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2 years ago

a beaker has mass of 125 g. what is the mass of liquid if the beaker plus liquid have a mass of 232 g?

Physics

1 answer:

MrRissso [65]2 years ago

7 0

Answer: 107g

Explanation:

Just subtract 125 from 232, like this: 232-125=107g.

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Please help on this one?

bezimeni [28]

Using the given equation you get:

E = 1.99x10^-25 / 9.0x10^-6

Divide 1.99 by 9.0: 1.99/9.0 = 0.22

For the scientific notation, when dividing subtract the two exponents:

25 -6 = 19

So you now have 0.22 x 10^-19

Now you need to change the 0.22 to be in scientific notation form:

2.2 x 10^-20

The answer is B.

3 0

2 years ago

A speeding Thunderbird left skid marks on the road that were 76.7 m long when it came to a stop. If the acceleration of the car

FrozenT [24]

Answer: 39.2 m/s

Explanation:

You can use the kinematic equation:

$v_f^2=v_i^2+2a*d$

We know the final velocity because it says it came to a stop. So now all we gotta do is plug in.

$0^2=v_i^2+2(-10)(76.7)\\v_i^2=1,534\\v_i=\sqrt{1534} \\=39.166 m/s$

4 0

2 years ago

An object on the moon feels lighter than the same object on earth. which statement explains this phenomenon?

natka813 [3]

Answer:

Explanation:

cause

7 0

2 years ago

What is the highest and lowest frequency a human being can hear

nata0808 [166]

Every person is different. But for a planet-wide overall average that roughly represents all human beings on Earth, the figures usually used are:

from 20 Hz to 20,000 Hz .

6 0

2 years ago

An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is

Andreas93 [3]

Answer:

$v=1.08\times 10^7\ m/s$

Explanation:

Initial speed of the electron, u = 0

The charge per unit area of each plate, $\dfrac{Q}{A}=1.69\times 10^{-7}\ C/m^2$

Separation between the plates, $d=1.75\times 10^{-2}\ m$

An electron is released from rest, u = 0

Using equation of kinematics,

$v^2-u^2=2ad$ ..........(1)

Acceleration of the electron in electric field, $a=\dfrac{qE}{m}$ ............(2)

Electric field, $E=\dfrac{\sigma}{\epsilon_o}$ ............(3)

From equation (1), (2) and (3) :

$v=\sqrt{\dfrac{2q\sigma d}{m\epsilon_o}}$

$v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 1.69\times 10^{-7}\times 1.75\times 10^{-2}}{9.1\times 10^{-31}\times 8.85\times 10^{-12}}}$

v = 10840393.1799 m/s

$v=1.08\times 10^7\ m/s$

So, the electron is moving with a speed of $1.08\times 10^7\ m/s$ before it reaches the positive plate. Hence, this is the required solution.

3 0

3 years ago