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3 years ago

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How do you do this? I know that the Density equation is D=m/v but these are in kilograms and Liters...so do I convert it to gram

s and ml? Or do I just solve the equation like I normally would?

1 answer:

BARSIC [14]3 years ago

8 0

Explanation:

The problem doesn't specify that the units have to be g/mL, so you can calculate the density in kg/L without converting the mass or volume.

Just make sure that either way, you write the units.

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Write a program that calculates the average of n integers. the program should prompt the user to enter the value for n and then

ikadub [295]

#include<studio.h>

int main( )

{

int n;

int a,b,c,d,x,y;

int avarage;

printf("enter value of n:\n");

scanf("%d",&n);

printf("enter value of a:\n,b:\n,c:\n,d:\n,x:\n,y:\n);

scan f("%d\%d\n%d\n%d\n%d\n%d\n",&a,b,c,d,x,y);

sum=(a+b+c+d+x+y);

avarage=(sum/n);

print f("%d",avarage);

if

{

n=positive interger

}

else

{

printf ("n must be positive");

}

return 0;

}

8 0

2 years ago

If the spring constant is doubled, what value does the period have for a mass on a spring? A. The period would double by square

antoniya [11.8K]

Answer:

The period would decrease by sqrt(2)

Explanation:

The restoring force is given by,

F = -kx

According to Newton's second law of motion,

ma = -kx

ma + kx = 0

The time period is given by,

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

Where $\omega$ is the angular velocity and it is given by,

$\omega$ = $\sqrt{\frac{k}{m} }$

Now if the spring constant is doubled then,

$k_{2} = 2k$

Thus,

$T_{2}$ = $\frac{2\pi }{\sqrt{\frac{2k}{m} } }$

$\frac{T_{2} }{T} = \frac{\frac{2\pi }{\sqrt{\frac{2k}{m} } }}{\frac{2\pi }{\sqrt{\frac{k}{m} } }}$

$\frac{T_{2} }{T} = \sqrt{\frac{k}{2k} } = \sqrt{\frac{1}{2} }$

$T_{2} = \frac{T}{\sqrt{2} }$

Thus, The period would decrease by sqrt(2).

Hence, option D is correct.

3 0

2 years ago

Read 2 more answers

Question C) needs to be answered, please help (physics)

Zarrin [17]

(a) Differentiate the position vector to get the velocity vector:

<em>r</em><em>(t)</em> = (3.00 m/s) <em>t</em> <em>i</em> - (4.00 m/s²) <em>t</em>² <em>j</em> + (2.00 m) <em>k</em>

<em>v</em><em>(t)</em> = d<em>r</em>/d<em>t</em> = (3.00 m/s) <em>i</em> - (8.00 m/s²) <em>t</em> <em>j</em>

<em></em>

(b) The velocity at <em>t</em> = 2.00 s is

<em>v</em> (2.00 s) = (3.00 m/s) <em>i</em> - (16.0 m/s) <em>j</em>

<em></em>

(c) Compute the electron's position at <em>t</em> = 2.00 s:

<em>r</em> (2.00 s) = (6.00 m) <em>i</em> - (16.0 m) <em>j</em> + (2.00 m) <em>k</em>

The electron's distance from the origin at <em>t</em> = 2.00 is the magnitude of this vector:

||<em>r</em> (2.00 s)|| = √((6.00 m)² + (-16.0 m)² + (2.00 m)²) = 2 √74 m ≈ 17.2 m

(d) In the <em>x</em>-<em>y</em> plane, the velocity vector at <em>t</em> = 2.00 s makes an angle <em>θ</em> with the positive <em>x</em>-axis such that

tan(<em>θ</em>) = (-16.0 m/s) / (3.00 m/s) ==> <em>θ</em> ≈ -79.4º

or an angle of about 360º + <em>θ</em> ≈ 281º in the counter-clockwise direction.

3 0

3 years ago

Eddie and Val observed the picture of an athlete running in a race.

dedylja [7]

Answer:

Your answer would be C <u><em>Hope this helps</em></u>

8 0

3 years ago

A record is dropped vertically onto a freely rotating (undriven) turntable. Frictional forces act to bring the record and turnta

alexgriva [62]

Answer:

The loss of initial Kinetic energy = 37.88 %

Explanation:

Given:

Rotational inertia of the turntable = $I_t$

Rotational inertia ( $I_r$ ) of the record = $0.61\times I_t$

According to the question:

<em>Frictional forces act to bring the record and turntable to a common angular speed.</em>

So,angular momentum will be conserved as it is an inelastic collision.

Considering the initial and final angular velocity of the turn table as $\omega _i\ ,\ \omega_f$ respectively.

Note :

Angular momentum $(L)$ = Product of moment of inertia $(I)$ and angular velocity $(\omega)$ .

Lets say,

⇒ initial angular momentum = final angular momentum

⇒ $L_i=L_f$

⇒ $(I_t)\times \omega_i = (I_t+I_r)\times \omega_f$

⇒ $\omega _f=\frac{I_t}{I_t+I_r} \times (\omega_i)$ ...equation (i)

Now we will find the ratio of the Kinetic energies.

⇒ $K_i=\frac{I_t\times \omega_i^2}{2}$ ⇒ $K_f=\frac{(I_r+I_t)\times \omega_f^2}{2}$

Their ratios:

⇒ $\frac{K_f}{K_i} =\frac{\frac{(I_t+I_r)\times \omega_f^2}{2} }{\frac{I_t\times \omega_i^2}{2} }$

⇒ $\frac{K_f}{K_i} = {\frac{(I_t+I_r)\times \omega_f^2}{2} } \times {\frac{2}{I_t\times \omega_i^2}}$

Plugging the values of $\omega _f^2$ as $\omega _f^2 =(\frac{I_t}{I_t+I_r} \times \omega_i\ )^2$ from equation (i) in the ratios of the Kinetic energies.

⇒ $\frac{K_f}{K_i} =\frac{(I_t+I_r)\times \frac{(I_t)^2}{(I_t+I_r)^2} \times \omega_i^2}{I_t\times \omega_i^2} =\frac{(I_t)^2}{(I_t+I_r)}\times \frac{1}{I_t}=\frac{I_t}{I_t+I_r}$

Now,

The Kinetic energy lost in fraction can be written as:

⇒ $\frac{K_f-K_i}{K_i}$

Now re-arranging the terms.

$\frac{K_f-K_i}{K_i} =(\frac{K_f}{K_i} -1)= \frac{I_t}{I_t+I_r} -1=\frac{I_t-I_t-I_r}{I_t+I_r} =\frac{-I_r}{(I_t+I_r)}$

Plugging the values of $I_r$ and $I_t$ .

⇒ $\frac{K_f}{K_i} = \frac{-0.61I_t}{0.61I_t+I_t} =\frac{-0.61}{1.61} =-0.3788$

To find the percentage we have to multiply it with $100$ and here negative means for loss of Kinetic energy.

⇒ $\frac{K_f}{K_i} = =-0.3788\times 100= 37.88$

So the percentage of the initial Kinetic energy lost is 37.88

4 0

2 years ago