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3 years ago

Which of the Moon’s positions will cause the highest or spring tides?

Physics

2 answers:

olya-2409 [2.1K]3 years ago

7 0

The answer is B. A and C. Spring tides occur during a new moon and full moon

Contact [7]3 years ago

5 0

Answer:

Positions A and C will cause the highest tides, called spring tides, because the gravity of the Moon and Sun will work together to create the two tidal bulges on opposite sides of Earth. So B.

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You push a 85 kg shopping cart from rest with a net force of 250 n for 5 seconds,at which point it flies off a cliff that is 100

Vikki [24]

m = mass of the cart = 85 kg

F = net force on the cart = 250 N

a = acceleration of the cart

acceleration of the cart is given as

a = F/m

a = 250/85

a = 2.94 m/s²

t = time for which the force is applied = 5 sec

v₀ = initial velocity of the cart = 0 m/s

v = final velocity of the cart just before it flies off the cliff = ?

using the equation

v = v₀ + a t

inserting the values

v = 0 + (2.94) (5)

v = 14.7 m/s

consider the motion of cart after it flies off the cliff in vertical direction :

v' = initial velocity in vertical direction = 0 m/s

a' = acceleration in vertical direction = g = acceleration due to gravity = 9.8 m/s²

t' = time taken for the cart to land = ?

Y' = vertical displacement of the cart = height of cliff = 100 m

using the kinematics equation

Y' = v' t' + (0.5) a' t'²

100 = (0) t' + (0.5) (9.8) t'²

t' = 4.52 sec

consider the motion of cart along the horizontal direction after it flies off the cliff

X = distance traveled from the base of cliff = ?

t' = time of travel = 4.52 sec

v = velocity along the horizontal direction = 14.7 m/s

distance traveled from the base of cliff is given as

X = v t'

X = 14.7 x 4.52

X = 66.4 m

3 0

3 years ago

A position vector in the first quadrant has an x-component of 18 m and a magnitude of 30 m. What is the value of its y-component

Snezhnost [94]

Answer:

The value is 24meters

Explanation:

Using

r= xi+yj

To get the magnitude of vector x

We say

/r/= √x²+y²

30²= √18² + y²

y= √576

Y= 24m

7 0

3 years ago

What kind of model is shown below?

Rudiy27

D. a foot model

btw this is a joke right cuz there ain’t no picture lol

7 0

3 years ago

Sometimes a person cannot clearly see objects close up or far away. To correct this type of vision, bifocals are often used. The

Rudik [331]

Answer:

1) P₁ = -2 D, 2) P₂ = 6 D

Explanation:

for this exercise in geometric optics let's use the equation of the constructor

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where f is the focal length, p and q are the distance to the object and the image, respectively

1) to see a distant object it must be at infinity (p = ∞)

$\frac{1}{f_1} = \frac{1}{q}$

q = f₁

2) for an object located at p = 25 cm

$\frac{1}{f_2} = \frac{1}{25} + \frac{1}{q}$

We can that in the two expressions we have the distance to the image, this is the distance where it can be seen clearly in general for a normal person is q = 50 cm

we substitute in the equations

1) f₁ = -50 cm

$\frac{1}{f_2} = \frac{1}{25} + \frac{1}{50}$

$\frac{1}{f_2}$ = 0.06

f₂ = 16.67 cm

the expression for the power of the lenses is

P = $\frac{1}{f}$

where the focal length is in meters

1) P₁ = 1/0.50

P₁ = -2 D

2) P₂ = 1 /0.16667

P₂ = 6 D

4 0

3 years ago

In a television set, electrons are accelerated from rest through a potential difference of 20 kV. The electrons then pass throug

Svetradugi [14.3K]

Answer: Fmax = 5.54*10^-12 N

Explanation: From the question, we have the potential difference (V) =20kv = 20,000v and strength of magnetic field (B) =0.41 T.

The maximum force experienced by a charge of magnitude (q) is given as

Fmax = qvB

Where v = velocity of electron.

The velocity of the electron can be gotten by using the work energy theorem.

The kinetic energy of the electron (mv²/2) equals the work done needed to accelerate it.

mv²/2 = qV.

Where m = mass of an electronic charge = 9.11×10^-31 kg, q = magnitude of an electronic charge = 1.609×10^-19 c, v = velocity of electron, V = potential difference = 20,000v.

By substituting the parameters, we have that

(9.11×10^-31 × v²)/2 = 1.609×10^-19 × 20000

(9.11×10^-31 × v²) = 1.609×10^-19 × 20000 ×2

v² = (1.609×10^-19 × 20000 ×2)/9.11×10^-31

v² = 64.36*10^(-16)/9.11×10^-31

v² = 7.0647×10^15

v = √7.0647×10^15

v = 8.40×10^7 m/s

Fmax = 1.609×10^-19 × 8.40×10^7 × 0.41

Fmax = 5.54*10^-12 N

7 0

3 years ago