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Artist 52 [7]
3 years ago
6

Two Atoms are isotopes of each other if they have the same number of_______ but a different number _________.

Physics
1 answer:
zloy xaker [14]3 years ago
4 0
The correct answer is #4. They have the same number of protons but a different number of neutrons.
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What will change the velocity of a periodic wave?
Georgia [21]
If it's a mechanical wave, then its speed depends on the physical characteristics of the medium.

If it's an electromagnetic wave, then its speed depends on the
electrical characteristics of the medium.

Either way, the properties of the medium determine the wave speed.
You want to change the speed ?  You have to change the properties
of the medium.
7 0
3 years ago
Christina drives his moped 7 kilometers North. She stops for lunch and then drives 5 kilometers East. What distance did she cove
Mashutka [201]

Answer:

She covers the distance is 12 km.

The magnitude of displacement is 8.6 km.

The direction of her displacement is north east.

Explanation:

Given that,

Christina drives his moped 7 kilometers North and stop for lunch and then drive 5 km east.

We need to calculate the total distance

Using formula of distance

d=d_{1}+d_{2}

Put the value into the formula

d=7+5

d=12\ km

We need to calculate the magnitude of displacement

Using formula of displacement

D=xi+yj

D=5i+7j

D=\sqrt{5^2+7^2}

D= 8.6\ km

The direction of her displacement is north east.

Hence, She covers the distance is 12 km.

The magnitude of displacement is 8.6 km.

The direction of her displacement is north east.

6 0
3 years ago
The current through a 10 ohm resistor connected to a 120 volt power supply is
Leno4ka [110]

Answer:I=12 A

Explanation:

Given

Resistance R=10 \Omega

Voltage V=120 V

According to ohm's law current through a conductor is directly proportional to the voltage applied.

V\propto I

V=IR

where V=Voltage

I=Current

R=resistance

I=\frac{V}{R}

I=\frac{120}{10}

I=12 A

4 0
3 years ago
One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has
Mars2501 [29]

Answer:

6.86 m/s

Explanation:

This problem can be solved by doing the total energy balance, i.e:

initial (KE + PE)  = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}

Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.

Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.

Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s

The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²

This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)

Now, 1/6m(v0)² = mgL ⇒ v0 = \sqrt{6gL}

Hence, v0 = 6.86 m/s

4 0
3 years ago
A mass is oscillating with amplitude A at the end of a spring.
Dmitry_Shevchenko [17]

A) x=\pm \frac{A}{2\sqrt{2}}

The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):

E=\frac{1}{2}kA^2 (1)

where k is the spring constant.

The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:

E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2 (2)

where x is the displacement, m the mass, and v the speed.

We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

U=\frac{1}{3}K

Using (2) we can rewrite this as

U=\frac{1}{3}(E-U)=\frac{1}{3}E-\frac{1}{3}U\\U=\frac{E}{4}

And using (1), we find

U=\frac{E}{4}=\frac{\frac{1}{2}kA^2}{4}=\frac{1}{8}kA^2

Substituting U=\frac{1}{2}kx^2 into the last equation, we find the value of x:

\frac{1}{2}kx^2=\frac{1}{8}kA^2\\x=\pm \frac{A}{2\sqrt{2}}

B) x=\pm \frac{3}{\sqrt{10}}A

In this case, the kinetic energy is 1/10 of the total energy:

K=\frac{1}{10}E

Since we have

K=E-U

we can write

E-U=\frac{1}{10}E\\U=\frac{9}{10}E

And so we find:

\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A

3 0
3 years ago
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