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4 years ago

7

Electric charges that are different

2 answers:

Naya [18.7K]4 years ago

8 0

A.) attract to each other

kupik [55]4 years ago

8 0

The awnser is B

I am just writing word because it says my awsner is too short

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The ______________________ of ______________________ in the cores of terrestrial worlds are responsible for ____________________

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2 years ago

How does moon appears to mov e across the night sky

trasher [3.6K]

Answer:

The moon moves because the earth rotates around its axis of rotation. Although the moon moves too around its orbit, the speed is too small to notice. In other words, the moon's position changes insignificantly, however, the earth rotates fast enough to notice.

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2 years ago

Jet001 [13]

Answer:

the answer is C

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3 years ago

Read 2 more answers

PLEASE HELP ASAP

77julia77 [94]

Answer:

<em>The first choice (32m/s) is the closest to the answer</em>

Explanation:

The magnitude of a vector is the distance between the initial and the end point of the vector.

Being Vx and Vy the horizontal and vertical components of the vector V respectively, the magnitude of V is calculated as:

$\mid \vec{V} \mid =\sqrt{V_x^2+V_y^2}$

The components of the velocity of the physics student's projectile launcher are Vx=28 m/s and Vy=15 m/s.

Calculate the magnitude of the velocity:

$\mid \vec{V} \mid =\sqrt{28^2+15^2}$

$\mid \vec{V} \mid =\sqrt{784+225}$

$\mid \vec{V} \mid =\sqrt{1009}$

$\mid \vec{V} \mid =31.8 \ m/s$

The first choice (32m/s) is the closest to the answer

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3 years ago

A place-kicker must kick a football from a point 36.0 m (about 40 yards) from the goal. half the crowd hopes the ball will clear

mixas84 [53]

Explanation:

Let's calculate the components of the football's velocity:

$v_{0x} = (23.9\:\text{m/s})\cos{51.5°} = 14.9\:\text{m/s}$

$v_{0y} = (23.9\:\text{m/s})\sin{51.5°} = 18.7\:\text{m/s}$

a) The time it takes for the football to travel 36.0 m horizontally is

$t = \dfrac{x}{v_{0x}} = \dfrac{36.0\:\text{m}}{14.9\:\text{m/s}} = 2.4\:\text{s}$

During this time, the y-displacement of the football is

$y = v_{0x}t - \frac{1}{2}gt^2$

$\:\:\:\:= (18.7\:\text{m/s})(2.4\:\text{s}) - \frac{1}{2}(9.8\:\text{m/s}^2)(2.4\:\text{s})^2$

$\:\:\:\:= 16.7\:\text{m}$

This means that the football cleared the crossbar by 16.7 m - 3.05 m = 13.7 m

b) To determine whether the football was rising or falling while clearing the crossbar, let's look at the y-component of its velocity after 2.4 s:

$v_y = v_{0y} - gt = 18.7\:\text{m/s} - (9.8\:\text{m/s}^2)(2.4\:\text{s})$

$\:\:\:\:\:\:= -4.82\:\text{s}$

Since its sign is negative, this means that the football was already on its way down.

3 0

3 years ago