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3 years ago

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Determine the values of mm and nn when the following mass of the Earth is written in scientific notation: 5,970,000,000,000,000,

000,000,000 kgkg.

1 answer:

nataly862011 [7]3 years ago

5 0

Answer:

"m" and "n" are 5.97 and 24 respectively.

Explanation:

Standard form which is a scientific notation [m × 10^n] can be regarded as way to reduce large figures to small one in a decimal firm and this is usually done for conviniency sake.

✓Let us find "m" , to do this we will shorten the long number so that it will be from 1-9. And this is 5.97, hence

m = 5.97

✓ " n" can be determined by counting the digits from our right hand then stop where the decimal point was put when we were finding our "m". Hence

n= 24

✓ if we input the values to the scientific expresion above, we have

m = 5.97 and n = 24

Hence, the Standard form = 5.97 × 10^24 kg which is the scientific notation.

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An insulating sphere is 8.00 cm in diameter and carries a 6.50 ÂµC charge uniformly distributed throughout its interior volume.

Kobotan [32]

Explanation:

(a) Formula to calculate the density is as follows.

$\rho = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.04)^{3}}$

= $2.42 \times 10^{-2} C/m^{3}$

Now, calculate the charge as follows.

$q_{in} = \rho(\frac{4}{3} \pi r^{3})$

= $2.42 \times 10^{-2} C/m^{3} \times 4.1762 \times (0.01)^{3}$

= $10.106 \times 10^{-8}$ C

or, = 101.06 nC

(b) For r = 6.50 cm, the value of charge will be calculated as follows.

$q_{in} = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.065)^{3}}$

= 7.454 $\mu C$

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3 years ago

Baseball player A bunts the ball by hitting it in such a way that it acquires an initial velocity of 1.3 m/s parallel to the gro

pashok25 [27]

Answer:

Explanation:

cSep 20, 2010

well, since player b is obviously inadequate at athletics, it shows that player b is a woman, and because of this, she would not be able to hit the ball. The magnitude of the initial velocity would therefore be zero.

Anonymous

Sep 20, 2010

First you need to solve for time by using

d=(1/2)(a)(t^2)+(vi)t

1m=(1/2)(9.8)t^2 vertical initial velocity is 0m/s

t=.45 sec

Then you find the horizontal distance traveled by using

v=d/t

1.3m/s=d/.54sec

d=.585m

Then you need to find the time of player B by using

d=(1/2)(a)(t^2)+(vi)t

1.8m=(1/2)(9.8)(t^2) vertical initial velocity is 0

t=.61 sec

Finally to find player Bs initial horizontal velocity you use the horizontal equation

v=d/t

v=.585m/.61 sec

so v=.959m/s

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3 years ago

Read 2 more answers

IMPORTANTT!!! SOLVEE!!! and EXPLAIN!!!

Pepsi [2]

My answer -

I believe that the answer is (A).

P.S

Happy to help you have an AWESOME!! day

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3 years ago

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MaRussiya [10]

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2 years ago

A wheel initially spinning at wo = 50.0 rad/s comes to a halt in 20.0 seconds. Determine the constant angular acceleration and t

svetlana [45]

Answer:

(I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

Explanation:

Given that,

Initial spinning = 50.0 rad/s

Time = 20.0

Distance = 2.5 m

Mass of pole = 4 kg

Angle = 60°

We need to calculate the angular acceleration

Using formula of angular velocity

$\omega=-\alpha t$

$\alpha=-\dfrac{\omega}{t}$

$\alpha=-\dfrac{50.0}{20.0}$

$\alpha=-2.5\ rad/s^2$

The angular acceleration is -2.5 rad/s²

We need to calculate the number of revolution

Using angular equation of motion

$\theta=\omega_{0}t+\dfrac{1}{2}\alpha t$

Put the value into the formula

$\theta=50\times20-\dfrac{1}{2}\times2.5\times20^2$

$\theta=500\ rad$

The number of revolution is 500 rad.

(II). We need to calculate the torque

Using formula of torque

$\vec{\tau}=\vec{r}\times\vec{f}$

$\tau=r\times f\sin\theta$

Put the value into the formula

$\tau=2.5\times4\times 9.8\sin60$

$\tau=84.87\ N-m$

Hence, (I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

8 0

3 years ago