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3 years ago

A baseball pitcher throws a 0.14 kg ball toward a batter who is 18 m away.

Physics

1 answer:

Olin [163]3 years ago

5 0

Answer:

A. 112 J

Explanation:

KE = ½mv² = ½(0.14)40² = 112 J

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4 years ago

Read 2 more answers

A length of copper wire carries a current of 14 A, uniformly distributed through its cross section. The wire diameter is 2.5 mm,

gavmur [86]

Answer:

a. ρ $_\beta=1.996J/m^3$

b. $U_E=9.445x10^{-15} J/m^3$

Explanation:

a. To find the density of magnetic field given use the gauss law and the equation:

$i=14A$ , $d=2.5mm$ , $R=3.3$ Ω, $l=1 km$ , $E_o=8.85x10^{-12}F/m$ , $u_o=4*x10^{-7}H/m$

ρ $_\beta=\frac{\beta^2}{2*u_o}$

ρ $_\beta=\frac{1}{2*u_o}*(\frac{u_o*i^2}{2\pi *r})^2$

ρ $_\beta=\frac{u_o*i^2}{8\pi*r}=\frac{4\pi *10^{-7}H/m*(14A)^2}{8\pi*(1.25x10^{-3}m)^2}$

ρ $_\beta=1.996J/m^3$

b. The electric field can be find using the equation:

$U_E=\frac{1}{2}*E_o*E^2$

$E=(\frac{i*R}{l})^2$

$U_E=\frac{1}{2}*8.85x10^{-12}*(\frac{14A*3.3}{1000m})^2$

$U_E=9.445x10^{-15} J/m^3$

4 0

3 years ago

Please help ! I’m stuck on this

bazaltina [42]

Answer:

When equal and opposite forces meet each other, it results in motion and they would repel away from each other, causing the asteroid to be sent away from the earth.

Explanation:

Newton's Third Law states that every action has an equal and opposite reaction.

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3 years ago

A 35 g steel ball is held by a ceiling-mounted electromagnet 4.0 m above the floor. A compressed-air cannon sits on the floor, 4

HACTEHA [7]

Answer:

7.9 m/s

Explanation:

When both balls collide, they have spent the same time for their motions.

Motion of steel ball

This is purely under gravity. It is vertical.

Initial velocity, u = 0 m/s

Distance, s = 4.0 m - 1.2 m = 2.8 m

Acceleration, a = g

Using the equation of motion

$s = ut+\frac{1}{2}at^2$

$2.8 \text{ m} = 0+\dfrac{gt^2}{2}$

$t = \sqrt{\dfrac{5.6}{g}}$

Motion of plastic ball

This has two components: a vertical and a horizontal.

The vertical motion is under gravity.

Considering the vertical motion,

Initial velocity, u = ?

Distance, s = 1.2 m

Acceleration, a = -g (It is going up)

Using the equation of motion

$s = ut+\frac{1}{2}at^2$

$1.2\text{ m} = ut-\frac{1}{2}gt^2$

Substituting the value of t from the previous equation,

$1.2\text{ m} = u\sqrt{\dfrac{5.6}{g}}-\dfrac{1}{2}\times g\times\dfrac{5.6}{g}$

$u\sqrt{\dfrac{5.6}{g}} = 4.0$

Taking g = 9.8 m/s²,

$u = \dfrac{4.0}{0.756} = 5.29 \text{ m/s}$

This is the vertical component of the initial velocity

Considering the horizontal motion which is not accelerated,

horizontal component of the initial velocity is horizontal distance ÷ time.

$u_h = \dfrac{4.4\text{ m}}{0.756\text{ s}} = 5.82\text{ m/s}$

The initial velocity is

$v_i = \sqrt{u^2+u_h^2} = \sqrt{(5.29\text{ m/s})^2+(5.82\text{ m/s})^2} = 7.9 \text{ m/s}$

4 0

4 years ago