In 1913 Niels Bohr formulated a method of calculating the different energy levels of the hydrogen atom.

A bird watcher meanders through the woods, walking 1.93 km due east, 1.03 km due south, and 3.84 km in a direction 52.8 ° north

Sedaia [141]

Answer:

Magnitude of displacement = 2.07 km

Magnitude of average velocity = 1.17 kmph

Explanation:

Let east represent positive x axis and north represent positive y axis.

A bird watcher meanders through the woods, walking 1.93 km due east, 1.03 km due south, and 3.84 km in a direction 52.8 ° north of west.

1.93 km due wast

s ₁ = 1.93 i km

1.03 km due south

s₂ = -1.03 j km

3.84 km in a direction 52.8 ° north of west

s₃ = -3.84 cos 52.8 i + 3.84 sin 52.8 j = -2.32 i + 3.06 j km

Total displacement

s = s ₁+ s₂+ s₃ = 1.93 i - 1.03 j -2.32 i + 3.06 j = -0.39 i + 2.03 j

Magnitude of displacement, $=\sqrt{(-0.39)^2+2.03^2}=2.07km$

Time taken = 1.771 hour

Magnitude of average velocity, $=\frac{2.07}{1.771}=1.17km/hr$

7 0

2 years ago

4. Define resistance and describe what would happen to a light bulb if the voltage increased but the resistance stayed the same.

Arlecino [84]

Current would increase proportionally to voltage. Power dissipation (heating) would increase with the square of the voltage. And resistance means, "the refusal to accept or comply with something"

4 0

2 years ago

A supply bag is dropped from a rescue plane. After the bag falls for 3.2 seconds , what is the velocity of the bag?

loris [4]

Answer: -31.36 m/s

Explanation:

This is a problem of motion in one direction (specifically vertical motion), and the equation that best fulfills this approach is:

$V_{f}=V_{o}+a.t$ (1)

Where:

$V_{f}$ is the final velocity of the supply bag

$V_{o}=0$ is the initial velocity of the supply bag (we know it is zero because we are told it was "dropped", this means it goes to ground in free fall)

$a=g=-9.8m/s^{2}$ is the acceleration due gravity (the negtive sign indicates the gravity is downwards, in the direction of the center of the Earth)

$t=3.2s$ is the time

Knowing this, let's solve (1):

$V_{f}=0+(-9.8m/s^{2})(3.2s)$ (2)

Finally:

$V_{f}=-31.36m/s$ Note the negative sign is because the direction of the bag is downwards as well.

8 0

2 years ago

Water of density 1000 kg/m3 falls without splashing at a rate of 0.373 L/s from a height of 40.5 m into a 0.64 kg bucket on a sc

Sphinxa [80]

Answer:

F_scale = 20.18 N

Explanation:

The scale reading corresponds to two factors, the first the weight of the water in the container and the second the force of the liquid that is falling at the moment of reading.

* Let's find the amount of liquid in the container for a time of t = 2.93 s

Let's use a direct proportion rule. If 0.373 l falls in one second at t = 2.93 s, how many liters are there

V_{water} = 2.93 s (0.373 l / 1s) = 1.09 l

V_{water} = 1.09 10⁻³ m³

the amount of water is

ρ = m / V

m = ρ V

m = 1000 1.09 10⁻³

m = 1.09 kg

so the weight of the liquid in the container for this time is

W = mg

W = 1.09 9.8

W = 10.68 N

* Let's look for the force of the falling jet

Let's use Bernoulli's equation, where the subscript 1 is for the container and the subscript 2 is for the water at a height h

P₁ + 1/2 ρ g v₁² + ρ g y₁ = P₂ + 1/2 ρ g v₂² + ρ g y₂

In this case, the water falls freely, so the external pressure is atmospheric.

P₂ = P_{atm}

since they indicate that the water falls, we assume that its initial velocity is zero v₂ = 0

let's use kinematics to find the speed of a drop when it reaches the container y = 0

v² = v₀² - 2 g (y-y₀)

v = $\sqrt{0 -2 g ( 0-y_o)}$

let's calculate

v = √(2 9.8 40.5)

v = 28.17 m / s

this is the speed in the container v₁ = 28.17 m / s

the height from where it falls is y₂ = 40.5 and reaches the container y₁ = 0

we substitute in Bernoulli's equation

P₁ +1/2 ρ g v₁² + 0 = P_{atm} + 0 + ρ g y₂

P₁ + ½ ρ g v₁² = P_{atm} + ρ g y₂

P₁ = P_{atm} + ρ g y₂ - ½ ρ g v₁²

P₁ = 1 10⁵ + 1000 9.8 40.5 - ½ 1000 28.17²

P₁ = 1 10⁵ + 3.97 10⁵ - 3.69 10⁵

P₁ = 1.28 10⁵ Pa

The definition of Pressure is

P = F / A

F = P A

We must suppose a time to carry out the reading suppose an average time of the modern equipment t = 0.1 s, in this time how much is now arriving

m₂ = 0.373 0.2 = 0.0746 l = 0.0746 10⁻³ m³

the volume is V = A l

if the length of l = 1 m

A = 0.0746 10⁻³ m³ = 7.45 10⁻⁵ m²

the force of this jet is

F = P A

F = 1.28 10⁵ 7.46 10⁻⁵

F = 9.5 N

with these data let's use the equilibrium equation

F_ scale -W - F = 0

F_scale = W + F

F_scale = 10.682 + 9.5

F_scale = 20.18 N

4 0

2 years ago

Speedy Sue, driving at 34.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 160 m ahead traveling at 5.20 m/s

Zielflug [23.3K]

Answer:

there will be collision

Explanation:

$v_{s}$ = speed of sue = 34 m/s

$v_{v}$ = speed of van = 5.20 m/s

$v_{sv}$ = speed of sue relative to van = $v_{s} - v_{v}$ = 34 - 5.20 = 28.8 m/s

$d_{s}$ = stopping distance after brakes are applied

$D$ = distance between sue and van = 160 m

$v_{f}$ = final speed of sue = 0 m/s

$a$ = acceleration = - 1.80 m/s²

Using the kinematics equation

$v_{f}^{2} = v_{o}^{2} + 2 a d_{s}$

$0^{2} = 28.8^{2} + 2 (1.80) d_{s}$

$d_{s} = 230.4$ m

Since $d_{s} < D$

hence there will be collision

7 0

2 years ago