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3 years ago

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which statement below is FALSE? A Gravity acts on any object with mass. B The closer objects are, the stronger the gravitational

force between them. C The farther you are from the center of the Earth, the less gravitational force you feel. D Gravitational force depends on what the object is made of.

1 answer:

GenaCL600 [577]3 years ago

5 0

Answer:

Explanation:

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A truck covers 40.0 m in 7.80 s while uniformly slowing down to a final velocity of 1.70 m/s. Please provide with explanation an

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a = -0.223 m/s

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3 years ago

Pulsars are neutron stars that emit X rays and other radiation in such a way that we on Earth receive pulses of radiation from t

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Answer:

(a) $a_{c} = 5.41\times 10^{9} m/s^{2}$

(b) $a_{t} = 2.99\times 10^{- 5} m/s^{2}$

Given:

Time period of Pulsar, $T_{P} = 33.085 ms == 33.085\times 10^{- 3} s$

Equatorial radius, R = 15 Km = 15000 m

Spinning time, $t_{s} = 9.50\times 10^{10}$

Solution:

(a) To calculate the value of the centripetal acceleration, $a_{c}$ on the surface of the equator, the force acting is given by the centripetal force:

$m\times a_{c} = \frac{mv_{c}^{2}}{R}$

$a_{c} = \frac{v_{c}^{2}}{R}$ (1)

where

$v_{c} = \frac{distance covered(i.e., circumference)}{ T}$

$v_{c} = \frac{2\pi R}{Time period, T}$ (2)

Now, from (1) and (2):

$a_{c} = R\frac({2\pi )^{2}}{T^{2}}$

$a_{c} = 15000\frac{2\pi )^{2}}{(33.085\times 10^{- 3})^{2}}$

$a_{c} = 5.41\times 10^{9} m/s^{2}$

(b) To calculate the tangential acceleration of the object :

The tangential acceleration of the object will remain constant and is given by the equation of motion as:

$v = u + a_{t}t_{s} = 0$

where

u = $v_{c}$

$a_{t} = - \frac{2\pi R}{Tt_{s}}$

$a_{t} = - \frac{2\pi 15000}{33.085\times 10^{- 3}\times 9.50\times 10^{10}}$

$a_{t} = 2.99\times 10^{- 5} m/s^{2}$

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3 years ago

A loaded ore car has a mass of 950 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine s

Ierofanga [76]

Answer:

a). P=11.04kW

b). Pmax=11.38 kW

c). Wt=6423.166kJ

Explanation:

The power of the motor when the speed is constant is the work in a determinate time.

$P=\frac{W}{t}$

The work is the force the is applicated in a distance so

W=F*d

replacing:

$P=F*\frac{d}{t}$ and $\frac{d}{t}$ determinate distance in time is velocity so

a).

$P=F*v$

$F=m*a\\F=m*g*sin(33.5)$

$P=950kg*9.8\frac{m}{s^{2}}*sin(33.5)*2.15\frac{m}{s}\\P=11047.846 W\\P=11.0478 kW$

b).

The maximum power must the motor provide, is the maximum force with the maximum speed of the motor in this case

The first step is find the acceleration so

$vi=0\frac{m}{s} \\vf=2.15 \frac{m}{s}\\vf=vi+a*t\\vf-vi=a*t\\ a=\frac{vf-vi}{t}= a=\frac{2.15\frac{m}{s}-0\frac{m}{s}}{13s}\\a=0.1653 \frac{m}{s^{2}}$

The maximum force is when the car is accelerating so

$Ft=Fa+Fg\\Ft=m*a+m*g*sin(33.5)\\Ft=950kg*0.1653\frac{m}{s^{2}}+950*9.8\frac{m}{s^{2}}*sin(33.5)\\Ft=5295.565 N$

so the maximum force is the maximum force by the maximum speed

$Pmax=Ft*v\\Pmax=5295.565N*2.15\frac{m}{s}\\Pmax=11385.46\\Pmax=11.3854kW$

c).

The total energy transfer without any friction is the weight move in the high axis y in this case, so is easy to know that distance

W=m*g*h

h=Length*sin(33.5)

W=m*g*Length*sin(33.5)

W=950 kg*9.8* 1250m*sin(33.5)

W=6423166.667 kJ

W=6423.166 kJ

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4 years ago

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-- loud sounds

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4 years ago

Which of the following best describes what we mean by the universe?

irina1246 [14]

Answer:

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Explanation:

The universe is defined as a closed system that contains all forms of energy and matter. Therefore, includes the so-called dark matter and dark energy, which make up most of the universe, since ordinary matter would only represent just over 5% of the total.

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4 years ago