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kakasveta [241]
3 years ago
9

A simple pendulum consisting of a bob of mass m attached to a string of length L swings with a period T. If the pendulum is take

n into the orbiting space station what will happen to the bob
Physics
1 answer:
Juli2301 [7.4K]3 years ago
5 0

To solve this problem we will use the definition of the period in a simple pendulum, which warns that it is dependent on its length and gravity as follows:

T =2\pi \sqrt{\frac{L}{g}}

Here,

L = Length

g = Acceleration due to gravity

We can realize that 2 \pi is a constant so it is proportional to the square root of its length over its gravity,

T \propto \sqrt{\frac{L}{g}}

Since the body is in constant free fall, that is, a point where gravity tends to be zero:

g \rightarrow 0 \Rightarrow T \rightarrow \infty

The value of the period will tend to infinity. This indicates that the pendulum will no longer oscillate because both the pendulum and the point to which it is attached are in free fall.

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A power plant uses Uranium<br> to produce energy
antoniya [11.8K]

Answer:

how is that a question?

Explanation:

yeah i dunno the answer cause thats not a question

7 0
3 years ago
A block is thrown with an initial velocity of 30.0 m/s at an angle of 25.0o above the horizontal. What is the highest elevation
Serga [27]

The highest elevation reached by the ball in its trajectory is 16.4 m.

To find the answer, we need to know about the maximum height reached in a projectile.

What's the mathematical expression of the maximum height reached in a projectile motion?

  • The maximum height= U²× sin²(θ)/g
  • U= initial velocity, θ= angle of projectile with horizontal and g= acceleration due to gravity

What's the maximum height reached by a block that is thrown with an initial velocity of 30.0 m/s at an angle of 25° above the horizontal?

  • Here, U = 30.0 m/s and θ= 25°
  • Maximum height= 30²× sin²(25)/9.8

= 16.4m

Thus, we can conclude that the highest elevation reached by the ball in its trajectory is 16.4 m.

Learn more about the projectile motion here:

brainly.com/question/24216590

#SPJ4

3 0
2 years ago
How does gravity affect the motions of Earth and the Moon?
RoseWind [281]

Answer:

The Sun's gravitational pull keeps our planet orbiting the Sun. The motion of the Moon is affected by the gravity of the Sun and Earth. Moon's gravity pulls on the Earth and makes the tides rise and fall.

8 0
2 years ago
Read 2 more answers
A cubical Gaussian surface surrounds two positive charges, each has a charge q 1 1 = + 3.90 × 10 − 12 3.90×10−12 C, and three ne
Masteriza [31]

Answer:

The electric flux is zero because charge is zero.

Explanation:

Given that,

Positive charge q_{1}=3.90\times10^{-12}\ C

Negative charge q_{2}=-2.60\times10^{-12}\ C

We need to calculate the total charged

Using formula of charge

Q_{enc}=2q_{1}+3q_{2}

Put the value into the formula

Q_{enc}=2\times3.90\times10^{-12}+3\times(-2.60\times10^{-12})

Q_{enc}=0

We need to calculate the electric flux

Using formula of electric flux

\phi=\dfrac{Q_{enc}}{\epsilon_{0}}

Put the value into the formula

\phi=\dfrac{0}{8.85\times10^{-12}}

Hence, The electric flux is zero because charge is zero.

7 0
3 years ago
a ballistic pendulum is used to measure the speed of high-speed projectiles. A 6 g bullet A is fired into a 1 kg wood block B su
Galina-37 [17]

Answer:

(a) v-bullet = 399.04 m/s

(b) I = 2.38 kg m/s

(c) T = 2.59 N

Explanation:

(a) To calculate the initial speed of the bullet, you first take into account that the kinetic energy of both wood block and bullet, just after the bullet impacts the block, is equal to the potential gravitational energy of block and bullet when the cord is at 60° respect to the vertical.

The potential energy is given by:

U=(M+m)gh       (1)

U: potential energy

M: mass of the wood block = 1 kg

m: mass of the bullet = 6g = 6.0*10^-3 kg

g: gravitational constant = 9.8m/s^2

h: distance to the ground

The distance to the ground is calculate d by using the information about the length of the cord and the degrees of the cord respect to the vertical:

h=l-lsin\theta\\\\h=2.2m-2,2m\ sin60\°=0.29m

The potential energy is:

U=(1kg+6*10^{-3}kg)(9.8m/s^2)(0.29m)=2.85J

Next, the potential energy is equal to kinetic energy of the block and the bullet at the beginning of its motion:

U=\frac{1}{2}(M+m)v^2\\\\v=\sqrt{2\frac{U}{M+m}}=\sqrt{2\frac{2.85J}{1kg+6*10^{-3}kg}}=2.38\frac{m}{s}

Next, you use the momentum conservation law, in order to calculate the speed of the bullet before the impact:

Mv_1+mv_2=(M+m)v    (2)

v1: initial velocity of the wood block = 0m/s

v2: initial speed of the bullet

v: speed of bullet and block = 2.38m/s

You solve the equation (2) for v2:

M(0)+mv_2=(M+m)v    

v_2=\frac{M+m}{m}v=\frac{1kg+6*10^{-3}kg}{6*10^{-3}kg}(2.38m/s)\\\\v_2=399.04\frac{m}{s}

The speed of the bullet before the impact with the wood block is 399.04 m/s

(b) The impulse is gibe by the change in the velocity of the block, multiplied by the mass of the block:

I=M\Delta v=M(v-v_1)=(1kg)(2.38m/s-0m/s)=2.38kg\frac{m}{s}

The impulse is 2.38 kgm/s

(c) The force on the cord after the impact is equal to the centripetal force over the block and bullet. That is:

T=F_c=(M+m)\frac{v^2}{l}=(1.006kg)\frac{(2.38m/s)^2}{2.2m}=2.59N    

The force on the cord after the impact is 2.59N

4 0
3 years ago
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