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marusya05 [52]
3 years ago
9

What is the following product √10 * √10 10 10√10 100 2√10

Physics
2 answers:
insens350 [35]3 years ago
8 0
√10(√10) = √10^2

The root and the square cancels:
√10^2 = 10

10 is your answer

hope this helps
qwelly [4]3 years ago
7 0

Answer:

The result of the product √10*√10 is 10

Explanation:

To solve this product, it is first appropriate to apply the property of root multiplication with the same index.  This property indicates that multiplying two roots with the same index is the same as multiplying on a single root with that index. In this case this property is applied as follows:

\sqrt{10} *\sqrt{10} =\sqrt{10*10}

Now you can think of the next step in two ways. One of them is to multiply within the root and then solve the root as follow:

\sqrt{10*10} =\sqrt{100} =10

Another way is to express the values ​​within the root as power, because they are the same number.

\sqrt{10*10} =\sqrt{10^{2} }

The root is the opposite operation to the power, then the root nullification property can be applied: if you have an elevated number or variable to an exponent that is within a root with the same index, the power with the root is voids:

\sqrt{10^{2} } =10

Note that in both cases the result is the same, so the method you choose to resolve this account is indistinct. <em><u>The result of the product √10*√10 is 10</u></em>.

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Marina CMI [18]

Explanation:

It is given that, the force of gravity pulls down on your school with a total force of 400,000 N.

The force of gravity is given by the formula as follows :

F = mg

m is mass and g is the acceleration due to gravity

Mass of an object remains same always. But if the mass of your school becomes twice of the initial mass, the force of gravity pulling down on your school would be exactly twice.

4 0
3 years ago
A football player, with a mass of 69.0 kg, slides on the ground after being knocked down. At the start of the slide, the player
White raven [17]

Answer:

(a) -472.305  J

(b) 1 m

Explanation:

(a)

Change in mechanical energy equals change in kinetic energy

Kinetic energy is given by0.5mv^{2}

Initial kinetic energy is 0.5\times 69\times 3.7^{2}=472.305 J

Since he finally comes to rest, final kinetic energy is zero because the final velocity is zero

Change in kinetic energy is given by final kinetic energy- initial kinetic energy hence

0-472.305  J=-472.305  J

(b)

From fundamental kinematic equation

v^{2}=u^{2}+2as

Where v and u are final and initial velocities respectively, a is acceleration, s is distance

Making s the subject we obtain

s=\frac {v^{2}-u^{2}}{-2a} but a=\mu g hence

s=\frac {v^{2}-u^{2}}{-2\mu g}=\frac {0^{2}-3.7^{2}}{-2*0.7*9.81}=0.996796272\approx 1 m

7 0
3 years ago
A stone is thrown vertically upward with a speed of 18 m/s. (a) How long does it take the stone to reach a height of 11 m? (b) h
bagirrra123 [75]

Answer:

a) It takes the stone 0.7743 s to reach a height of 11 m for the first time on its way up and 2.899 s to reach again that height on its way down.

b) At t = 0.7743 s the velocity is 10.41 m/s and at t = 2.899 s the velocity is -10.41 m/s.

c) There are two answers because the stone reaches the height of 11 m one time on its way up and one more time again on its way down.

Please, see the attached figures and the explanation for a description of the figures.

Explanation:

Hi there!

The equations for the height and velocity of the stone are as follows:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height

y0 = initial height

v0 = initial velocity

t = time

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive)

v = velocity at time t

a) Let´s calculate the time it takes the stone to reach a height of 11 m. The origin of the frame of reference is at the throwing point so that y0 = 0:

y = y0 + v0 · t + 1/2 · g · t²        

11 m = 18 m/s · t - 1/2 · 9.8 m/s² · t²    

0 = -4.9 m/s² · t² + 18 m/s · t - 11 m

Solving the quadratic equation:

t = 0.7743 s and t = 2.899 s

(Notice that I have used more significant figures to avoid error by rounding)

The stone will be two times at a height of 11 m, one on its way up (at 0.7743 s) and one on its way down  (at 2.899 s). Then, it takes the stone 0.7743 s to reach a height of  11 m for the first time.

b)  Let´s use the equation of velocity:

v = v0 + g · t

at t = 0.77443 s

v = 18 m/s - 9.8 m/s² · 0.77443 s

v = 10.41 m/s

at t = 2.899 s

v = 18 m/s - 9.8 m/s² · 2.899 s

v = - 10.41 m/s

(Both velocities have to be of the same magnitude but of different sign, that´s why I haven´t rounded the time.)

c) There are two answers because the stone reaches the height of 11 m one time on its way up and one more time again on its way down. On its way up, the velocity is 10.41 m/s and on its way down it is -10.41 m/s.

Figures

The functions to plot are the following:

height in function of time (figure 1, x-axis: time. y-axis: height)

y = -4.9t² + 18t

velocity in function of time (figure 2, x-axis: time. y-axis velocity)

v = -9.8t + 18

Acceleration in function of time (figure 3, x-axis: time. y-axis: acceleration)

a = -9.8

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Answer:

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