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3 years ago

What changes depending on location in the universe, mass or weight. Explain

Physics

2 answers:

valentinak56 [21]3 years ago

7 0

Answer:

Weight

Explanation:

The difference between mass and weight is that mass is the amount of matter in a material, while weight is a measure of how the force of gravity acts upon that mass. The amount of matter in a material is a characteristic of a given object. But, the force of gravity that acts upon the object's mass depends of where the object is placed. For example, placed on Earth, some force is corroborated, but on the Moon, the force of gravity would be different because Moon has a different acceleration of gravity.

Black_prince [1.1K]3 years ago

3 0

Weight changes. Mass should remain the same. Weight is the effect of mass in a gravity field, so on Earth, you'll with 180 pounds, but on the moon, only 30 - while your mass remains the same in both places.

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The first law states that “objects at rest and objects in motion remain in motion in a straight line unless acted upon by an unbalanced force”. Keeping the ice smooth will make sure there is not friction, friction would slow the puck down

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3 years ago

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Analyzing a blood sample usually involves which type of separation method?

schepotkina [342]

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3 years ago

Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 kg), each having (a) a speed o

Ad libitum [116K]

Answer:

Part A:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 6.05x10^{-14}m$ < $\lambda_{electron} = 1.10x10^{-10}m$

Part B:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 1.29x10^{-13}m$ < $\lambda_{electron} = 5.525x10^{-12}m$

Explanation:

The wavelength of each particle can be determined by means of the De Broglie equation.

$\lambda = \frac{h}{p}$ (1)

Where h is the Planck's constant and p is the momentum.

$\lambda = \frac{h}{mv}$ (2)

Part A

Case for the electron:

$\lambda = \frac{6.624x10^{-34} J.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

But $J = Kg.m^{2}/s^{2}$

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

$\lambda = 1.10x10^{-10}m$

Case for the proton:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(6.55x10^{6}m/s)}$

$\lambda = 6.05x10^{-14}m$

Hence, the proton has a smaller wavelength than the electron.

Part B 

For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:

$KE = \frac{1}{2}mv^{2}$

$2KE = mv^{2}$

$v^{2} = \frac{2KE}{m}$

$v = \sqrt{\frac{2KE}{m}}$ (3)

Case for the electron:

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{9.11x10^{-31}Kg}}$

but $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{9.11x10^{-31}Kg}}$

$v = 1.316x10^{8}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(1.316x10^{8}m/s)}$

$\lambda = 5.525x10^{-12}m$

Case for the proton :

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{1.67x10^{-27}Kg}}$

But $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{1.67x10^{-27}Kg}}$

$v = 3.07x10^{6}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(3.07x10^{6}m/s)}$

$\lambda = 1.29x10^{-13}m$

Hence, the proton has a smaller wavelength than the electron.

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3 years ago

Which of the following depends on an object's displacement not distance from the starting point?

spayn [35]

The answer is A) velocity, because velocity is speed and displacement.
Hope this helps

6 0

3 years ago

If μs is greater than some critical value, the woman cannot start the crate moving no matter how hard she pushes. calculate this

fiasKO [112]

weight = mg acts downwards 
normal force = N acts upwards.
and force F acts at an angle θ below the horizontal.
(Let us assume that the woman pushes from the left, so F is acted towards the right, which is below the horizontal)
so that, Frictional force, f=us*N acts towards the left

Now we balance the forces along x and y directions:
y direction: N = mg + F sinΘ
x direction: us * N = F cosΘ

We let the value of µs be equal to a value such that any F will not be able to move the crate. Then, if we increase F by an amount F', then the force pushing the crate towards the right also increases by F' cosΘ. Additionally, the frictional force f must raise by exactly this amount.
Since f can’t exceed us*N, so the normal force must increase by F' cosΘ/us.
Also, from the y direction equation, the normal force exceeds by F' sin Θ.

These two values must be the same, therefore:
us = cot θ

4 0

3 years ago