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3 years ago

A cannon launches a 4.0kg bowling ball with 50J of kinetic energy what is the bowling ball’s speed

Physics

2 answers:

AleksAgata [21]3 years ago

7 0

Answer:

5 m/s

Explanation:

50=1/2*4v^2

4*1/2=2

25*2=50

so...

square rood of 25 is 5

answer 5 m/s

sorry if that didn't make since

kobusy [5.1K]3 years ago

4 0

Explanation:

I will try to find the formula

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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, $r=0.5\ m$

<u>During the motion of a charge the magnetic force and the centripetal forces are balanced:</u>

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

$m.v=q.B.r$

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.

<u>We firstly find the relativistic mass as:</u>

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

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

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

now kinetic energy:

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

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

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

6 0

3 years ago

Resistance increases with the length of the wire

Marta_Voda [28]

If the length of the wire increases, then the amount of resistance will also increase.

1. Take a long piece of wire and cut it 10 pieces. Those pieces should all be different sizes, one should be 5___ (units in meter, cm, inches, etc.), and the next should be 5 ___ (units in meter, cm, inches, etc.) more than the one before.
2. Take one piece of wire and measure the resistance using ___ and record the results in the data table.
3. Repeat the previous step with all the pieces of wire.
4. Compare and contrast the results you have found.

I hope this helps a bit :)

4 0

3 years ago

When a trebuchet counterweight is hoisted by soldiers, what form of energy is

vladimir1956 [14]

10. A safe place to use the trebuchet would be away from other buildings and people. A good example of a place would be a large field with no nearby structures.
14. Many factors need to be kept consistent throughout the experiment. One example of a variable that would need to be consistent is the weight and size of the projectile.
15. It is important to do many trials so that you can make sure that the results of each trial are nearly the same. If they are all vastly different, then it means that something has gone wrong.

Sorry I was only able to answer a few questions, but I hope these few answers help! :)

4 0

3 years ago

Read 2 more answers

A circular disc of mass 20kg and radius 15cm is mounted in an horizontal cylindrical axle of radius

disa [49]

Using the concepts of energy, rotational Newton's second law and rotational kinematics we can find the kinematic energy of the system formed by the disk and the cylindrical axis

KE = 0.23 J

given parameters

Disk radius R = 15 cm = 0.15 m
Cylinder radius r = 1.5 cm = 0.0015 m
Disk mass M = 20 kg
Time t = 1.2 s
Force F = 12 N

to find

Kinetic energy (KE)

This exercise must be solved in parts:

1st part. Endowment kinetic energy is the energy due to the circular motion of an object and is described by the equation

KE = ½ I w²

Where KE is the kinetic energy, I the moment of inertia and w the angular velocity

The moment of inertia is a magnitude that measures the inertia for rotational movement, it is a scalar quantity, therefore it is additive. In this system it is composed of two bodies, the disk and the cylindrical axis, for which the total moment of inertia it is

I_{ total} = I_{ disk} + I_{ cylinder}

the moments of inertia with respect to an axis passing through the center of mass are tabulated

disk I_{disk} = ½ M R²

cylinder I_{cylinder} = ½ m r²

where M and m are the masses of the disk and cylinder respectively, R and r their radii

I_{total} = ½ (M R² + m r²) = ½ M R² ( $1 + \frac{m}{M} \ (\frac{r}{R})^2$ )