All categories

3 years ago

Right answer gets brainlist

Physics

1 answer:

ira [324]3 years ago

3 0

Numberrrr 1 Inertiaaaaa

You might be interested in

Suman climb a 5 metre high ladder in 10 second calculate his power the mass of his body is 50 kg

nadezda [96]

Answer:

245W

Explanation:

Given parameters:

Mass of his body = 50kg

Height of climb = 5m

Time = 10s

Unknown:

Power = ?

Solution:

Power is defined as the rate at which work is done;

Mathematically,

Power = $\frac{work done}{time}$

Work done = mgh

m is the mass

g is the acceleration due to gravity

h is the height

Work done = 50 x 9.8 x 5 = 2450J

Power = $\frac{2450}{10}$ = 245W

7 0

3 years ago

A horizontal spring-mass system has low friction, spring stiffness 205 N/m, and mass 0.6 kg. The system is released with an init

Andreyy89

Answer:

(a). The maximum stretch during the motion is 20.7 cm

(b). The maximum speed during the motion is 3.84 m/s.

(c). The energy is 0.060 Watt.

Explanation:

Given that,

Spring stiffness = 205 N/m

Mass = 0.6 kg

Compression of spring = 13 cm

Initial speed = 3 m/s

(a). We need to calculate the maximum stretch during the motion

Using conservation of energy

$E_{initial}=E_{final}$

$\dfrac{1}{2}kx_{c}^2+\dfrac{1}{2}mv^2=\dfrac{1}{2}kx_{m}^2$

Put the value into the formula

$\dfrac{1}{2}\times205\times(13\times10^{-2})^2+\dfrac{1}{2}\times0.6\times3^2=\dfrac{1}{2}\times205\times x_{m}^2$

$x_{m}=\sqrt{\dfrac{4.43\times2}{205}}$

$x_{m}=20.7\ cm$

(b). Maximum speed comes when stretch is zero.

We need to calculate the maximum speed during the motion

Using conservation of energy

$E_{initial}=E_{final}$

$\dfrac{1}{2}kx_{c}^2+\dfrac{1}{2}mv^2=\dfrac{1}{2}mv'^2$

Put the value into the formula

$\dfrac{1}{2}\times205\times(13\times10^{-2})^2+\dfrac{1}{2}\times0.6\times3^2=\dfrac{1}{2}\times0.6\times v'^2$

$v'=\sqrt{\dfrac{4.43\times2}{0.6}}$

$v'=3.84\ m/s$

(c). Now suppose that there is energy dissipation of 0.02 J per cycle of the spring-mass system

We need to calculate the time period

Using formula of time period

$T=2\pi\sqrt{\dfrac{m}{k}}$

Put the value into the formula

$T=2\pi\sqrt{\dfrac{0.6}{205}}$

$T=0.33\ sec$

We need to calculate the energy

Using formula of energy

$E=\dfrac{P}{t}$

Put the value into the formula

$E=\dfrac{0.02}{0.33}$

$E=0.060\ Watt$

Hence, (a). The maximum stretch during the motion is 20.7 cm

(b). The maximum speed during the motion is 3.84 m/s.

(c). The energy is 0.060 Watt.

5 0

3 years ago

In what ways are meteorites different from meteors? What is the probable origin of each?

borishaifa [10]

Answer:

Meteorites have comparatively huge size than the size of the meteors

The meteorites are originated as the debris from an a comet, asteroid, or meteoroid due to a heavy impact. Whereas, the meteors are the dust formed from an asteroids.

Explanation:

The difference between the meteorites from the meteors is that, the meteorites have comparatively huge size than the size of the meteors. Also, the meteorites are capable enough to survive moving through the earth's atmosphere. while the meteors burns up due to heat produced by the friction while moving in the earth's atmosphere.

The meteorites are originated as the debris from an a comet, asteroid, or meteoroid due to a heavy impact. Whereas, the meteors are the dust formed from an asteroids.

6 0

4 years ago

How much ancient plant debris is necessary to form one foot of coal?

jenyasd209 [6]

It is said that ten feet of ancient plant debris was needed in order to make one foot of coal. This is true because fossil fuels were formed from plants and animals that lived 300 million years ago in primordial swamps and oceans. <span>Eventually, many of the seas receded and left dry land with fossil fuels like the one you are talking about, coal</span>

3 0

4 years ago

Un depósito de gran superficie se llena de agua hasta una altura de 0,3 m. En el fondo del depósito hay un orificio de 5 cm2 de

ahrayia [7]

Answer:

a) El caudal de salida del chorro es $1.213\times 10^{-3}\,\frac{m^{3}}{s}$ .

Explanation:

a) Asúmase que el tanque se encuentra a presión atmósferica y que la sima del tanque tiene una altura de 0 metros. La rapidez de salida del chorro del depósito se determined a partir del Principio de Bernoulli, cuya línea de corriente entre la cima y la sima del tanque queda descrita por la siguiente ecuación:

$\Delta z = \frac{v_{out}^{2}}{2\cdot g}$

Donde:

$\Delta z$ - Diferencia de altura, medida en metros.

$g$ - Constante gravitacional, medida en metros por segundo al cuadrado.

$v_{out}$ - Rapidez de salida del chorro, medida en metros por segundo.

Se despeja la rapidez de salida del chorro:

$v_{out} = \sqrt{2\cdot g \cdot \Delta z}$

Si $g = 9.807\,\frac{m}{s^{2}}$ y $\Delta z = 0.3\,m$ , entonces la rapidez de salida del chorro es:

$v_{out} = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.3\,m)}$

$v_{out} \approx 2.426\,\frac{m}{s}$

Ahora, la cantidad de líquido que sale del depósito por unidad de tiempo se obtiene al multiplicar la rapidez de salida del chorro por el área transversal del orificio. Esto es:

$\dot V_{out} = v_{out}\cdot A_{t}$