All categories

3 years ago

A sprinter accelerates at 7.5 m/s from rest in 2.0 s, what distance did she go? (15 m)

Physics

1 answer:

a_sh-v [17]3 years ago

8 0

Answer:

<em>The sprinter traveled a distance of 7.5 m</em>

Explanation:

<u>Motion With Constant Acceleration </u>

It's a type of motion in which the rate of change of the velocity of an object is constant.

The equation that rules the change of velocities is:

$v_f=v_o+at\qquad\qquad [1]$

Where:

a = acceleration

vo = initial speed

vf = final speed

t = time

The distance traveled by the object is given by:

$\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]$

Using the equation [1] we can solve for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

The sprinter travels from rest (vo=0) to vf=7.5 m/s in t=2 s. Computing the acceleration:

$\displaystyle a=\frac{7.5-0}{2}$

$a=3.75\ m/s^2$

Now calculate the distance:

$\displaystyle x=0*2+\frac{3.75*2^2}{2}$

$\displaystyle x=7.5\ m$

The sprinter traveled a distance of 7.5 m

You might be interested in

An inductor in an LC circuit has a maximum current of 2.4 A and a maximum energy of 56 mJ.

Harrizon [31]

Answer:

The energy stored in the capacitor, when the current in the inductor is 1.2 A, is 41.6 mJ.

Explanation:

In a LC oscillating circuit, the energy is stored in the electric field (between the plates of the capacitor) and in the magnetic field (surrounding the wires of the inductor).

At any time, the sum of both energies can be expressed as follows:

E = 1/2 Q² / C + 1/2 L I²

In this type of circuit, energy oscillates, which means that it is exchanging between both fields all time.

When the capacitor is completely discharged, all the energy is stored in the magnetic field, and at that time, the current is maximum.

The total energy, when I is maximum, can be written as follows:

E = 1/2 L I² (1)

In our case, when I= 2.4A, E= 56 mJ.

So, we can find out the value of L, which will allow us to know the value of the magnetic energy at any time, having the value of the instantaneous current.

Solving for L in (1):

L = 2 *.56 mJ / (2.4)² A² = 20 mH

The next step is getting the value of the energy stored in the inductor, when I = 1.2 A, as follows:

Em = 1/2 *20 mH.* (1.2)² A² = 14.4 mJ

As the total energy must be always the same, i.e., 56 mJ, the energy stored in the capacitor, assuming no losses, must be the difference between the total energy and the one stored in the magnetic field:

Ec = 56 mJ - 14.4 mJ = 41.6 mJ

3 0

3 years ago

Derive the equation of motion of the block of mass m1 in terms of its displacement x. The friction between the block and the sur

Alenkasestr [34]

Answer:

the equivalent mass : $m_e = m_1+m_2+\frac{I}{R^2}$

the equation of the motion of the block of mass $m_1$ in terms of its displacement is = $(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)$

Explanation:

Let use m₁ to represent the mass of the block and m₂ to represent the mass of the cylinder

The radius of the cylinder be = R

The distance between the center of the pulley to center of the block to be = x

Also, the angles of inclinations of the cylinder and the block with respect to the ground to be $\phi$ and $\beta$ respectively.

The velocity of the block to be = v

The equivalent mass of the system = $m_e$

In the terms of the equivalent mass, the kinetic energy of the system can be written as:

$K.E = \frac{1}{2} m_ev^2$ --------------- equation (1)

The angular velocity of the cylinder = $\omega$ : &

The inertia of the cylinder about its center to be = I

The angular velocity of the cylinder can be written as:

$v = \omega R$

$\omega =\frac{v}{R}$

The kinetic energy of the system in terms of individual mass can be written as:

$K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I\omega^2$

By replacing $\omega$ with $\frac{v}{R}$ ; we have:

$K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I(\frac{v}{R})^2$

$K.E = \frac{1}{2}(m_1+ m_2+ \frac{I}{R} )v^2$ ------------------ equation (2)

Equating both equation (1) and (2); we have:

$m_e = m_1+m_2+\frac{I}{R^2}$

Therefore, the equivalent mass : $m_e = m_1+m_2+\frac{I}{R^2}$ which is read as;

The equivalent mass is equal to the mass of the block plus the mass of the cylinder plus the inertia by the square of the radius.

The expression for the force acting on equivalent mass due to the block is as follows:

$f_{block }=m_1gsin \beta$

Also; The expression for the force acting on equivalent mass due to the cylinder is as follows:

$f_{cylinder} = m_2gsin \phi$

Equating the above both equations; we have the equation of motion of the equivalent system to be

$m_e \bar x = f_{cylinder}-f_{block}$

which can be written as follows from the previous derivations

$(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)$

Finally; the equation of the motion of the block of mass $m_1$ in terms of its displacement is = $(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)$

8 0

3 years ago

If the car has a mass of 0.4 kg, the ratio of height to width of the ramp is 19/85, the initial displacement is 1.9 m, and the c

Rasek [7]

Answer:

final displacement lf = 0.39 m

Explanation:

from change in momentum equation:

$\delta p = m \sqrt(2g * y/x)* [\sqrt li + \sqrt lf]$

given: m = 0.4kg, y/x = 19/85, li = 1.9 m,

\delta p = 1.27 kg*m/s.

putting all value to get the final displacement value

$1.27 = 0.4\sqrt(2*9.81 *(19/85))* [\sqrt 1.9 + \sqrt lf]$

final displacement lf = 0.39 m

5 0

3 years ago

Two planets A and B, where B has twice the mass of A, orbit the Sun in elliptical orbits. The semi-major axis of the elliptical

lozanna [386]

Answer:

2.83

Explanation:

Kepler's discovered that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit, that is called Kepler's third law of planet motion and can be expressed as:

$T=\frac{2\pi a^{\frac{3}{2}}}{\sqrt{GM}}$ (1)

with T the orbital period, M the mass of the sun, G the Cavendish constant and a the semi major axis of the elliptical orbit of the planet. By (1) we can see that orbital period is independent of the mass of the planet and depends of the semi major axis, rearranging (1):

$\frac{T}{a^{\frac{3}{2}}}=\frac{2\pi}{\sqrt{GM}}$

$\frac{T^{2}}{a^{3}}=(\frac{2\pi }{\sqrt{GM}})^2$ (2)

Because in the right side of the equation (2) we have only constant quantities, that implies the ratio $\frac{T^{2}}{a^{3}}$ is constant for all the planets orbiting the same sun, so we can said that:

$\frac{T_{A}^{2}}{a_{A}^{3}}=\frac{T_{B}^{2}}{a_{B}^{3}}$

$\frac{T_{B}^{2}}{T_{A}^{2}}=\frac{a_{B}^{3}}{a_{A}^{3}}$

$\frac{T_{B}}{T_{A}}=\sqrt{\frac{a_{B}^{3}}{a_{A}^{3}}}=\sqrt{\frac{(2a_{A})^{3}}{a_{A}^{3}}}$

$\frac{T_{B}}{T_{A}}=\sqrt{\frac{2^3}{1}}=2.83$

6 0

3 years ago

Read 2 more answers

The particles that make up protons and neutrons are called

Sphinxa [80]

Answer:

Quark

Explanation:

In the past the proton,neutron and electron were believed to be indivisible. Quarks were later discovered to be particles that make up the proton and neutron. Quarks are known as fast moving points of energy and are usually joined to one another through gluons. Neutron has two down quarks and one up quark while proton has two up quarks and one down quark.

6 0

3 years ago