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3 years ago

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A pendulum is swinging back and forth with a period of 2.0 seconds here on Earth. This pendulum is then brought to the Moon, whe

re the acceleration due to gravity is much smaller. What will happen to the period of the pendulum, assuming everything else about it (mass, length, initial swing height, etc) remains exactly the same? Explain your answer.

1 answer:

kompoz [17]3 years ago

7 0

Answer:

A greater period.

Explanation:

The period of a simple pendulum is represented by the following formula:

$T = 2\pi\cdot \sqrt{\frac{l}{g} }$

Where:

$l$ - Length of the pendulum

$g$ - Gravity constant

Given that gravity constant in the Moon is less than on Earth, period shall be greater.

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An electron moves along the z-axis with vz=4.1Ã107m/s. As it passes the origin, what are the strength and direction of the magne

Artist 52 [7]

Answer:

a

The value of magnetic field is $B_a=6.56*10^{-19} T$

In the direction of the positive x - axis

b

The value of magnetic field is $B_a=6.56*10^{-19} T$

In the direction of the positive z - axis

Explanation:

From the question we are told that

The velocity of the electron is $v_z = 4.1*10^{7}m/s$

Considering the first position

The equation for the magnetic field is

$\= B = \frac{\mu_0}{ 4 \pi} \frac{q \=v * \= r}{r^2}$

Now $r = \sqrt{i^2 + j^2 + k^2}$

substituting values

$r = \sqrt{1^2 +0^2 +0^2}$

$= 1$

$\= r = \frac{\r r}{r}$

$\r r = 1 i + 0j + 0k$

Therefore $\= r = \frac{1i + 0j + 0k}{1}$

$= i$

So Substituting $4 \pi *10^ {-7}$ for $\mu_o$ , $1.602 *10^{-19}$ for q

$\= B_a = \frac{4\pi *10^{-7}}{4 \pi} \frac{1.602 *10^{-19} * 4.1*10^{7} * i}{1^2}$

$\= B_a=6.56*10^{-19} T (i)$

Considering the second position

Here

$r = \sqrt{0^2 + 0^2 + 2^2}$

$=2$

$\= r = \frac{\r r}{r}$

$\r r = 0 i + 0j + 2k$

$\= r = \frac{0i + 0j + 2k}{2}$

$= k$

So Substituting $4 \pi *10^ {-7}$ for $\mu_o$ , $1.602 *10^{-19}$ for q

$\= B_a = \frac{4\pi *10^{-7}}{4 \pi} \frac{1.602 *10^{-19} * 4.1*10^{7} * k}{1^2}$

$\= B_a=6.56*10^{-19} T (k)$

6 0

4 years ago

Which law explains why your head would hurt if you bumped it against the door?

eimsori [14]

Newtons law of motion

3 0

3 years ago

Two strong permanent magnets are placed one inch apart. Which two fundamental forces will most affect them?

Anuta_ua [19.1K]

The answer i got was c. gravity and strong
i hope it helps

3 0

3 years ago

A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle

aalyn [17]

Answer:

4.56 s

Explanation:

Let t (seconds) be the time it takes from the moment when the motorcycle starts to accelerate until it catches up with the car. Since prior to this they are traveling at constant speed, they would have maintained a distance of 52 m before accelerating.

The distance traveled by the car, with respect the motorcycle position when it start accelerating is

$s_c = s_0 + v_ct = 52 + 19t$

The distance traveled by the motorcycle after accelerating, with respect the motorcycle position when it start accelerating is

$s_m = v_mt + at^2/2 = 19t + 5t^2/2$

When the motorcycle catches up to the car, their position are at the same

$s_c = s_m$

$52 + 19t = 19t + 5t^2/2$

$52 = 5t^2/2$

$t^2 = 20.8$

$t = \sqrt{20.8} = 4.56 s$

5 0

3 years ago

A stationary 500 kg tank fires a 20 kg miegile at 100 m/s. What is the velocity of the tank after the missile is fired? Assume t

dedylja [7]

Answer:

v₁ = 4 [m/s].

Explanation:

This problem can be solved by using the principle of conservation of linear momentum. Where momentum is preserved before and after the missile is fired.

$P=m*v$

where:

P = linear momentum [kg*m/s]

m = mass [kg]

v = velocity [m/s]

$(m_{1}*v_{1})=(m_{2}*v_{2})$

where:

m₁ = mass of the tank = 500 [kg]

v₁ = velocity of the tank after firing the missile [m/s]

m₂ = mass of the missile = 20 [kg]

v₂ = velocity of the missile after firing = 100 [m/s]

$(500*v_{1})=(20*100)\\v_{1}=2000/500\\v_{1}=4[m/s]$

8 0

3 years ago