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lord [1]
3 years ago
11

An electron is released from rest in a region of space with a nonzero electric field.

Physics
1 answer:
enyata [817]3 years ago
3 0

Answer:

Decreases

Explanation:

When an electron is released from rest in a region of space with nonzero electric field. And when the electrons move the electric potential energy of the system decreases because the kinetic energy increases and for the total energy to remain constant the potential energy must reduce.

An electron is accelerated in an electric field meaning its kinetic energy increases.

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Please help! the mass of an object is measured on a pan balance with a precision of 0.005 g and the recorded value of 128.01 g.
inysia [295]

Yes, these two objects have different masses.

<h3>How can we calculate that this statement is right ?</h3>

To calculate the precision of mass we are using the formula,

Precision(P) = \frac{m_1}{m_1+\triangle m}

Or,\trianglem=\frac{m_1}{P}-m₁

For the first case we are given,

m₁= The recorded value of mass

= 128.01 g

P= Precision of the mass

=0.005g

So, according to the formula, \trianglem will be,

\trianglem=  \frac{128.01}{0.005}-128.01

Or,\trianglem=25,473.99 g

Or,\trianglem=25.47 Kg

For the first case \trianglem is 25.47 Kg..

For the second case we are given,

m₁= The recorded value of mass

= 0.13 Kg

P= Precision of the mass

=0.005 Kg

So, according to the formula, \trianglem will be,

\trianglem= \frac{0.13}{0.005}-0.13

Or,\trianglem= 25.87 kg

For the second case \trianglem is 25.87 Kg.

For the two cases  \trianglem has different values, 25.47 Kg≠25.87 Kg.

Therefore we can conclude that, these two objects have different masses.

Learn more about Mass:

brainly.com/question/3187640

#SPJ4

7 0
2 years ago
A revolutionary war cannon, with a mass of 2260 kg, fires a 15.5 kg ball horizontally. The cannonball has a speed of 109 m/s aft
Anton [14]

Answer:

The gain in velocity is 0.37m/s

Explanation:

We need solve this problem though the conservation of momentum. That is,

m_1 v_1 = m_2 v_2

m_1=2260Kg\\m_2=15.5Kg\\v_2= 109m/s

Using the equation to find v_1,

v_1=\frac{m_2 v_2}{m_1}\\v_1=\frac{15.5*109}{2260}\\v_1= 0.7475

Using the conservation of energy equation, we have,

KE= \frac{1}{2}m*v^2

KE_{cball}=\frac{1}{2}(15.5)(109)^2=92077.75J

KE_{cannon}=\frac{1}{2}(2260)(0.7475)^2=631.39J

Total KE= 92077.75+13425530=92708.9J

Now this energy over the cannonball

KE=\frac{1}{2}m*v_2^2

92708.9=\frac{1}{2}15.5v_2^2

V_2 = 109.37m/s

The gain in velocity is 0.37m/s

4 0
3 years ago
Yolanda is performing a neutralization reaction using hydrochloric acid and sodium hydroxide. Since both are clear liquids, what
Bad White [126]
I should think that Yolanda should use litmus paper throughout the reaction as its color will tell when the mixture is neutral and therefore then, when she can stop adding either the hydrochloric acid or the sodium hydroxide or if one or the other needs to be added to make the solution approach neutral. Blue litmus paper stays blue in a neutral solution, but will turn red in an acidic solution.
5 0
3 years ago
When a falling meteoroid is at a distance above the Earth's surface of 3.40 times the Earth's radius, what is its acceleration d
Alchen [17]

Answer:

g = 0.85 ms^{-2}

Explanation:

g = \frac{GM}{h^{2} }

were; g is the acceleration due to Earth's gravity, G is Newton's gravitation constant (6.674 x 10^{-11} Nm^{2}kg^{-2}), M is the mass of the earth (5.972 x 10^{24} kg), and h is the distance of meteoroid to the earth.

h = 3.40 x R

  = 3.40 x 6371 km

h = 21661.4 km

  = 21661400 m

Thus,

g = \frac{6.674*10^{-11}*5.972*10^{24}  }{(21661400)^{2} }

  = \frac{3.9857 *10^{14} }{4.6922*10^{14} }

  = 0.84944

g = 0.85 ms^{-2}

The acceleration due to the Earth's gravitation is 0.85 ms^{-2}.

6 0
3 years ago
One billiard ball is shot east at 2.2 m/s. A second, identical billiard ball is shot west at 0.80 m/s. The balls have a glancing
Leona [35]

Answer:

(a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

Explanation:

Given that,

Velocity of one ball u₁= 2.2i m/s

Velocity of second ball u₂=- 0.80i m/s

Final velocity of the second ball v₂= 1.36j m/s

The mass of the identical balls are

m = m_{1}=m_{2}

(a). We need to calculate the speed of the first ball after the collision

Using law of conservation of momentum

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

Along X- axis

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}

v_{1}=u_{1}+u_{2}

Put the value into the formula

v_{1}=2.2i-0.80i

v_{1}=1.4i\ m/s

Along Y-axis

0=m_{1}v_{1}+m_{2}v_{2}

m_{1}v_{1}=-m_{2}v_{2}

v_{1}=-v_{2}

Put the value into the formula

v_{1}=-1.36j\ m/s

Then the final speed of the first ball

v_{1}=\sqrt{(1.4)^2+(1.36)^2}

v_{1}=1.95\ m/s

(b) We need to calculate the direction of the first ball after the collision

Using formula of direction

\tan\theta=\dfrac{v_{2}}{v_{1}}

\tan\theta=\dfrac{-1.36}{1.4}

\theta=\tan^{-1}\dfrac{-1.36}{1.4}

\theta=-44.16^{\circ}

Negative sign shows the direction of first ball .

Hence, (a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

7 0
3 years ago
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