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Dmitry_Shevchenko [17]
3 years ago
5

An alpha particle(the nucleus of a helium atom) has a mass of 6.64*10^-27kg and a charge of +2e. What are the magnitude and dire

ction of the electric field that will balance the gravitational force on it? Mp= 1.67*10^-27kg
Physics
2 answers:
emmasim [6.3K]3 years ago
8 0
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

Weight = electric force 
<span>mg = qE </span>

<span>6.64x10^-27 x 9.81 = (2 x 1.60x10^-19) E 
</span>qE =mg, 
<span>E = mg/q = 6.64•10^-27•9.8/2•1.6•10^-19 =2.03•10^-7 V/m</span>
Rainbow [258]3 years ago
3 0

Answer:

Magnitude of electric field E= 2.03×10^-7NC^-1

The charge is positive,the electric force is in the same direction as the electric field according to F= qE

Therefore,the direction of the electric field is upward.

Explanation:

Fg=Fe

Where Fe is the force of the electric field

Fg is force of the gravitational field.

Mg= we

Mg= 2eE

E=mg/2e

E= (6.64×10^-27)(9.8)/2(1.6×10^-19)

E=( 6.50×10^-26)/(3.2×10^-19)

E= 2.03×10^-7NC^-1

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A 0.0382-kg bullet is fired horizontally into a 3.78-kg wooden block attached to one end of a massless, horizontal spring (k = 8
wel

Answer:

280.87 ms⁻¹

Explanation:

Consider the motion of the bullet-block combination after collision

m = mass of the bullet = 0.0382 kg

M = mass of wooden block = 3.78 kg

V = velocity of the bullet-block combination after collision

k = spring constant of the spring = 833 N m⁻¹

A = Amplitude of oscillation = 0.190 m

Using conservation of energy

Kinetic energy of  bullet-block combination after collision = Spring potential energy gained due to compression of spring

(0.5)(m + M)V^{2} = (0.5)kA^{2}

(0.0382 + 3.78)V^{2} = (833)(0.190)^{2}

V = 2.81 ms⁻¹

v_{o} = initial velocity of the bullet before striking the block

Using conservation of momentum for the collision between bullet and block

m v_{o} = (m + M) V

(0.0382) v_{o} = (0.0382 + 3.78) (2.81)

v_{o} = 280.87 ms⁻¹

7 0
3 years ago
Railroad cars are loosely coupled so that there is a noticeable time delay from the time the first and last car is moved from re
olga nikolaevna [1]

Answer:

Without this slack, a locomotive might simply sit still and spin its wheels. The loose coupling enables a longer time for the entire train to gain momentum, requiring less force of the locomotive wheels against the track. In this way, the overall required impulse is broken into a series of smaller impulses. (This loose coupling can be very important for braking as well).

Explanation:

8 0
3 years ago
Calculate the amount of momentum of an object with a mass of 10 kg travelling al a
irina [24]

Answer:

50kg.m/s

Explanation:

In order to find momentum you must use the formula P=mv

p= momentum

m=mass

v= velocity

so in other words, momentum= mass times velocity

or in this case, momentum= 10 times 5  :)

7 0
3 years ago
In your own words, explain what a field is in science. Use specific examples to support your explanation.
tresset_1 [31]

Answer:

Field, In physics, a region in which each point is affected by a force.

Explanation:

4 0
3 years ago
A 0.560 kg snowball is fired from a cliff 14.2 m high with an initial velocity of 13.3 m/s, directed 26.0° above the horizontal.
enot [183]

Answer:

a) v = 21.34 m/s

b) v = 21.34 m/s

c) v = 21.34 m/s

Explanation:

Mass of the snowball, m = 0.560 kg

Height of the cliff, h = 14.2 m

Initial velocity of the ball, u = 13.3 m/s

θ = 26°

The speed of the slow ball as it reaches the ground, v = ?

The initial Kinetic energy of the snow ball, KE_{0}  = 0.5 mu^{2}

Potential energy of the snow ball at the given height, PE = mgh

Final Kinetic energy of the ball as it reaches the ground, KE_{f} = 0.5mv^{2}

a) Using the principle of energy conservation,

KE_{0} + PE = KE_{f} \\0.5mu^{2} + mgh = 0.5mv^{2}\\v^{2} =2( 0.5u^{2} + gh)\\v^{2} =u^{2} + 2gh\\v = \sqrt{u^{2} + 2gh} \\v = \sqrt{13.3^{2} + 2*9.8*14.2}\\v = 21.34 m/s

b) The speed remains v = 21.34 m/s since it is not a function of the angle of launch

c)The principle of energy conservation used cancels out the mass of the object, therefore the speed is not dependent on mass

v = 21. 34 m/s

7 0
3 years ago
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