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3 years ago

Multi-Select Question

Physics

1 answer:

dalvyx [7]3 years ago

7 0

Answer:

Correct Options :

Acceleration on Earth: 1.6 m/s

Mass on the Moon: 100 kg

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A small 12.00 g plastic ball is suspended by a string in a uniform, horizontal electric field. If the ball is in equilibrium whe

ArbitrLikvidat [17]

Answer: Here is the complete question:

A small 12.00g plastic ball is suspended by a string in a uniform, horizontal electric field with a magnitude of 103 N/C. If the ball is in equilibrium when the string makes a 30 angle with the vertical, what is the net charge on the ball?

Answer: The charge on the ball is 5.71 × 10^-4 C

Explanation:

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3 years ago

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kicyunya [14]

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3 years ago

how is the position of electrons involved in metallic bonding different from the position of electrons that form ionic and coval

Yuri [45]

While ionic bonds join metals to nonmetals, and covalent bonds join nonmetals to nonmetals, metallic bonds are responsible for the bondingbetween metal atoms. In metallic bonds, the valence electrons from the s and p orbitals of the interacting metal atoms delocalize.

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3 years ago

A football player who weighs 550 N stands indoors wearing her football boots. The boot’s

EleoNora [17]

Answer:

183333 Pa

Explanation:

The weight of the football player is : 550 N ,thus the force the player exerts on the floor is 550 N

The area of blades in contact with the floor is = 30cm² = 0.003 m²

Pressure = Force / Area

Pressure = 550 / 0.003

Pressure = 183333 Pa

3 0

2 years ago

Green light (λ = 518 nm) strikes a single slit at normal incidence. What width slit will produce a central maximum that is 3.00

MariettaO [177]

Answer:

6.9066 × 10⁻⁵ m

Explanation:

For constructive interference, the expression is:

$d\times sin\theta=m\times \lambda$

Where, m = 1, 2, .....

d is the distance between the slits.

The formula can be written as:

$sin\theta=\frac {\lambda}{d}\times m$ ....1

The location of the bright fringe is determined by :

$y=L\times tan\theta$

Where, L is the distance between the slit and the screen.

For small angle , $sin\theta=tan\theta$

So,

Formula becomes:

$y=L\times sin\theta$

Using 1, we get:

$y=L\times \frac {\lambda}{d}\times m$

Thus, the distance between the central maximum is 3.00 cm

First bright fringe , m = 1 occur at 3.00 / 2 = 1.50 cm

Since,

1 cm = 0.01 m

y = 0.0150 m

Given L = 2.00 m

λ = 518 nm

Since, 1 nm = 10⁻⁹ m

So,

λ = 518 × 10⁻⁹ m

Applying the formula as:

$0.0150\ m=2.00\ m\times \frac {518\times 10^{-9}\ m}{d}\times 1$

⇒ d, distance between the slits = 6.9066 × 10⁻⁵ m

7 0

2 years ago