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Sveta_85 [38]
3 years ago
10

What is the length of AC?

Physics
1 answer:
Grace [21]3 years ago
7 0
D: 16 AB is half of AC and since you’re given AB=8 to get the full length you just multiply it by 2
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A girl swings a 0.250 kg rock attached to a taut string in a circle around her head. Her hand holds the end of the string above
nydimaria [60]

Complete Question

The diagram of with this question is shown on the first uploaded image

Answer:

The value is  v = -6.543  \^  i + 9.47 \^ j + 0 \^ k

Explanation:

From the question we are told that

   The mass of the rock is  m = 0.250 \ kg

    The length of the string is  L = 0.75 \  m

    The angle the string makes horizontal is  \theta  = 11.9^o

     The angle which the projection of the string onto  the xy -plane makes with the positive x-axis is  \phi = 34.6^o

    The angular velocity of the rock is  w = 2.50 rev/s  = 2.50 * 2\pi  =15.7 \ rad/s

Generally the radius of the circle made by the length of the string is mathematically represented as

               r = L cos(\theta )

=>            r = 0.75  cos(11.9 )

=>            r = 0.734 \ m

Generally the resultant tangential velocity is mathematically represented as

      v__{R}}  = w * r

=>  v__{R}}  = 15.7  *0.734

=>  v__{R}}  =  11.5 \ m/s

Generally the tangential velocity along the x-axis is  

      v_x  = -v__{R}} *  sin(\phi)

=>   v_x  =- 11.5 *  sin(34.6)

=>   v_x  = -6.543 \ m/s

The negative sign show that the velocity is directed toward the negative x-axis

Generally the tangential velocity along the y-axis is  

      v_y = v__{R}} *  cos(\phi)

=>   v_y  = 11.5 *  cos(34.6)

=>   v_y  = 9.47 \ m/s

Generally the tangential velocity along the y-axis is  

      v_z = v__{R}} *  cos(90)

=>   v_z = 0 \ m/s

Generally the tangential velocity at that instant is mathematically represented as

       v = -6.543  \^  i + 9.47 \^ j + 0 \^ k

8 0
3 years ago
When an object is fully converted into energy the amount of energy liberated is
EleoNora [17]

Answer:

Mass, m = 4 kg

Explanation:

<u>Given the following data;</u>

Energy = 3.6 * 10^17 Joules

We know that the speed of light is equal to 3 * 10⁸ m/s.

To find the mass of the substance;

The theory of special relativity by Albert Einstein gave birth to one of the most famous equation in science.

The equation illustrates, energy equals mass multiplied by the square of the speed of light.

Mathematically, the theory of special relativity is given by the formula;

E = mc^{2}

Where;

  • E is the energy possessed by a substance.
  • m is the mass.
  • c is the speed of light.

Substituting into the formula, we have;

3.6 * 10^{17} = m * 300000000^{2}

3.6 * 10^{17} = m * 9*10^{16}

m = \frac {3.6 * 10^{17}}{9*10^{16}}

Mass, m = 4 kg

8 0
3 years ago
A barrel 1 m tall and 60 cm in diameter is filled to the top with water. What is the pressure it exerts on the floor beneath it?
allsm [11]

Answer:

The pressure on the ground is about 9779.5 Pascal.

The pressure can be reduced by distributing the weight over a larger area using, for example, a thin plate with an area larger than the circular area of the barrel's bottom side.  See more details further below.

Explanation:

Start with the formula for pressure

(pressure P) = (Force F) / (Area A)

In order to determine the pressure the barrel exerts on the floor area, we need the calculate the its weight first

F_g = m \cdot g

where m is the mass of the barrel and g the gravitational acceleration. We can estimate this mass using the volume of a cylinder with radius 30 cm and height 1m, the density of the water, and the assumption that the container mass is negligible:

V = h\pi r^2=1m \cdot \pi\cdot 0.3^2 m^2\approx 0.283m^3

The density of water is 997 kg/m^3, so the mass of the barrel is:

m = V\cdot \rho = 0.283 m^3 \cdot 997 \frac{kg}{m^3}= 282.151kg

and so the weight is

F_g = 282.151kg\cdot 9.8\frac{m}{s^2}=2765.08N

and so the pressure is

P = \frac{F}{A} = \frac{F}{\pi r^2}= \frac{2765.08N}{\pi \cdot 0.3^2 m^2}\approx 9779.5 Pa

This answers the first part of the question.

The second part of the question asks for ways to reduce the above pressure without changing the amount of water. Since the pressure is directly proportional to the weight (determined by the water) and indirectly proportional to the area, changing the area offers itself here. Specifically, we could insert a thin plate (of negligible additional weight) to spread the weight of the barrel over a larger area. Alternatively, the barrel could be reshaped (if this is allowed) into one with a larger diameter (and smaller height), which would achieve a reduction of the pressure.  

7 0
3 years ago
How does Newton's first law affect human motions answer
vazorg [7]
The First Law states, "A body at rest will remain at rest, and a body in motion will remain in motion unless it is acted upon by an external force." This simply means that things cannot start, stop or change direction all by themselves. It requires some force acting on them from the outside to cause such a change. Hope this helps! Mark brainly please!
8 0
3 years ago
German physicist Werner Heisenberg related the uncertainty of an object's position ( Δ x ) to the uncertainty in its velocity (
Ierofanga [76]

Answer:

The uncertainty in the position of the electron is 5.79x10^{-9}m

Explanation:

The Heisenberg uncertainty principle is defined as:

\Lambda p\Lambda x ≥ \frac{h}{4 \pi}  (1)

Where \Lambda p is the uncertainty in momentum, \Lambda x is the uncertainty in position and h is the Planck's constant.

The momentum is defined as:

p =mv  (2)

Therefore, equation 2 can be replaced in equation 1

\Lambda (mv) \Lambda x ≥ \frac{h}{4 \pi}

Since, the mass of the electron is constant, v will be the one with an associated uncertainty.

m \Lambda v \Lambda x ≥ \frac{h}{4 \pi} (3)

Then, \Lambda x can be isolated from equation 3

\Lambda x ≥ \frac{h}{m \Lambda v 4 \pi}  (4)

\Lambda x = \frac{6.626x10^{-34}J.s}{(9.11x10^{-31} kg)(0.01x10^{6}m/s) 4 \pi}

But 1J = Kg.m^{2}/s^{2}

\Lambda x = \frac{(6.624x10^{-34} Kg.m^{2}/s^{2}.s)}{(9.11x10^{-31} kg)(0.01x10^{6}m/s) (4 \pi)}

\Lambda x = 5.79x10^{-9}m

Hence, the uncertainty in the position of the electron is  5.79x10^{-9}m

7 0
3 years ago
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