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3 years ago

Which statement is true about nuclear fusion? it is caused by the same process that causes nuclear fission.

2 answers:

6 0

The correct statement about nuclear fusion is the last one: "it produces nearly all the elements that are heavier than helium". Nuclear fusion is the process of combination of two smaller atoms into a larger one. It requires a high activation energy, which is reached in the stars. Elements from He to Iron were formed in the Sun, but larger elements require a higher energy and are formed inside supernovas.

san4es73 [151]3 years ago

5 0

Answer:

It produces nearly all the elements that are heavier than helium.

Explanation:

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One ring of radius a is uniformly charged with charge +Q and is placed so its axis is the x-axis. A second ring with charge –Q i

kati45 [8]

Answer:

The force exerted on an electron is $7.2\times10^{-18}\ N$

Explanation:

Given that,

Charge = 3 μC

Radius a=1 m

Distance = 5 m

We need to calculate the electric field at any point on the axis of a charged ring

Using formula of electric field

$E=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

$E_{1}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times3\times10^{-6}\times5}{(1^2+5^2)^{\frac{3}{2}}}$

$E_{1}=1.0183\times10^{3}\ N/C$

Using formula of electric field again

$E_{2}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times(-3\times10^{-6})\times5}{((0.5)^2+5^2)^{\frac{3}{2}}}$

$E_{2}=-1.064\times10^{3}\ N/C$

We need to calculate the resultant electric field

Using formula of electric field

$E=E_{1}+E_{2}$

Put the value into the formula

$E=1.0183\times10^{3}-1.064\times10^{3}$

$E=-0.045\times10^{3}\ N/C$

We need to calculate the force exerted on an electron

Using formula of electric field

$E = \dfrac{F}{q}$

$F=E\times q$

Put the value into the formula

$F=-0.045\times10^{3}\times(-1.6\times10^{-19})$

$F=7.2\times10^{-18}\ N$

Hence, The force exerted on an electron is $7.2\times10^{-18}\ N$

8 0

2 years ago

A torque of 36.5 N · m is applied to an initially motionless wheel which rotates around a fixed axis. This torque is the result

vivado [14]

Answer:

$21.6\ \text{kg m}^2$

$3.672\ \text{Nm}$

$54.66\ \text{revolutions}$

Explanation:

$\tau$ = Torque = 36.5 Nm

$\omega_i$ = Initial angular velocity = 0

$\omega_f$ = Final angular velocity = 10.3 rad/s

t = Time = 6.1 s

I = Moment of inertia

From the kinematic equations of linear motion we have

$\omega_f=\omega_i+\alpha_1 t\\\Rightarrow \alpha_1=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha_1=\dfrac{10.3-0}{6.1}\\\Rightarrow \alpha_1=1.69\ \text{rad/s}^2$

Torque is given by

$\tau=I\alpha_1\\\Rightarrow I=\dfrac{\tau}{\alpha_1}\\\Rightarrow I=\dfrac{36.5}{1.69}\\\Rightarrow I=21.6\ \text{kg m}^2$

The wheel's moment of inertia is $21.6\ \text{kg m}^2$

t = 60.6 s

$\omega_i$ = 10.3 rad/s

$\omega_f$ = 0

$\alpha_2=\dfrac{0-10.3}{60.6}\\\Rightarrow \alpha_1=-0.17\ \text{rad/s}^2$

Frictional torque is given by

$\tau_f=I\alpha_2\\\Rightarrow \tau_f=21.6\times -0.17\\\Rightarrow \tau=-3.672\ \text{Nm}$

The magnitude of the torque caused by friction is $3.672\ \text{Nm}$

Speeding up

$\theta_1=0\times t+\dfrac{1}{2}\times 1.69\times 6.1^2\\\Rightarrow \theta_1=31.44\ \text{rad}$

Slowing down

$\theta_2=10.3\times 60.6+\dfrac{1}{2}\times (-0.17)\times 60.6^2\\\Rightarrow \theta_2=312.03\ \text{rad}$

Total number of revolutions

$\theta=\theta_1+\theta_2\\\Rightarrow \theta=31.44+312.03=343.47\ \text{rad}$

$\dfrac{343.47}{2\pi}=54.66\ \text{revolutions}$

The total number of revolutions the wheel goes through is $54.66\ \text{revolutions}$ .

3 0

2 years ago

What is the acceleration of a Ford Mustang GT that can go from 0.00 to 27.8<br> m/s in 5.15 seconds?

bixtya [17]

Finding acceleration= final speed-initial speed/time taken (or A=V-U\T)

Finial speed= 27.8s
Initial speed= 0s
Time taken= 5.15

So..

27.8-0/5.15= 5.40m/s (rounded to two decimal places)

4 0

2 years ago

A wheel has a constant angular acceleration of 4.5 rad/s2. during a certain 5.0 s interval, it turns through an angle of 128 rad

dalvyx [7]

The solution for this problem is through this formula:Ø = w1 t + 1/2 ã t^2
where:Ø - angular displacement w1 - initial angular velocity t - time ã - angular acceleration
128 = w1 x 4 + ½ x 4.5 x 5^2 128 = 4w1 + 56.254w1 = -128 + 56.25 4w1 = 71.75w1 = 71.75/4
w1 = 17.94 or 18 rad s^-1
w1 = wo + ãt
w1 - final angular velocity
wo - initial angular velocity
18 = 0 + 4.5t t = 4 s

3 0

3 years ago

The six statements below represent Newton's three laws of motion and Kepler's three laws of planetary motion. Match each stateme

mote1985 [20]

Answer:

1. Force = mass x acceleration - Newton

2. A planet moves faster in the part of its orbit nearer the Sun and slower when farther from the Sun, sweeping out equal areas in equal times - Kepler

3. For any force, there is an equal and opposite reaction force - Newton .

4. An object moves at constant velocity if there is no net force acting upon it - Newton

5. The orbit of each planet about the Sun is an ellipse with the Sun at one focus - Kepler.

6. More distant planets orbit the Sun at slower average speeds, obeying the precise mathematical relationship p2-a3 - Kepler.

Explanation:

The three laws of planetary motion formulated by Johannes Kepler or Kepler's laws of planetary motion:

The first law states that the planets move around the Sun in an elliptical orbit with the Sun at one of the foci.
The second law states that the line segment joining a planet to the Sun sweeps out equal areas in equal time.
The third law states that the square of the orbital period (p) of a planet is directly proportional to the cube of the mean distance (a) from the Sun (or semi-major axis of its orbit) i.e., p² is proportional to a³.

The three laws of motion formulated by Sir Isaac Newton or Newton's laws of motion:

The first law, also known as the law of inertia states that an object at rest or moves at a constant velocity will remain at rest or keep moving at a constant velocity unless it is acted upon by a force.
The second law states that the total force (F) applied on an object is directly related to the acceleration (a) of that object produced by the applied force and the mass (m) of the object, i.e., F = ma (assuming the mass m is constant).
The third law, also known as the law of action and reaction states that when an object exerts a force on another object, then the latter exerts a force equal in magnitude and opposite in direction on the former object i.e., for every action, there is an equal and opposite reaction. The example includes the recoiling of a gun when it fires a bullet forward.

5 0

3 years ago