 4H

4 0
3 years ago
The initial and final velocities of two blocks experiencing constant acceleration are respectively −7.45 m/s and 14.9 m/s. (a) T
ikadub [295]

Answer:

a) a_{1}=3.7 m/s^{2}

b) a_{2}=3.68 m/s^{2}

Explanation:

a) The displacement of the first object is 22.5 m, so we can use the next equation:

v_{f}^{2}=v_{i}^{2}+2a\Delta x

a=\frac{v_{f}^{2}-v_{i}^{2}}{2x}

a=\frac{14.9^{2}-(-7.45)^{2}}{2*22.5}

a_{1}=3.7 m/s^{2}

positive acceleration.

b) Using the same equation we can find the second value of the acceleration:

a=\frac{v_{f}^{2}-v_{i}^{2}}{2x}

a=\frac{14.9^{2}-(-7.45)^{2}}{2*22.6}

a_{2}=3.68 m/s^{2}

positive acceleration.

I hope it helps you!

8 0
3 years ago
g In 1956, Frank Lloyd Wright proposed the construction of a mile-high building in Chicago. Suppose the building had been constr
Lorico [155]

To solve this problem it is necessary to apply the concepts related to acceleration due to gravity, as well as Newton's second law that describes the weight based on its mass and the acceleration of the celestial body on which it depends.

In other words the acceleration can be described as

a = \frac{GM}{r^2}

Where

G = Gravitational Universal Constant

M = Mass of Earth

r = Radius of Earth

This equation can be differentiated with respect to the radius of change, that is

\frac{da}{dr} = -2\frac{GM}{r^3}

da = -2\frac{GM}{r^3}dr

At the same time since Newton's second law we know that:

F_w = ma

Where,

m = mass

a =Acceleration

From the previous value given for acceleration we have to

F_W = m (\frac{GM}{r^2} ) = 600N

Finally to find the change in weight it is necessary to differentiate the Force with respect to the acceleration, then:

dF_W = mda

dF_W = m(-2\frac{GM}{r^3}dr)

dF_W = -2(m\frac{GM}{r^2})(\frac{dr}{r})

dF_W = -2F_W(\frac{dr}{r})

But we know that the total weight (F_W) is equivalent to 600N, and that the change during each mile in kilometers is 1.6km or 1600m therefore:

dF_W = -2(600)(\frac{1.6*10^3}{6.37*10^6})

dF_W = -0.3N

Therefore there is a weight loss of 0.3N every kilometer.

4 0
3 years ago
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