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My name is Ann [436]

2 years ago

14

A force of 185 n is needed to keep a small boat moving at 2.39 m/s. what is the power required to keep the boat moving at the st

eady speed?

1 answer:

mash [69]2 years ago

5 0

We know, P = F . ΔV

P = 185 * 2.39 = 442.15 watt

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Un móvil se encuentra en la posición inicial x0 = 22 m, y se mueve con velocidad 5 m/s. Calcula la posición en la que se encuent

Brilliant_brown [7]

Answer:

La posición en la que se encuentra el móvil en el instante t = 30 s es 172 m.

Explanation:

El movimiento rectilíneo uniforme (MRU) es el movimiento que describe un cuerpo o partícula a través de una línea recta a velocidad constante.

La distancia recorrida, x , por un móvil que tiene un MRU con un velocidad v durante el intervalo de tiempo t es:

x= x0 + v*t

donde x0 es la posición inicial.

En este caso:

x0= 22 m
v= 5 m/s
t= 30 s

Reemplazando:

x= 22 m + 5 m/s* 30 s

Resolviendo:

x= 22 m + 150 m

x= 172 m

<u><em>La posición en la que se encuentra el móvil en el instante t = 30 s es 172 m.</em></u>

8 0

2 years ago

Determine the stopping distances for a car with an initial speed of 88 km/h and human reaction time of 2.0 s for the following a

seropon [69]

Explanation:

Given that,

Initial speed of the car, u = 88 km/h = 24.44 m/s

Reaction time, t = 2 s

Distance covered during this time, $d=24.44\times 2=48.88\ m$

(a) Acceleration, $a=-4\ m/s^2$

We need to find the stopping distance, v = 0. It can be calculated using the third equation of motion as :

$s=\dfrac{v^2-u^2}{2a}$

$s=\dfrac{-(24.44)^2}{2\times -4}$

s = 74.66 meters

s = 74.66 + 48.88 = 123.54 meters

(b) Acceleration, $a=-8\ m/s^2$

$s=\dfrac{v^2-u^2}{2a}$

$s=\dfrac{-(24.44)^2}{2\times -8}$

s = 37.33 meters

s = 37.33 + 48.88 = 86.21 meters

Hence, this is the required solution.

4 0

3 years ago

A swimming pool is 50 ft wide and 100 ft long and its bottom is an inclined plane, the shallow end having a depth of 4 ft and th

Nina [5.8K]

Explanation:

We define force as the product of mass and acceleration.

F = ma

It means that the object has zero net force when it is in rest state or it when it has no acceleration. However in the case of liquids. just like the above mentioned case, the water is at rest but it is still exerting a pressure on the walls of the swimming pool. That pressure exerted by the liquids in their rest state is known as hydro static force.

Given Data:

Width of the pool = w = 50 ft

length of the pool = l= 100 ft

Depth of the shallow end = h(s) = 4 ft

Depth of the deep end = h(d) = 10 ft.

weight density = ρg = 62.5 lb/ft

Solution:

a) Force on a shallow end:

$F = \frac{pgwh}{2} (2x_{1}+h)$

$F = \frac{(62.5)(50)(4)}{2}(2(0)+4)$

$F = 25000 lb$

b) Force on deep end:

$F = \frac{pgwh}{2} (2x_{1}+h)$

$F = \frac{(62.5)(50)(10)}{2} (2(0)+10)$

$F = 187500 lb$

c) Force on one of the sides:

As it is mentioned in the question that the bottom of the swimming pool is an inclined plane so sum of the forces on the rectangular part and triangular part will give us the force on one of the sides of the pool.

1) Force on the Rectangular part:

$F = \frac{pg(l.h)}{2}(2(x_{1} )+ h)$

$x_{1} = 0\\h_{s} = 4ft$

$F = \frac{(62.5)(100)(2)}{2}(2(0)+4)$

$F =25000lb$

2) Force on the triangular part:

$F = \frac{pg(l.h)}{6} (3x_{1} +2h)$

here

h = h(d) - h(s)

h = 10-4

h = 6ft

$x_{1} = 4ft\\$

$F = \frac{62.5 (100)(6)}{6} (3(4)+2(6))$

$F = 150000 lb$

now add both of these forces,

F = 25000lb + 150000lb

F = 175000lb

d) Force on the bottom:

$F = \frac{pgw\sqrt{l^{2} + ((h_{d}) - h(s)) } (h_{d}+h_{s}) }{2}$

$F = \frac{62.5(50)\sqrt{100^{2}(10-4) } (10+4) }{2}$

$F = 2187937.5 lb$

7 0

3 years ago

A photon of wavelength 192 nm strikes an aluminum surface along a line perpendicular to the surface and releases a photoelectron

alex41 [277]

Answer:

$KE=3.529\times10^{−27}\ J$

Explanation:

Given that

Wavelength λ=192 nm

So energy of photon,E

$E=\dfrac{hC}{\lambda }$

Now by putting the values

$h=6.6\times 10^{-34}\ m^2.kg/s$

$C=3\times 10^{8}\ m/s$

$E=\dfrac{6.6\times 10^{-34}\times 3\times 10^{8}}{192\times 10^{-9} }$

$E=1.03\times 10^{-18} J$

We know that

Kinetic energy given as

$KE=\dfrac{P^2}{2m}$

$KE=\dfrac{E^2}{2mC^2}$

$KE=\dfrac{(1.03\times 10^{-18})^2}{2\times 1.67\times 10^{-27}(3\times 10^8)^2}$

$KE=3.529\times10^{−27}\ J$

5 0

2 years ago

Photographs of many young stars show long jets of material apparently being ejected from their poles.

Rudiy27

Answer:

A) True

Explanation:

Researchers have detected numerous jets of gas ejected from poles of young stars and planetary nebulae.

By examining images of hydrogen molecules excited at infrared wavelengths, scientists have been able to see through the gas and dust in the Milky Way, in order to observe the most distant targets. These goals are normally hidden from view and many of them have never been seen before.

The entire study area covers approximately 1,450 times the size of the full moon, or the equivalent of an image of 95 gigapixels. The survey reveals jets emanating from proto-stars and planetary nebulas, as well as remnants of supernovae, the illuminated edges of vast clouds of gas and dust, and the warm regions that surround massive stars and their associated groups of smaller stars.

7 0

3 years ago