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3 years ago

Vector A = 50 m, 20°. Vector -3A would be equal to

Physics

1 answer:

Kruka [31]3 years ago

3 0

Answer:

Many familiar physical quantities can be specified completely by giving a single number and the appropriate unit. For example, “a class period lasts 50 min” or “the gas tank in my car holds 65 L” or “the distance between two posts is 100 m.” A physical quantity that can be specified completely in this manner is called a scalar quantity. Scalar is a synonym of “number.” Time, mass, distance, length, volume, temperature, and energy are examples of scalar quantities.

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If an atom has twelve electrons, how many electrons are in the n=1 energy level? how many electrons are in the n=2 energy level?

Sav [38]

The shells correspond with the principal quantum numbers (n = 1, 2, 3, 4 ...) Each shell can contain only a fixed number of electrons. The first shell can hold up to two electrons, the second shell can hold up to eight electrons, the third shell can hold up to 18 and so on.

3 0

3 years ago

At what distance does a 100 Watt lightbulb deliver the same power per unit surface area as a 75 Watt lightbulb produces 10 m awa

ch4aika [34]

Answer:

At 11.5 m

Explanation:

The power per unit area corresponds to the intensity, which is given by

$I=\frac{P}{4\pi r^2}$

where

P is the power

$4\pi r^2$ is the area irradiated at a distance r from the source (it corresponds to the surface area of a sphere of radius r)

Here we want the intensity of the two light bulbs to be the same, so

$I_1 = I_2\\\frac{P_1}{4 \pi r_1^2}=\frac{P_2}{4\pi r_2^2}$

where we have

P1 = 100 W is the power of the first light bulb

P2 = 75 W is the power of the second light bulb

r2 = 10 m is the distance from the second light bulb

Solving for r1, we find

$r_1 = r_2 \sqrt{\frac{P_1}{P_2}}= (10 m) \sqrt{\frac{100 W}{75 W}} = 11.5 m$

4 0

3 years ago

A Young'sdouble-slit interference experiment is performed with monochromatic light. The separation between the slits is 0.44 mm.

Artyom0805 [142]

Answer:

Explanation:

The double slit interference phonemene is described for the case of constructive interference

d sin θ= m λ (1)

let's use trigonometry to find the sinus

tan θ = y / L

in general in interference phenomena the angles are small

tan θ = sin θ / cos θ = sin θ

The double slit interference phonemene is described for the case of constructive interference

d sin θ = m lam (1)

let's use trigonometry to find the sinus

tan θ = y / L

in general in interference phenomena the angles are small

tan θ = sin θ / cos θ = sin θ

we substitute

sin θ = y / L

we substitute in equation 1

d y / L = m λ

λ = dy / L m

let's reduce the magnitudes to the SI system

d = 0.44 mm = 0.44 10⁻³ m

y = 5.5 mm = 5.5 10⁻³ m

L = 4.2m

m = 1

let's calculate

λ = 0.44 10⁻³ 5.5 10⁻³ / (4.2 1)

λ = 5.76190 10-7 m

let's reduce to num

lam = 5.56190 10-7 m (109 nm / 1m)

lam = 556,190 nmtea

we substitute

without tea = y / L

we substitute in equation 1

d y / L = m lam

lam = dy / L m

let's reduce the magnitudes to the SI system

d = 0.44 me = 0.44 10-3 m

y = 5.5 mm = 5.5 10-3

L = 4.2m

m = 1

let's calculate

lam = 0.44 10⁻³ 5.5 10⁻³ / (4.2 1)

lam = 5.76190 10⁻⁷ m

let's reduce to num

lam = 5.56190 10⁻⁷ m (109 nm / 1m)

lam = 556,190 nm

6 0

4 years ago

A lightweight string is wrapped several times around the rim of a small hoop. If the free end of the string is held in place and

MariettaO [177]

Answer:

Explanation:

Let T be the tension

For linear motion of hoop downwards

mg -T = ma , m is mass of the hoop . a is linear acceleration of CG of hoop .

For rotational motion of hoop

Torque by tension

T x R , R is radius of hoop.

Angular acceleration be α,

Linear acceleration a = α R

So TR = I α

= I a / R

a = TR² / I

Putting this value in earlier relation

mg -T = m TR² / I

mg = T ( 1 + m R² / I )

T = mg / ( 1 + m R² / I )

mg / ( 1 + R² / k² )

Tension is less than mg or weight because denominator of the expression is more than 1.

5 0

3 years ago

A 6 kg block initially at rest is pulled to East along a horizontal surface with coefficient of kinetic friction μk=0.15 by a co

ozzi

Answer:

1.8 m/s

Explanation:

Draw a free body diagram of the block. There are four forces:

Normal force Fn up.

Weight force mg down.

Applied force F to the east.

Friction force Fn μ to the west.

Sum the forces in the y direction:

∑F = ma

Fn − mg = 0

Fn = mg

Sum the forces in the x direction:

F − Fn μ = ma

F − mg μ = ma

a = (F − mg μ) / m

a = (12 N − 6 kg × 9.8 m/s² × 0.15) / 6 kg

a = 0.53 m/s²

Given:

Δx = 3 m

v₀ = 0 m/s

a = 0.53 m/s²

Find: v

v² = v₀² + 2aΔx

v² = (0 m/s)² + 2 (0.53 m/s²) (3 m)

v = 1.8 m/s

8 0

3 years ago