Which statement accurately describes which subatomic particles are involved in nuclear and chemical reactions?

A 20 kilogram ball rolls down a 10 meter ramp at the rate of 15 meters per second. The kinetic energy is joules

kobusy [5.1K]

Answer:2250J

Explanation:

mass(m)=20kg

velocity(v)=15m/s

Kinetic energy=(m x v^2)/2

Kinetic energy =(20 x 15^2)/2

Kinetic energy =(20x15x15)/2

Kinetic energy=4500/2

Kinetic energy=2250J

3 0

3 years ago

If you were trying to kill a fish witha a spear, woukd you ain where you see the fish?

patriot [66]

No, you can't aim where you see the fish, because refraction will change the position of the fish and make it appear in a different position from where it actually is.

Hope this helps!

8 0

3 years ago

Read 2 more answers

A nonprofit organization posted an article in the local newspaper about a recent marathon. The marathon raised money for vaccina

mr Goodwill [35]

Answer:

An original funding source.

Explanation:

Non profit organization planned a marathon through which the money was raised for the vaccinations help, hence; here they are showing the original funding source, that is the number of runners and the amount raised by them.

3 0

3 years ago

Read 2 more answers

A new planet has been discovered and given the name Planet X . The mass of Planet X is estimated to be one-half that of Earth, a

harina [27]

Answer:

vₐ = v_c $( \ 1 + \frac{1}{2} ( \frac{\Delta M}{M} - \frac{\Delta R}{R}) \ )$

Explanation:

To calculate the escape velocity let's use the conservation of energy

starting point. On the surface of the planet

Em₀ = K + U = ½ m v_c² - G Mm / R

final point. At a very distant point

Em_f = U = - G Mm / R₂

energy is conserved

Em₀ = Em_f

½ m v_c² - G Mm / R = - G Mm / R₂

v_c² = 2 G M (1 /R - 1 /R₂)

if we consider the speed so that it reaches an infinite position R₂ = ∞

v_c = $\sqrt{\frac{2GM}{R} }$

now indicates that the mass and radius of the planet changes slightly

M ’= M + ΔM = M ( $1+ \frac{\Delta M}{M}$ )

R ’= R + ΔR = R ( $1 + \frac{\Delta R}{R}$ )

we substitute

vₐ = $\sqrt{\frac{2GM}{R} } \ \frac{\sqrt{1+ \frac{\Delta M}{M} } }{ \sqrt{1+ \frac{ \Delta R}{R} } }$

let's use a serial expansion

√(1 ±x) = 1 ± ½ x +…

we substitute

vₐ = v_ c ( $(1 + \frac{1}{2} \frac{\Delta M}{M} ) \ ( 1 - \frac{1}{2} \frac{\Delta R}{R} )$ )

we make the product and keep the terms linear

vₐ = v_c $( \ 1 + \frac{1}{2} ( \frac{\Delta M}{M} - \frac{\Delta R}{R}) \ )$

5 0

2 years ago

A proton moving at 8.9 × 106 m/s through a magnetic field of 0.96 T experiences a magnetic force of magnitude 3.8 × 10−13 N. Wha

goblinko [34]

Answer: 15.66 °

Explanation: In order to solve this proble we have to consirer the Loretz force for charge partcles moving inside a magnetic field. Thsi force is given by:

F=q v×B = qvB sin α where α is teh angle between the velocity and magnetic field vectors.

From this expression and using the given values we obtain the following:

F/(q*v*B) = sin α

3.8 * 10^-13/(1.6*10^-19*8.9*10^6* 0.96)= 0.27

then α =15.66°

8 0

3 years ago