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3 years ago

A ship leaves port and travels 50 miles at a standard position of 45⁰. The ship

Physics

1 answer:

laiz [17]3 years ago

8 0

Answer:

t = 4.468 h

Explanation:

For this exercise let's start by calculating the distance where the boat is

the first trip is 50 miles at 45 °, then 90 miles at 80 °,

to find the total distance let's find the distance of each displacement

cos 45 = x₁ / 50

sin 45 = y₁ / 50

x₁ = 50 cos 45 = 35.35 miles

y₁ = 50 cos 45 = 35.35 miles

cos 80 = x₂ / 90

sin 80 = y₂ / 90

x₂ = 90 cos 80 = 15.63 miles

y₂ = 90 sin 80 = 88.63 miles

let's find the total displacement in each axis

x_total = x₁ + x₂

x_total = 35.35 + 15.63

x_total = 50.98 miles

y_total = y₁ + y₂

y_total = 35.35 + 88.63

y_total = 123.98 miles

Let's use the Pythagorean theorem to find the modulus of the displacement

R = √ (x_total² + y_total²)

R = √ (50.98² + 123.98²)

R = 134.05 miles

The boat goes at a constant speed,

v = R / t

t = R / v

let's calculate

t = 134.05 / 30

t = 4.468 h

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The chemical formula for sodium fluoride is A) Na7F B) NaF C) NaF2 D) NaF7

ikadub [295]

The correct answer would be the letter B.) NaF this attracts opposite charges. The final formula of sodium fluoride would be NaF it’s compound is formed by the complete transfer of electrons from a metal to a nonmetal.

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3 years ago

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The viewing screen in a double-slit experiment with monochromatic light. Fringe C is the central maximum. The fringe separation

makvit [3.9K]

Answer:

Part A:

a) If the wavelength of the light is decreased the fringe spacing Δy will decrease.

Part B:

b) If the spacing between the slits is decreased the fringe spacing Δy will increase.

Part C:

a) If the distance to the screen is decreased the fringe spacing will decrease.

Part D:

The dot in the center of fringe E is $920\ x\ 10^{-9} m$ farther from the left slit than from the right slit.

Explanation:

In the double-slit experiment there is a clear contrast between the dark and bright fringes, that indicate destructive and constructive interference respectively, in the central peak and then is less so at either side.

The position of bright fringes in the screen where the pattern is formed can be calculated with

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

$m=0,\pm 1,\pm 2,\pm 3,.....$

m is the order number.
$\lambda$ is the wavelength of the monochromatic light.
L is the distance between the screen and the two slits.
d is the distance between the slits.

Part A: a) In the above equation for the position of bright fringes we can see that if the wavelength of the light $\lambda$ is decreased the overall effect will be that the fringes are going to be closer. That means that the fringe spacing Δy will decrease.

Part B: b) In the above equation for the position of bright fringes we can see that if the spacing between the slits d is decreased the fringes are going to be wider apart. That means the fringe spacing Δy will increase.
Part C: a) In the above equation we can see that if the distance to the screen L is decreased the fringes are going to be closer. That means the fringe spacing Δy will decrease.

Part D: We are told that the central maximum is the fringe C that corresponds with m=0. That means that fringe E corresponds with the order number m=2 if we consider it to be the second maximum at the rigth of the central one. To calculate how much farther from the left slit than from the right slit is a dot located at the center of the fringe E in the screen we use the condition for constructive interference. That says that the path length difference Δr between rays coming from the left and right slit must be $\vartriangle r=m \lambda$

We simply replace the values in that equation :

$\vartriangle r= m \lambda =2.\ 460\ nm$

$\vartriangle r= 920\ x\ 10^{-9} m$

The dot in the center of fringe E is $920\ x\ 10^{-9}m$ farther from the left slit than from the right slit.

3 0

3 years ago

A projectile of mass 1.800 kg approaches a stationary target body at 4.800 m/s. The projectile is deflected through an angle of

Artyom0805 [142]

Answer:

P = 5.22 Kg.m/s

Explanation:

given,

mass of the projectile = 1.8 Kg

speed of the target = 4.8 m/s

angle of deflection = 60°

Speed after collision = 2.9 m/s

magnitude of momentum after collision = ?

initial momentum of the body = m x v

= 1.8 x 4.8 = 8.64 kg.m/s

final momentum after collision

momentum along x-direction

P_x = m v cos θ

P_x = 1.8 x 2.9 x cos 60°

P_x = 2.61 kg.m/s

momentum along y-direction

P_y = m v sin θ

P_y = 1.8 x 2.9 x sin 60°

P_y = 4.52 kg.m/s

net momentum of the body

$P = \sqrt{P_x^2 + P_y^2}$

$P = \sqrt{4.52^2 + 2.61^2}$

P = 5.22 Kg.m/s

momentum magnitude after collision is equal to P = 5.22 Kg.m/s

5 0

3 years ago

When air resistance equals the weight of an object, the object has reached

MissTica

Answer:

When air resistance equals the weight of an object, the object has reached free fall.

Explanation:

When an object has only force acting on it as gravity then, it experiences free fall.
During free fall all the forces except gravity is balanced by one another.
In the question, object's weight is balanced by air resistance so it is in the state of free fall.
At the null point of free fall, object experiences weightlessness i.e. it feels like object is not attracted by any force.

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3 years ago

A magic medallion is suspended from a string inside a compartment of Hogwarts Express which is running straight westwards on hor

Serggg [28]

Answer:

a = 1 m/s² and

Explanation:

The first two parts can be seen in attachment

We use Newton's second law on each axis

Y axis

Ty - W = 0

Ty = w

X axis

Tx = m a

With trigonometry we find the components of tension

Sin θ = Ty / T

Ty = T sin θ

Cos θ = Tx / T

Tx = T cos θ

We calculate the acceleration with kinematics

Vf = Vo + a t

a = (Vf -Vo) / t

a = (20 -10) / 10

a = 1 m/s²

We substitute in Newton's equations

T Sin θ = mg

T cos θ = ma

We divide the two equations

Tan θ = g / a

θ = tan⁻¹ (g / a)

θ = tan⁻¹ (9.8 / 1)

θ = 84º

We see that in the expression of the angle the mass does not appear therefore you should not change the angle

4 0

2 years ago