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3 years ago

Please help meh its due

Physics

2 answers:

vova2212 [387]3 years ago

6 0

Answer:

-0.000393025 AU

Explanation:

Jupiter’s distance from the sun in AU is 0.000080045 light years. Neptune’s distance from the sun in AU is 0.00047307 light years. 0.000080045-0.00047307=-0.000393025.

Fed [463]3 years ago

3 0

-0.000393025 light years

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A hockey player uses a leaf blower to apply a constant force on a hockey puck. The puck accelerates in a straight line across th

blagie [28]

I will slow down and eventually stop

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3 years ago

Jane has a mass of 55 kg and his body covers 700 nails with a surface area of 1.00 mm,

Oksana_A [137]

We have,

Jane mass is 55 kg
His body covered with 700 nails all of them having a surface area of 1.00 mm² each = 700 × 1 = 700 mm² = 700/1000000 = 7/10000

We know that,

Pressure = Force/Area

Let's calculate force as we already have area;

F = ma
F = 55 × 9.8 { Acceleration due to gravity }
F = 539 N

Now, if should she would be on 700 nails then pressure will be;

P = F/A
P = 539/7 × 10000
P = 5390000/7
P = 770,000 Pascal

And if should would be on a 1 nail only,

P = F/A
P = 539/1 × 1000000
P = 539000000 Pascal

As, you can notice yourself that the pressure will be so high with only 1 nail and because of this, nail will pass through jane's body.

6 0

2 years ago

A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's in

german

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

$\Sigma F = -f = m\cdot a$ (Eq. 1)

Where:

$f$ - Kinetic friction force, measured in newtons.

$m$ - Mass of the box, measured in kilograms.

$a$ - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight ( $W = m\cdot g$ ) and uniform accelerated motion ( $v = v_{o}+a\cdot t$ ), we expand the previous expression:

$-f = \left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$

And the magnitude of the friction force exerted on the box is calculated by this formula:

$f = -\left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$ (Eq. 1b)

Where:

$W$ - Weight, measured in newtons.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$t$ - Time, measured in seconds.

If we know that $W = 52.4\,N$ , $g = 9.807\,\frac{m}{s^{2}}$ , $v_{o} = 1.37\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ and $t = 2.8\,s$ , the magnitud of the kinetic friction force exerted on the box is: