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3 years ago

What equation gives the position at a specific time for an object with constant acceleration?

Physics

1 answer:

melisa1 [442]3 years ago

6 0

Answer:

$x = x_{0}+v_{0}t+\frac{1}{2}at^{2}$ equation gives the position at a specific time for an object with constant acceleration

Explanation:

$x = x_{0}+v_{0}t+\frac{1}{2}at^{2}$ equation gives the position at specific time for an object having constant acceleration. Constant acceleration is referred as the change of velocity with respect to the time is known as the acceleration, but when the velocity changes occurs at constant rate this rate is termed as the constant acceleration. The constant acceleration can never be zero. The velocity changes but that change occurs in the consistently. The acceleration is affected by the mass.

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What % of an object’s mass is above the water line if the object’s density is 0.82g/ml?

Sladkaya [172]

To develop this problem we will apply the Archimedes model. As well as the definitions of Weight based on mass and acceleration. The first in turn will be considered under the relationship of Density and Volume. From the values given we have to:

$\rho_w =1 g/mL \rightarrow \text{Water Density}$

$\rho_o = 0.82g/mL \rightarrow \text{Object density}$

Since it is in equilibrium, the weight of the object will have a reaction from the water, which will cause the sum of forces between the two objects to be zero, therefore

$\sum F= 0$

$F_w-F_o = 0$

$F_w = F_o$

$m_w g = m_og$

$\rho_w V_w g = \rho_o V_o g$

The value of gravity is canceled because it is a constant

$\frac{V_w}{V_o} = \frac{\rho_o}{\rho_w}$

$\frac{V_w}{V_o} = \frac{0.82}{1}$

$\frac{V_w}{V_o} = 0.82$

The portion of the object that is submerged corresponds to 82%, while the portion that is visible, above the water level will be 18%

4 0

3 years ago

Ree is 27 inches tall on her first birthday. She grows around 6 inches per year after that. What is the rate of change for the s

RSB [31]

Answer:

Explanation:

The rate of change in this scenario corresponds to the number of inches that Ree grows every year. Basically, Ree's height can be written as a linear equation as follows:

$y=mx+q$

where

x is the number of years after the first birthday

y is the height in inches

q = 27 is the height of Ree on her first birthday

m = 6 is the inches gained by Ree at each birthday

So the equation can also be rewritten as

y = 6x + 27

4 0

4 years ago

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A mass of 4kg suspended by a light string 2m long and at rest is projected horizontally with a velocity of 1.5 m/s. find the ang

Dafna11 [192]

Answer:

19.5°

Explanation:

The energy of the mass must be conserved. The energy is given by:

1) $E=\frac{1}{2}mv^2+mgh$

where m is the mass, v is the velocity and h is the hight of the mass.

Let the height at the lowest point of the be h=0, the energy of the mass will be:

2) $E=\frac{1}{2}mv^2$

The energy when the mass comes to a stop will be:

3) $E=mgh$

Setting equations 2 and 3 equal and solving for height h will give:

4) $h=\frac{v^2}{2g}$

The angle ∅ of the string with the vertical with the mass at the highest point will be given by:

5) $cos\phi=\frac{l-h}{l}$

where l is the lenght of the string.

Combining equations 4 and 5 and solving for ∅:

6) $\phi={cos}^{-1}(\frac{l-h}{l})={cos}^{-1}(1-\frac{h}{l})={cos}^{-1}(1-\frac{v^2}{2gl})$

8 0

4 years ago

The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi

Feliz [49]

To solve this problem we will use the Ampere-Maxwell law, which describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

$\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}$

Where,

B= Magnetic Field

l = length

$\mu_0$ = Vacuum permeability

$\epsilon_0$ = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

$B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}$

Recall that the speed of light is equivalent to

$c^2 = \frac{1}{\mu_0 \epsilon_0}$

Then replacing,

$B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}$

$B = \frac{r}{2C^2} \frac{dE}{dt}$

Our values are given as

$dE = 2150N/C$

$dt = 5s$

$C = 3*10^8m/s$

$D = 0.440m \rightarrow r = 0.220m$

Replacing we have,

$B = \frac{r}{2C^2} \frac{dE}{dt}$

$B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}$

$B =5.25*10^{-16}T$

Therefore the magnetic field around this circular area is $B =5.25*10^{-16}T$

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3 years ago

an object sinks to the bottom of a container filled with water.what is the relationship between the weight of the object and the

mars1129 [50]

If the object sinks, then it must be heavier than the weight of the water
it displaces ... heavier than the buoyant force acting on it.

If the buoyant force were equal or greater than the object's weight, then
the object would rise to the surface in water.

6 0

3 years ago

Read 2 more answers