Assume the motions and currents mentioned are along the x axis and fields are in the y direction. (a) does an electric field exe
1 answer:
<span> (a) does an electric field exert a force on a stationary charged object?
Yes. The force exerted by an electric field of intensity E on an object with charge q is
</span>
<span>As we can see, it doesn't depend on the speed of the object, so this force acts also when the object is stationary.
</span><span>(b) does a magnetic field do so?
No. In fact, the magnetic force exerted by a magnetic field of intensity B on an object with charge q and speed v is
</span>
where
is the angle between the direction of v and B.
As we can see, the value of the force F depends on the value of the speed v: if the object is stationary, then v=0, and so the force is zero as well.
<span>(c) does an electric field exert a force on a moving charged object?
Yes, The intensity of the electric force is still
</span>
<span>as stated in point (a), and since it does not depend on the speed of the charge, the electric force is still present.
</span><span>(d) does a magnetic field do so?
</span>Yes. As we said in point b, the magnetic force is
And now the object is moving with a certain speed v, so the magnetic force F this time is different from zero.
<span>(e) does an electric field exert a force on a straight current-carrying wire?
Yes. A current in a wire consists of many charges traveling through the wire, and since the electric field always exerts a force on a charge, then the electric field exerts a force on the charges traveling through the wire.
</span><span>(f) does a magnetic field do so?
Yes. The current in the wire consists of charges that are moving with a certain speed v, and we said that a magnetic field always exerts a force on a moving charge, so the magnetic field is exerting a magnetic force on the charges that are traveling through the wire.
</span><span>(g) does an electric field exert a force on a beam of moving electrons?
Yes. Electrons have an electric charge, and we said that the force exerted by an electric field is
</span>
<span>So, an electric field always exerts a force on an electric charge, therefore on an electron beam as well.
</span><span>(h) does a magnetic field do so?
Yes, because the electrons in the beam are moving with a certain speed v, so the magnetic force
</span>
<span>is different from zero because v is different from zero.</span>
Yes. The force exerted by an electric field of intensity E on an object with charge q is
</span>
<span>As we can see, it doesn't depend on the speed of the object, so this force acts also when the object is stationary.
</span><span>(b) does a magnetic field do so?
No. In fact, the magnetic force exerted by a magnetic field of intensity B on an object with charge q and speed v is
</span>
where
As we can see, the value of the force F depends on the value of the speed v: if the object is stationary, then v=0, and so the force is zero as well.
<span>(c) does an electric field exert a force on a moving charged object?
Yes, The intensity of the electric force is still
</span>
<span>as stated in point (a), and since it does not depend on the speed of the charge, the electric force is still present.
</span><span>(d) does a magnetic field do so?
</span>Yes. As we said in point b, the magnetic force is
And now the object is moving with a certain speed v, so the magnetic force F this time is different from zero.
<span>(e) does an electric field exert a force on a straight current-carrying wire?
Yes. A current in a wire consists of many charges traveling through the wire, and since the electric field always exerts a force on a charge, then the electric field exerts a force on the charges traveling through the wire.
</span><span>(f) does a magnetic field do so?
Yes. The current in the wire consists of charges that are moving with a certain speed v, and we said that a magnetic field always exerts a force on a moving charge, so the magnetic field is exerting a magnetic force on the charges that are traveling through the wire.
</span><span>(g) does an electric field exert a force on a beam of moving electrons?
Yes. Electrons have an electric charge, and we said that the force exerted by an electric field is
</span>
<span>So, an electric field always exerts a force on an electric charge, therefore on an electron beam as well.
</span><span>(h) does a magnetic field do so?
