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3 years ago

What is the weight of a 24.52kg Television dropped on pluto(acceleration of 0.59m/s/s)?

Physics

2 answers:

jolli1 [7]3 years ago

6 0

W=ma=24.52*0.59 = ...

ella [17]3 years ago

6 0

Answer:

14.47 Newtons

Explanation:

Hello

The weight of an object is the force of gravity on the object and can be defined as the product of mass by the acceleration of gravity, w = mg

for example

a person with mass of 80 kilograms on earth where the acceleration of gravity is 9.8 m/s2 weigh.

w=mg

W=80kg*9.8 m/s2

W=784 Newtons

Now, let see the data we have about the question

$W=mg\\W=24.52 kg* 0.59 \frac{m}{s^{2} } \\W=14.47\ Newtons$

the weight is 14.47 N

Have a great day

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The potential in a region of space due to a charge distribution is given by the expression V = ax2z + bxy − cz2 where a = −3.00

Elis [28]

Answer:

$\vec{E} ( (0, -8.00 \ m, - 8.00 \ m) ) = ( 48.00 \frac{V}{m} , 0 , - 144.00 \frac{V}{m} )$

Explanation:

We know that the relationship between the electric field $\vec{E}(\vec{r})$ and the potential $V(\vec{r})$ is given by

$\vec{E} ( \vec{r}) = - \vec{\nabla} V(\vec{r})$

So, for our potential:

$V(r) = a x^2 z + b x y - c z^2$

the electric field is :

$\vec{E} ( \vec{r}) = - \vec{\nabla} ( a x^2 z + b x y - c z^2 )$

$\vec{E} ( \vec{r}) = - ( \frac{ \partial }{\partial x}, \frac{ \partial }{\partial y} , \frac{ \partial }{\partial z}) ( a x^2 z + b x y - c z^2 )$

$\vec{E} ( \vec{r}) = - ( \frac{ \partial }{\partial x} ( a x^2 z + b x y - c z^2 ) , \frac{ \partial }{\partial y} ( a x^2 z + b x y - c z^2 ) , \frac{ \partial }{\partial z} ( a x^2 z + b x y - c z^2 ))$

$\vec{E} ( \vec{r}) = - ( 2 a x z + b y , b x , a x^2 - 2 c z )$

This is the our electric field. At vector point

$\vec{r} = (0, -8.00 \ m, - 8.00 \ m)$

$\vec{E} ( (0, -8.00 \ m, - 8.00 \ m) ) = - ( 2 a * 0 * (-8.00 \ m) + b (-8.00 \ m) , b * 0 , a 0^2 - 2 c (-8.00 \ m) )$

$\vec{E} ( (0, -8.00 \ m, - 8.00 \ m) ) = - ( 0 - b 8.00 \ m , 0 , 0 + 2 c 8.00 \ m )$

$\vec{E} ( (0, -8.00 \ m, - 8.00 \ m) ) = ( b 8.00 \ m , 0 , - 2 c 8.00 \ m )$

Knowing

$b= 6.00 \frac{V}{m^2}$

and

$c=9.00 \frac{V}{m^2}$

the electric field is

$\vec{E} ( (0, -8.00 \ m, - 8.00 \ m) ) = ( 6.00 \frac{V}{m^2} * 8.00 \ m , 0 , - 9.00 \frac{V}{m^2} * 2* 8.00 \ m )$

$\vec{E} ( (0, -8.00 \ m, - 8.00 \ m) ) = ( 48.00 \frac{V}{m} , 0 , - 144.00 \frac{V}{m} )$

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Nastasia [14]

D.by dots that are all the same length
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3 years ago

If the acceleration due to gravity at the surface of planet x is double the value of earth's, how does planet x's mass compare t

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3 years ago

A 1400 kg car is traveling east on the highway at 31 m/s and collides into the rear of a slower moving pickup truck of 2400 kg,

zloy xaker [14]

Answer: 31 m/s due east

Explanation: this question can be solved using the law of conservation of linear momentum.

This law states that in a closed or isolated system, during collision, the vector sum of momentum before collision equals the vector sum of momentum after collision.

Momentum = mass × velocity

From our question, our parameters before collision are given below as

Mass of car = mc = 1400kg

Speed of car =vc = 31 m/s (due east)

Mass of truck = mt = 2400kg

Velocity of truck = vt = 25 m/s ( due east )

After collision

Velocity of car = ?

Velocity of truck = 34 m/s ( due east )

Vector sum of momentum before collision is given as

1400 (31) + 2400 (25) = 43400 + 60000 = 103400 kgm/s

After collision the truck is seen to move faster (v = 34 m/s) which implies that the car also moves due east .

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3 years ago

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Furkat [3]

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Do not forget about w = angular speed in rad/s and $w1 = 1 revolution/sec = 2Pi (rad/s)$
Now we can go to proportion
$v1=v2$
$w1*r1 = w2r2$ $w2 = w1 * r1/r2 = 2w1 = 4Pi (rad/s)$
$w2 = w3 (which is the angular velocity of the rear wheel) 
$
SOLVING FOR A : $v3 = w3 * r3 = 4pi * 14 (inch/s) = 14.66 ft/sec$
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3 years ago