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3 years ago

Which is larger 15mm or 0.15cm

Physics

1 answer:

kompoz [17]3 years ago

4 0

Answer:

1 cm = 10mm

Explanation:

hey there both are equal

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particles q1, q2, and q3 are in a straight line. particles q1 = -1.60 x10^-19 C, q2 = +1.60 x10^-19 C, q3 = -1.60 x 10^-19. .001

lozanna [386]

Q2 is 1.60x10^-19.

because if you round up or add, you will come with q2

3 0

3 years ago

Read 2 more answers

A What quantity is equal to

natulia [17]

Answer:

Option B is correct

Explanation:

acceleration is equal to rate of change of velocity.

6 0

3 years ago

Use Newton’s method to solve the equation0 =12+14x2−xsin(x)−12cos(2x),withx0=π2.Iterate using Newton’s method until an accuracy

Charra [1.4K]

Answer:

given function is

$f(x)=12+14x^2-xsin(x)-12cos(2x)$

$x_{0} =\pi ^2$

formula for newton's method is

$x_{n+1} =x_{n} -f(x_{n})/f'(x_{n} )$

so derivative of function is

$f'(x)=28x-sin(x)-xcos(x)+24sin(2x)$

now put values and solve

or you can also use MATLAB code to solve

i.e

function p= newton(x)

e=0.001;

for i=1:100

if abs(d(x))>e

if abs(k(x))>0

xm=x-(k(x)/d(x));

x=xm;

else

end

break;

end

disp(x)

disp(k(x))

return;

7 0

3 years ago

If zak's speed is 3.00 m/s when he starts to slide, what distance d will he slide before stopping? express your answer in meters

OverLord2011 [107]

That will depend on the coefficient of friction between the sliding surfaces, and also on Zak's weight. We don't have any of that information.

8 0

3 years ago

An atom of uranium 238 emits an alpha particle (an atom of He) and recoils with a velocity of 1.895 * 10^ 5 m/sec . With velocit

lora16 [44]

<u>Answer:</u> The velocity of released alpha particle is $1.127\times 10^7m/s$

<u>Explanation:</u>

According to law of conservation of momentum, momentum can neither be created nor be destroyed until and unless, an external force is applied.

For a system:

$m_1v_1=m_2v_2$

where,

$m_1\text{ and }v_1$ = Initial mass and velocity

$m_2\text{ and }v_2$ = Final mass and velocity

We are given:

$m_1=238u\\v_1=1.895\times 10^{5}m/s\\m_2=4u\text{ (Mass of }\alpha \text{ -particle)}\\v_2=?m/s$

Putting values in above equation, we get:

$238\times 1.895\times 10^5=4\times v_2\\\\v_2=\frac{238\times 1.895\times 10^5}{4}=1.127\times 10^7m/s$

Hence, the velocity of released alpha particle is $1.127\times 10^7m/s$

4 0

3 years ago