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2 years ago

How to find instantaneous velocity physics?

1 answer:

8 0

You should have the velocity as a function of time either given explicitly or implicitly (a graph)

v = ds/dt (differentiating the position vector)

integrating the acceleration.

you can use impulse or work and energy principle and also newton law of motion to find acceleration then velocity

NOT SURE IF THAT WHAT YOU WANT.

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What never change despite what planet you go to

laiz [17]

Your mass will never change despite if you go to Jupiter, Uranus, Mars, Earth, or any planet.

7 0

3 years ago

a parallel circuit is sometimes called a because the current splits up among all the resistors in the parallel circuit

Ahat [919]

A parallel circuit is sometimes called a current divider because current splits up among all the resistors in the parallel circuit. In addition, the current through the branches is inversely proportional to the resistance of the branch. If the resistance in each branch is kept constant but the voltage is decreased, the current will decrease.

8 0

3 years ago

Carlos pushes a 3 kg box with a force of 9 newtons. The force of friction on the box is 3 newtons in the opposite direction. Wha

Ivenika [448]

The 'net' force acting on the box is (9 - 3) = 6 newtons
in the direction that Carlos is pushing.

Force = (mass) x (acceleration)

6 = (3) x (acceleration)

Divide each side by 3 :

<em>2 m/s² = acceleration</em>

4 0

3 years ago

Read 2 more answers

Two small children decide it would be fun to toss a couple of large cats at each other. Cat A (7kg) is thrown at 7m/s and cat B

Alex777 [14]

Answer:

V=4.7m/s

Explanations:

Let Ma mass of cat A=7kg

Va velocity of cat A=7m/s

Mb mass of cat b=6.1kg

VB velocity of cat b=2m/s

From conservation of linear momentum

MaVa+MbVb=(Ma+Mb)V

7*7+6.1*2=(7+6.1)V

61.2=13.1V

V=4.7m/s

3 0

3 years ago

A disk with radius R and uniform positive charge density s lies horizontally on a tabletop. A small plastic sphere with mass M a

Yanka [14]

Answer:

a. F = Qs/2ε₀[1 - z/√(z² + R²)] b. h = (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

Explanation:

a. What is the magnitude of the net upward force on the sphere as a function of the height z above the disk?

The electric field due to a charged disk with surface charge density s and radius R at a distance z above the center of the disk is given by

E = s/2ε₀[1 - z/√(z² + R²)]

So, the net force on the small plastic sphere of mass M and charge Q is

F = QE

F = Qs/2ε₀[1 - z/√(z² + R²)]

b. At what height h does the sphere hover?

The sphere hovers at height z = h when the electric force equals the weight of the sphere.

So, F = mg

Qs/2ε₀[1 - z/√(z² + R²)] = mg

when z = h, we have

Qs/2ε₀[1 - h/√(h² + R²)] = mg

[1 - h/√(h² + R²)] = 2mgε₀/Qs

h/√(h² + R²) = 1 - 2mgε₀/Qs

squaring both sides, we have

[h/√(h² + R²)]² = (1 - 2mgε₀/Qs)²

h²/(h² + R²) = (1 - 2mgε₀/Qs)²

cross-multiplying, we have

h² = (1 - 2mgε₀/Qs)²(h² + R²)

expanding the bracket, we have

h² = (1 - 2mgε₀/Qs)²h² + (1 - 2mgε₀/Qs)²R²

collecting like terms, we have

h² - (1 - 2mgε₀/Qs)²h² = (1 - 2mgε₀/Qs)²R²

Factorizing, we have

[1 - (1 - 2mgε₀/Qs)²]h² = (1 - 2mgε₀/Qs)²R²

So, h² = (1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]

taking square-root of both sides, we have

√h² = √[(1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]]

h = (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

4 0

2 years ago