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3 years ago

Julio hit a baseball. What caused the ball to change direction at the time of impact?

2 answers:

7 0

The answer is c. the force of his swing

At the time of the impact, there is a collision between two bodies moving in opposite directions.

The force exerted on the ball causes the change of velocity.

Vikki [24]3 years ago

3 0

Answer: The force of his swing

Explanation: When Julio hit the baseball, the ball changed the direction after the impact. The contact force exerted by the bat on the baseball when it swing caused the change in direction of the ball. The force exerted in time t is causes change in momentum.

the force of gravity acts on the ball all the time. The speed of the ball is not a factor to change direction. The weight of the air exerts a drag on the ball which is also constant. Hence, only the force of swing caused the changed in the direction of the ball.

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The action force is the balloon pushing the air out. What is the magnitude of the reaction force of the air pushing on the ballo

aleksley [76]

Understanding Newtons second law that everything has an equal and opposite reaction. The reaction force from a balloons air being pushed out is the preasurized air it had to push out into the open air.

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4 years ago

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In the aftermath of an intense earthquake, the earth as a whole "rings" with a period of 54 minutes.

lawyer [7]

The frequency of the oscillation in hertz is calculated to be 0.00031 Hz.

The frequency of a wave is defined as the number of cycles completed per second while the period refers to the time taken to complete a cycle. The frequency is the inverse of period.

So;

Period(T) = 54 minutes or 3240 seconds

Frequency (f) = T-1 = 1/T = 1/3240 seconds = 0.00031 Hz

Learn more: brainly.com/question/14588679

6 0

2 years ago

Points A (-5,6), B (2,-2), and C (-6,-3) are placed in three different quadrants of a Cartesian coordinate system. Convert each

AURORKA [14]

Answer: A ( $\sqrt{61}$ ,309.8°)

B (2 $\sqrt{2}$ , 315°)

C ( $3\sqrt{5}$ , 26.56°)

Explanation: To transform rectangular coordinates into polar coordinates use:

$r=\sqrt{x^{2}+y^{2}}$ and $\theta=tan^{-1}(\frac{y}{x})$

For point A:

$r=\sqrt{(-5)^{2}+6^{2}}$

$r=\sqrt{61}$

$\theta=tan^{-1}(\frac{6}{-5})$

$\theta=tan^{-1}(-1.2)$

$\theta=-50.2$ °

Point A is in the II quadrant, so we substract the angle for 360° since it is in degrees:

$\theta=360-50.2$

$\theta=$ 309.8°

Polar coordinates for point A is ( $\sqrt{61}$ , 309.8°)

For point B:

$r=\sqrt{2^{2}+(-2)^{2}}$

$r=\sqrt{8}$

$r=2\sqrt{2}$

$\theta=tan^{-1}(\frac{-2}{2} )$

$\theta=tan^{-1}(1)$

$\theta=-45$ °

Point B is in IV quadrant, so:

$\theta=360-45$

$\theta=$ 315°

Polar coordinates for point B is ( $2\sqrt{2}$ , 315°)

For point C:

$r=\sqrt{(-6)^{2}+(-3)^{2}}$

$r=\sqrt{45}$

$r=3\sqrt{5}$

$\theta=tan^{-1}(\frac{-3}{-6} )$

$\theta=tan^{-1}(0.5)$

$\theta=$ 26.56°

Polar coordinates for point C is ( $3\sqrt{5}$ , 26.56°)

3 0

3 years ago

A very long uniform line of charge has charge per unit length λ1 = 4.68 μC/m and lies along the x-axis. A second long uniform li

Kitty [74]

Answer:

$E_{net} = 6.44 \times 10^5 N/C$

Explanation:

As we know that electric field due to infinite line charge distribution at some distance from it is given as

$E = \frac{2k \lambda}{r}$

now we need to find the electric field at mid point of two wires

So here we need to add the field due to two wires as they are oppositely charged

Now we will have

$E_{net} = \frac{2k\lambda_1}{r} + \frac{2k\lambda_2}{r}$

now plug in all data

$\lambda_1 = 4.68 \muC/m$

$\lambda_2 = 2.48 \mu C/m$

$r = 0.200 m$

now we have

$E_{net} = \frac{2k}{r}(4.68 + 2.48)$

$E_{net} = \frac{2(9\times 10^9)}{0.200}(7.16 \times 10^{-6})$

$E_{net} = 6.44 \times 10^5 N/C$

8 0

3 years ago

A neutral solid metal sphere of radius 0.1 m is at the origin, polarized by a point charge of 2 × 10−8 C at location m. At locat

liraira [26]

Answer: E = $1.8 *10 ^{4} N$

Explanation: The formulae for intensity of an electric field of a solid metal sphere relative to a point is given below

$E =\frac{Kq}{r^{2} }$ r

where $k=9* 10^{9}N/m^{2}$ , $q=2 *10 ^{-8} c$ , r = 0.1m r = is the position vector of the charge.

it has been stated in the question that the charge is placed at the center thus it has no position vector.

$E=\frac{9 * 10^{9}* 2* 10^{-8} }{0.1^{2} }\\ =\frac{18* 10^{1} }{0.01} \\=\frac{18* 10^{1} }{1 *10^{-2} } = 1.8*10^{4} N$

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3 years ago