Yes, because the electrons in the beam are moving with a certain speed v, so the magnetic force
</span>
<span>is different from zero because v is different from zero.</span>
You might be interested in
<h2>Answer:</h2>
(a) v(t) = [10.0t - 12.0t²] m/s and a(t) = [10.0 - 24.0t ] m/s² respectively
(b) -28.0m/s and -38.0m/s² respectively
(c) 0.83s
(d) 0.83s
(e) x(t) = 1.1573 m [where t = 0.83s]
<h2>Explanation:</h2>The position equation is given by;
x(t) = 5.0t² - 4.0t³ m --------------------(i)
(a) Since velocity is the time rate of change of position, the velocity, v(t), of the particle as a function of time is calculated by finding the derivative of equation (i) as follows;
v(t) = dx(t) / dt = =
[ 5.0t² - 4.0t³ ]
v(t) = 10.0t - 12.0t² --------------------------------(ii)
Therefore, the velocity as a function of time is v(t) = 10.0t - 12.0t² m/s
Also, since acceleration is the time rate of change of velocity, the acceleration, a(t), of the particle as a function of time is calculated by finding the derivative of equation (ii) as follows;
a(t) = dx(t) / dt = =
[ 10.0t - 12.0t² ]
a(t) = 10.0 - 24.0t --------------------------------(iii)
Therefore, the acceleration as a function of time is a(t) = 10.0 - 24.0t m/s²
(b) To calculate the velocity at time t = 2.0s, substitute the value of t = 2.0 into equation (ii) as follows;
=> v(t) = 10.0t - 12.0t²
=> v(2.0) = 10.0(2) - 12.0(2)²
=> v(2.0) = 20.0 - 48.0
=> v(2.0) = -28.0m/s
Also, to calculate the acceleration at time t = 2.0s, substitute the value of t = 2.0 into equation (iii) as follows;
=> a(t) = 10.0 - 24.0t
=> a(2.0) = 10.0 - 24.0(2)
=> a(2.0) = 10.0 - 48.0
=> a(2.0) = -38.0 m/s²
Therefore, the velocity and acceleration at t = 2.0s are respectively -28.0m/s and -38.0m/s²
(c) The time at which the position is maximum is the time at which there is no change in position or the change in position is zero. i.e dx / dt = 0. It also means the time at which the velocity is zero. (since velocity is dx / dt)
Therefore, substitute v = 0 into equation (ii) and solve for t as follows;
=> v(t) = 10.0t - 12.0t²
=> 0 = 10.0t - 12.0t²
=> 0 = ( 10.0 - 12.0t ) t
=> t = 0 or 10.0 - 12.0t = 0
=> t = 0 or 10.0 = 12.0t
=> t = 0 or t = 10.0 / 12.0
=> t = 0 or t = 0.83s
At t=0 or t = 0.83s, the position of the particle will be maximum.
To get the more correct answer, substitute t = 0 and t = 0.83 into equation (i) as follows;
<em>Substitute t = 0 into equation (i)</em>
x(t) = 5.0(0)² - 4.0(0)³ = 0
At t = 0; x = 0
<em>Substitute t = 0.83s into equation (i)</em>
x(t) = 5.0(0.83)² - 4.0(0.83)³
x(t) = 5.0(0.6889) - 4.0(0.5718)
x(t) = 3.4445 - 2.2872
x(t) = 1.1573 m
At t = 0.83; x = 1.1573 m
Therefore, since the value of x at t = 0.83s is 1.1573m is greater than the value of x at t = 0 which is 0m, then the time at which the position is at maximum is 0.83s
(d) The velocity will be zero when the position is maximum. That means that, it will take the same time calculated in (c) above for the velocity to be zero. i.e t = 0.83s
(e) The maximum position function is found when t = 0.83s as shown in (c) above;
Substitute t = 0.83s into equation (i)
x(t) = 5.0(0.83)² - 4.0(0.83)³
x(t) = 5.0(0.6889) - 4.0(0.5718)
x(t) = 3.4445 - 2.2872
x(t) = 1.1573 m [where t = 0.83s]
Answer:
Explanation:
In case of gas , work done
W = ∫ p dV , p is pressure and dV is small change in volume
the limit of integration is from Vi to Vf .
= ∫ p dV
= ∫ p₀ dV
= p₀ / (
)
= - 5p₀
Taking limit from Vi to Vf
W = - 5 p₀ ( ) ltr- atm.
Answer:
Explanation:
To solve this problem we use the Momentum's conservation Law, before and after the girl catch the ball:
(1)
At the beginning the girl is stationary:
(2)
If the girl catch the ball, both have the same speed:
(3)
We replace (2) and (3) in (1):
We can now solve the equation for v_{f}: