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3 years ago

15

A soccer player takes a free kick from a spot that is 20 m from the goal. The ball leaves his foot at an angle of 38 ∘, and it e

ventually hits the crossbar of the goal, which is 2.4 m from the ground. At what speed did the ball leave his foot?

1 answer:

Sergio039 [100]3 years ago

7 0

This dude can kick the ball

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A car travels along a straight line at a constant speed of 53.0 mi/h for a distance d and then another distance d in the same di

Shalnov [3]

A distance of d is covered with 53 mile/hr initially. Time taken to cover this distance t1 = d/53 hour Next distance of d is covered with x mile hours. Time taken to cover this distance t2 = d/x hours. We have average speed = 26.5 mile / hour

= Total distance traveled/ total time taken

= $\frac{2d}{\frac{d}{53}+\frac{d}{x}} = \frac{2}{\frac{1}{53}+\frac{1}{x} } = \frac{106x}{x+53}$

$26.5 = \frac{106x}{x+53} \\ \\ 79.5 x = 1404.5\\ \\ x = 17.67 miles/hour$

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3 years ago

The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi

Feliz [49]

To solve this problem we will use the Ampere-Maxwell law, which describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

$\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}$

Where,

B= Magnetic Field

l = length

$\mu_0$ = Vacuum permeability

$\epsilon_0$ = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

$B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}$

Recall that the speed of light is equivalent to

$c^2 = \frac{1}{\mu_0 \epsilon_0}$

Then replacing,

$B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}$

$B = \frac{r}{2C^2} \frac{dE}{dt}$

Our values are given as

$dE = 2150N/C$

$dt = 5s$

$C = 3*10^8m/s$

$D = 0.440m \rightarrow r = 0.220m$

Replacing we have,

$B = \frac{r}{2C^2} \frac{dE}{dt}$

$B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}$

$B =5.25*10^{-16}T$

Therefore the magnetic field around this circular area is $B =5.25*10^{-16}T$

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3 years ago

A 0.5 kg cheeseburger is lobbed at a particularly unhappy customer with a force of 10 N.

aleksley [76]

The acceleration that the cheeseburger experienced is 20 m/s^2.

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3 years ago

How does the Coriolis effect influence the direction of the Trade Winds in the Northern Hemisphere? Does it have the same effect

scoray [572]

Answer:

Part A

Coriolis effect is used to describe how objects which are not fixed to the ground are deflected as they travel over long distances due to the rotation of the Earth relative to the 'linear' motion of the objects

Due to the Coriolis effect the wind flowing towards the Equator from high pressure belts in the subtropical regions in both the Northern and Southern Hemispheres are deflected towards the western direction because the Earth rotates on its axis towards the east

Part B

In the Northern Hemispheres, the winds are known as northeasterly trade winds and in the Southern Hemisphere, they are known as the southeasterly trade wind. Therefore, Coriolis effect has the same effect on the direction of the Trade Winds in the Southern Hemisphere as it does in the Northern Hemisphere

Explanation:

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Elements from which two groups in the periodic table would most likely combine

Katarina [22]

The answer is actually 1 & 17, maybe you meant to put 1&17 as option b

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