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3 years ago

Which description accurately describes the tide represented by the image below?

Physics

2 answers:

igomit [66]3 years ago

8 0

The gravitational pull of the sun and moon combined
create larger than normal tides.

luda_lava [24]3 years ago

7 0

Its the first statement:

The gravitational pull of the sun and moon combined create larger than normal tides.
-It's especially in this event, when the sun and moon are aligned, that the tides rise higher

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If a baseball is in space and someone hits it with a bat what does the ball do

Darina [25.2K]

Answer:

It's will fly like float

Explanation:

Since in space there is no gravity that everything can float

3 0

3 years ago

A 30.0-kg girl in a swing is pushed to one side and held at rest by a horizontal force \vec{F} F ⃗ so that the swing ropes

Virty [35]

Answer:

169.74 N

Explanation:

Given,

Mass of the girl = 30 Kg

angle of the rope with vertical, θ = 30°

equating the vertical component of the tension

vertical component of the tension is equal to the weight of the girl.

T cos θ = m g

T cos 30° = 30 x 9.8

T = 339.48 N

Tension on the two ropes is equal to 339.48 N

Tension in each of the rope = T/2

= 339.48/2 = 169.74 N

Hence, the tension in each of the rope is equal to 169.74 N

7 0

3 years ago

A wire with a resistance of 20 is connected to a 12 battery. What is the current flowing through the wire?

Dima020 [189]

15.49 should be the answer if that is 12 watt battery.

3 0

3 years ago

A long solenoid that has 810 turns uniformly distributed over a length of 0.380 m produces a magnetic field of magnitude 1.00 10

IRINA_888 [86]

Answer:

0.037 A

Explanation:

Magnetic field = B = 1.00 e-4 T

Length = L = 0.380 m

Number of turns = 810

B = μ₀ N I / L

⇒ Current = I = B L / μ₀ N = ( 1 e-4) ( 0.380) / (4π × 10⁻⁷)(810)

= 0.037 A = 37.3 mA

5 0

3 years ago

Suppose you first walk 12.0 m in a direction 20? west of north and then 20.0 m in a direction 40.0? south of west. how far are y

Gnesinka [82]

The representation of this problem is shown in Figure 1. So our goal is to find the vector $\overrightarrow{R}$ . From the figure we know that:

$\left | \overrightarrow{A} \right |=12m \\ \\ \left | \overrightarrow{B} \right |=20m \\ \\ \theta_{A}=20^{\circ} \\ \\ \theta_{B}=40^{\circ}$

From geometry, we know that:

$\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}$

Then using vector decomposition into components:

$For \ A: \\ \\ A_x=-\left | \overrightarrow{A} \right |sin\theta_A=-12sin(20^{\circ})=-4.10 \\ \\ A_y=\left | \overrightarrow{A} \right |cos\theta_A=12cos(20^{\circ})=11.27 \\ \\ \\ For \ B: \\ \\ B_x=-\left | \overrightarrow{B} \right |cos\theta_B=-20cos(40^{\circ})=-15.32 \\ \\ B_y=-\left | \overrightarrow{B} \right |sin\theta_B=-20sin(40^{\circ})=-12.85$

Therefore:

$R_x=A_x+B_x=-4.10-15.32=-19.42m \\ \\ R_y=A_y+B_y=11.27-12.85=-1.58m$

So if you want to find out how far are you from your starting point you need to know the magnitude of the vector $\overrightarrow{R}$ , that is:

$\left | \overrightarrow{R} \right |=
\sqrt{R_x^2+R_y^2}=\sqrt{(-19.42)^2+(-1.58)^2}=\boxed{19.48m}$

Finally, let's find the compass direction of a line connecting your starting point to your final position. What we are looking for here is an angle that is shown in Figure 2 which is an angle defined with respect to the positive x-axis. Therefore:

 $\theta_R=180^{\circ}+tan^{-1}(\frac{\left | R_y \right |}{\left | R_x \right |}) \\ \\ \theta_R=180^{\circ}+tan^{-1}(\frac{1.58}{19.42}) \\ \\ \theta_R=180^{\circ}+4.65^{\circ}=185.85^{\circ}$

6 0

3 years ago