100 point good luck answering this 2+2-4+8+-5+???

A baseball thrown by a pitcher is hit by a batter. At the moment when the ball hits the bat, the force exerted on the bat by the

Slav-nsk [51]

Newton's Laws of motion describe the motion of an object based on the applied force

The correct option that gives the relationships between the forces is the the option;

$\underline{F_{ball}}$ on bat = $\underline{F_{bat}}$ on ball

(Either option A or B without the minus symbol before $F_{bat}$ , likely typographical error)

Reason:

Force exerted on the bat by the ball = $F_{ball}$

Force exerted on the ball by the bat = $F_{bat}$

Given that the batter hits the ball with the bat, the force exerted by the bat on the ball, $F_{bat}$ , is reacted to by the force of the ball acting on the bat, $F_{ball}$

According to Newton's third law of motion, action and reaction are equal and opposite. Therefore, at the moment when the ball hits the bat, we have;

$\underline{F_{ball}}$ on bat = $\underline{F_{bat}}$ on ball

Where the force of the bat is high, the ball is accelerated to travel at high speed

Learn more Newton's Laws of motion here:

brainly.com/question/24522313

8 0

3 years ago

What is the magnitude of the magnetic field at a point midway between them if the top one carries a current of 19.5 A and the bo

Phantasy [73]

Answer:

The magnetic field will be $\large{\dfrac{1.4 \times 10^{-4}}{d}} T$ , '2d' being the distance the wires.

Explanation:

From Biot-Savart's law, the magnetic field ( $\large{\overrightarrow{B}}$ ) at a distance ' $r$ ' due to a current carrying conductor carrying current ' $I$ ' is given by

$\large{\overrightarrow{B} = \dfrac{\mu_{0}I}{4 \pi}} \int \dfrac{\overrightarrow{dl} \times \hat{r}}{r^{2}}}$

where ' $\overrightarrow{dl}$ ' is an elemental length along the direction of the current flow through the conductor.

Using this law, the magnetic field due to straight current carrying conductor having current ' $I$ ', at a distance ' $d$ ' is given by

$\large{\overrightarrow{B}} = \dfrac{\mu_{0}I}{2 \pi d}$

According to the figure if ' $I_{t}$ ' be the current carried by the top wire, ' $I_{b}$ ' be the current carried by the bottom wire and ' $2d$ ' be the distance between them, then the direction of the magnetic field at 'P', which is midway between them, will be perpendicular towards the plane of the screen, shown by the $\bigotimes$ symbol and that due to the bottom wire at 'P' will be perpendicular away from the plane of the screen, shown by $\bigodot$ symbol.

Given $\large{I_{t} = 19.5 A}$ and $\large{I_{B} = 12.5 A}$

Therefore, the magnetic field ( $\large{B_{t}}$ ) at 'P' due to the top wire

$B_{t} = \dfrac{\mu_{0}I_{t}}{2 \pi d}$

and the magnetic field ( $\large{B_{b}}$ ) at 'P' due to the bottom wire

$B_{b} = \dfrac{\mu_{0}I_{b}}{2 \pi d}$

Therefore taking the value of $\mu_{0} = 4\pi \times 10^{-7}$ the net magnetic field ( $\large{B_{M}}$ ) at the midway between the wires will be

$\large{B_{M} = \dfrac{4 \pi \times 10^{-7}}{2 \pi d} (I_{t} - I_{b}) = \dfrac{2 \times 10^{-7}}{d} = \dfrac{41.4 \times 10 ^{-4}}{d}} T$

5 0

3 years ago

A student attaches a block of mass M to a vertical spring so that the block-spring system will oscillate if the block-spring sys

Vlada [557]

WAAAAAAAAAAAAAAAAAAAAAAAAARRRRR

7 0

3 years ago

What kind of organisms are able to convert sunlight into usable energy?

Tamiku [17]

Answer:

Photosynthetic organisms

Explanation:

The electromagnetic energy of sunlight is converted to chemical energy in the chlorophyll-containing cells of photosynthetic organisms. In eukaryotic cells these reactions occur in the organelle known as the chloroplast

Hope this helps! :)

4 0

3 years ago

Read 2 more answers

A very humble bumble bee is flying horizontally due North at a constant speed of 3.11 m/s. At the current location of the bumble

Reil [10]

To solve this problem we will apply the concepts of the Magnetic Force. This expression will be expressed in both the vector and the scalar ways. Through this second we can directly use the presented values and replace them to obtain the value of the magnitude. Mathematically this can be described as,

$\vec{F_B} = q(\vec{V}\times \vec{B})$