How much power is needed to push a 95 kg chest at 0.67 m/s along a horizontal floor where the coefficient of friction is 0.77?

A square wave has amplitude 0 V for the low voltage and 4 V for the high voltage. Calculate the average voltage by integrating o

Margarita [4]

Answer:

V_{average} = $\frac{1}{2} V_o$ , V_{average} = 2 V

Explanation:

he average or effective voltage of a wave is the value of the wave in a period

V_average = ∫ V dt

in this case the given volage is a square wave that can be described by the function

V (t) = $\left \{ {{V=V_o \ \ \ t< \tau /2} \atop {V=0 \ \ \ \ t> \tau /2 } } \right.$

to substitute in the equation let us separate the into two pairs

V_average = $\int\limits^{1/2}_0 {V_o} \, dt + \int\limits^1_{1/2} {0} \, dt$

V_average = $V_o \ \int\limits^{1/2}_0 {} \, dt$

V_{average} = $\frac{1}{2} V_o$

we evaluate V₀ = 4 V

V_{average} = 4 / 2)

V_{average} = 2 V

6 0

2 years ago

The girl makes a microscope with a 3.0 cm focal length objective and a 5.0 cm eyepiece. The microscope tube length is 10 cm. Use

saul85 [17]

To solve this problem we will use the concepts related to Magnification. Magnification is the process of enlarging the apparent size, not physical size, of something. This enlargement is quantified by a calculated number also called "magnification".

The overall magnification of microscope is

$M = \frac{Nl}{f_ef_0}$

Where

N = Near point

l = distance between the object lens and eye lens

$f_0$ = Focal length

$f_e$ = Focal of eyepiece

Given that the minimum distance at which the eye is able to focus is about 25cm we have that N = 25cm

Replacing,

$M = \frac{25*10}{3*5}$

$M = 16.67\approx 17\\$

Therefore the correct answer is C.

3 0

3 years ago

One of the waste products of a nuclear reactor is plutonium-239 . This nucleus is radioactive and decays by splitting into a hel

Gekata [30.6K]

Answer:

a) $v_{U-235} = 2.68 \cdot 10^{5} m/s$

$v_{He-4} = -1.57 \cdot 10^{7} m/s$

b) $E_{He-4} = 8.23 \cdot 10^{-13} J$

$E_{U-235} = 1.41 \cdot 10^{-14} J$

Explanation:

Searching the missed information we have:

E: is the energy emitted in the plutonium decay = 8.40x10⁻¹³ J

m(⁴He): is the mass of the helium nucleus = 6.68x10⁻²⁷ kg

m(²³⁵U): is the mass of the helium U-235 nucleus = 3.92x10⁻²⁵ kg

a) We can find the velocities of the two nuclei by conservation of linear momentum and kinetic energy:

Linear momentum:

$p_{i} = p_{f}$

$m_{Pu-239}v_{Pu-239} = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}$

Since the plutonium nucleus is originally at rest, $v_{Pu-239} = 0$ :

$0 = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}$

$v_{He-4} = -\frac{m_{U-235}v_{U-235}}{m_{He-4}}$ (1)

Kinetic Energy:

$E_{Pu-239} = \frac{1}{2}m_{He-4}v_{He-4}^{2} + \frac{1}{2}m_{U-235}v_{U-235}^{2}$

$2*8.40 \cdot 10^{-13} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}$

$1.68\cdot 10^{-12} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}$ (2)

By entering equation (1) into (2) we have:

$1.68\cdot 10^{-12} J = m_{He-4}(-\frac{m_{U-235}v_{U-235}}{m_{He-4}})^{2} + m_{U-235}v_{U-235}^{2}$

$1.68\cdot 10^{-12} J = 6.68 \cdot 10^{-27} kg*(-\frac{3.92 \cdot 10^{-25} kg*v_{U-235}}{6.68 \cdot 10^{-27} kg})^{2} +3.92 \cdot 10^{-25} kg*v_{U-235}^{2}$