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2 years ago

I don't want the answer i just need help on figuring out how to solve this!

Physics

1 answer:

Karo-lina-s [1.5K]2 years ago

3 0

Answer:

Yes, it will be a Home Run

Explanation:

(Note : It is assumed that the collision between bat and ball is perfectly elastic.)

( Also I don't know anything about baseball, so I checked the dimensions online, which are minimum 325 ft (left and right field foul pole) and minimum 400 ft centre field fence)).

STEP 1

Find Horizontal Range of Projectile

$R = \frac{ {u}^{2} \sin(2 \alpha ) }{g} = \frac{(49.1)^{2} \sin(44°) }{g}$

$R = 170.8 \: m \: = 560.3 \: ft$

Therefore, the ball will travel 560.3 ft horizontally well beyond 400 ft (largest dimension of the field), and it'll be a Home Run.

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What is the potential energy of two charges of +4.6 μC and +1.0 μC that are separated by a distance of 10.0 cm?

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Answer:

U = 0.413 J

Explanation:

the potential energy between two charges q1 and q2 is given by the following formula:

$U=k\frac{q_1q_2}{r}$ (1)

k: Coulomb's constant = 8.98*10^9 NM^2/C^2

q1: first charge = 4.6 μC = 4.6*10^-6 C

q2: second charge = 1.0 μC*10^-6 C

r: distance between charges = 10.0 cm = 0.10 m

You replace the values of all variables in the equation (1):

$U=(8.98*10^9Nm^2/C^2)\frac{(4.6*10^{-6}C)(1.0*10^{-6}C)}{0.10m}=0.413\ J$

Hence, the energy between charges is 0.413 J

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3 years ago

A toy car has a momentum of 3 kilogram meters per second south. The car has a 1-kilogram mass. Which is the velocity of the car?

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3 years ago

When 2 forces are applied in opposite directions, how do you calculate the net force?

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3 years ago

Alpha particles, each having a charge of +2e and a mass of 6.64 ×10-27 kg, are accelerated in a uniform 0.50 T magnetic field to

sergij07 [2.7K]

Answer:

$KE=1.2036\times 10^{-12}\ J$

Explanation:

Given:

charge on the alpha particle, $q=2e=3.2\times 10^{-19}\ C$

mass of the alpha particle, $m=6.64\times 10^{-27}\ kg$

strength of a uniform magnetic field, $B=0.5\ T$

radius of the final orbit, $r=0.5\ m$

During the motion of a charge the magnetic force and the centripetal forces are balanced:

$q.v.B=m.\frac{v^2}{r}$

$m.v=q.B.r$

where:

v = velocity of the alpha particle

$v=\frac{q.B.r}{m}$

$v=\frac{3.2\times 10^{-19}\times 0.5\times 0.5}{6.64\times 10^{-27}}$

$v=1.2048\times 10^{7}\ m.s^{-1}$

Here we observe that the velocity of the aprticle is close to the velocity of light. So the kinetic energy will be relativistic.

We firstly find the relativistic mass as:

$m'=\frac{1}{\sqrt{1-\frac{v^2}{c^2} } } \times m$

$m'=\frac{6.64\times 10^{-27}}{\sqrt{1-\frac{(1.2048\times 10^7)^2}{(3\times 10^8)^2} } }$

$m'=6.6533\times10^{-27}\ kg$

now kinetic energy:

$KE=m'.c-m.c$

$KE=6.6533\times 10^{-27}\times (3\times 10^8)^2-6.64\times 10^{-27}\times (3\times 10^8)^2$

$KE=1.2036\times 10^{-12}\ J$

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3 years ago

I really have no idea where to start with this. Could someone please exaplin the answer and how they got it? Theres a lot of the

LuckyWell [14K]

I believe the answer should be the last option. upon interaction, both objects should have the same charge after the electrons are transferred.

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3 years ago

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