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3 years ago

Hello,Plz Help Which conditions are necessary for rain to form?

1 answer:

5 0

Vas happenin!
The third one makes no since because the clouds carry the rain. It isn’t always cold when it’s going to rain

The fourth one is a good one

The second one again it’s not always cold when it’s raining

The first one could be it also

Hmmm I would go with the last one

Sorry if it’s wrong

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In one of the classic nuclear physics experiments at the beginning of the 20th century, alpha particles were accelerated toward

Murljashka [212]

Answer:

Explanation:

In the whole process , electric potential energy is converted into kinetic energy .

Kinetic energy = 3.83 MeV

= 3.83 x 1.6 x 10⁻¹⁶ J

= 6.128 x 10⁻¹⁶ J .

Let the closest distance of approach be r .

Electric potential energy = k Q q / r , Q is charge on nucleus , q is charge on alpha particle , r is closest distance .

Electric potential energy = 9 x 10⁹ x 79 x 1.6 x 10⁻¹⁹ x 2 x 1.6 x 10⁻¹⁹ / r

= 3640.32 x 10⁻²⁹ / r

So,

6.128 x 10⁻¹⁶= 3640.32 x 10⁻²⁹ / r

r = 3640.32 x 10⁻²⁹ / 6.128 x 10⁻¹⁶

= 594.05 x 10⁻¹³

= 59.405 x 10⁻¹²

= 59.405 pm .

6 0

3 years ago

During a rockslide, a 670 kg rock slides from rest down a hillside that is 740 m along the slope and 240 m high. The coefficient

ElenaW [278]

a) $1.58\cdot 10^6 J$

b) $1.15\cdot 10^6 J$

c) $0.43\cdot 10^6 J$

d) 35.8 m/s

Explanation:

The gravitational potential energy of an object is the energy possessed by the object due to its location with respect to the ground.

It is given by:

$U=mgh$

where

m is the mass of the object

g is the acceleration due to gravity

h is the height of the object, relative to a reference level

Here, the reference level is taken at the bottom of the hill (where the potential energy is zero).

So, we have:

m = 670 kg is the mass of the rock

$g=9.8 m/s^2$

h = 240 m is the initial height of the rock

So, the potential energy of the rock just before the slide is

$U=(670)(9.8)(240)=1.58\cdot 10^6 J$

The energy transferred to thermal energy during the slide is equal to the work done by friction, which is:

$W=F_f d$

where

$F_f$ is the force of friction

d = 740 m is the displacement of the rock along the ramp

The force of friction is given by:

$F_f=-\mu mg cos \theta$

where

$\mu=0.25$ is the coefficient of friction

m = 670 kg is the mass of the rock

$\theta$ is the angle of the ramp

Since we know the lenght of the ramp (d = 740 m) and the height (h = 240 m), we can find the angle:

$\theta=sin^{-1}(\frac{h}{d})=sin^{-1}(\frac{240}{740})=18.9^{\circ}$

Therefore, the work done by friction is:

$W=-\mu m g cos \theta d =-(0.25)(670)(9.8)(cos 18.9^{\circ})(740)=-1.15\cdot 10^6 J$

So, the energy transferred to thermal energy is $1.15\cdot 10^6 J$ .

According to the law of conservation of energy, the kinetic energy of the rock as it reaches the bottom of the hill will be equal to the initial potential energy (at the top) minus the energy transformed into thermal energy.

Therefore, we have:

$K_f = U_i -E_t$

where here we have:

$U_i=1.58\cdot 10^6 J$ is the potential energy of the rock at the top of the hill

$E_t=1.15\cdot 10^6 J$ is the energy converted into thermal energy

Substituting, we find

$K_f=1.58\cdot 10^6-1.15\cdot 10^6=0.43\cdot 10^6 J$

So, this is the kinetic energy of the rock at the bottom of the hill.

The kinetic energy of the rock at the bottom of the hill can be rewritten as

$K_f=\frac{1}{2}mv^2$

where

m is the mass of the rock

v is its final speed

In this problem, we have:

$K_f=0.43\cdot 10^6 J$ is the final kinetic energy of the hill

m = 670 kg is the mass of the rock

Therefore, the final speed of the rock is:

$v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(0.43\cdot 10^6)}{670}}=35.8 m/s$

7 0

3 years ago

An alpha particle can be produced in certain radioactive decays of nuclei and consists of two protons and two neutrons. The part

Varvara68 [4.7K]

Answer:

Explanation:

charge, q = 2e = 2 x 1.6 x 10^-19 C = 3.2 x 10^-19 C

mass, m = 4 u = 4 x 1.661 x 10^-27 kg = 6.644 x 10^-27 kg

Radius, r = 4.5 cm = 0.045 m

Magnetic field, B = 1.20 T

(a) Let the speed is v.

$v=\frac{Bqr}{m}$

$v=\frac{1.20\times 3.2\times 10^{-19}\times 0.045}{6.644\times 10^{-27}}$

v = 2.6 x 10^6 m/s

(b) Let T be the period of revolution

$T=\frac{2\pi r}{v}$

$T=\frac{2\times 3.14\times 0.045}{2.6\times 10^{6}}$

T = 1.09 x 10^-7 s

$K=\frac{B^{2}\times q^{2}\times r^{2}}{2m}$

$K=\frac{\left ( 1.20\times 3.2 \times 10^{-19}\times 0.045 \right )^{2}}{2\times 6.644\times 10^{-27}}$

K = 2.25 x 10^-14 J

(d) Let the potential difference is V.

K = qV

$V = \frac{K}{q}$

$V= \frac{2.25\times 10^-14}{3.2\times 10^{-19}}$

V = 70312.5 V

5 0

3 years ago

Is this right? Please tell me why its wrong or right

madam [21]

B,A,D,C u can check this by using formula of momentum P=mv..

8 0

3 years ago

A car traveling at 26 m/s starts to decelerate steadily. It comes to a complete stop in 6 seconds. What is its acceleration?

saul85 [17]

Alright let's start this off with our basic equation!

Accerlation (a) = ?

Initial velocity (V1)= 26 m/s
Final velocity (V2) = 0 ft/s because the car comes to a complete stop!

Time (t) = 6 seconds

The equation for acceleration is below!
a = $\frac{Final velocity - Initial Velocity}{time}$

So now, just plug in the values!
a = $\frac{0 - 26}{6}$
a = $\frac{-26}{6}$
a = -4.33 m/s²

Therefore, your acceleration is -4.33 m/s²!! Hope this helped and was one of the branliest answers :')

5 0

3 years ago