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3 years ago

As an tempature decreases what happens to the rate of radiation?

1 answer:

5 0

As the temperature decreases, the rate of radiation goes down, but the radiation exists as long as the temperature is above the absolute zero, which is actually 0 Kelvin. 0 Kelvin equals -273°C or -460°F. All objects in the world radiate if above that temperature.

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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

3 years ago

A 3 kg rock sits on a 0.8 meter ledge. If it is pushed off, how fast will it be going at the bottom?

andrey2020 [161]

As long as it sits on the shelf, its potential energy
relative to the floor is . . .

 Potential energy = (mass) x (gravity) x (height) =

 (3 kg) x (9.8 m/s²) x (0.8m) = 23.52 joules .

If it falls from the shelf and lands on the floor, then it has exactly that
same amount of energy when it hits the floor, only now the 23.52 joules
has changed to kinetic energy.

 Kinetic energy = (1/2) x (mass) x (speed)²

 23.52 joules = (1/2) x (3 kg) x (speed)²

Divide each side by 1.5 kg : 23.52 m²/s² = speed²

Take the square root of each side: speed = √(23.52 m²/s²) = 4.85 m/s  (rounded)

6 0

3 years ago

A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa

bearhunter [10]

Explanation:

Formula which holds true for a leans with radii $R_{1}$ and $R_{2}$ and index refraction n is given as follows.

$\frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

Since, the lens is immersed in liquid with index of refraction $n_{1}$ . Therefore, focal length obeys the following.

$\frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

$\frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

and, $\frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}$

or, $f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}$

$f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}$

= 32.4 cm

Using thin lens equation, we will find the focal length as follows.

$\frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}$

Hence, image distance can be calculated as follows.

$\frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}$

$s_{i} = \frac{fs_{o}}{s_{o} - f}$

$s_{i} = \frac{32.4 \times 100}{100 - 32.4}$

= 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

4 0

3 years ago

A 1000 kg elevator accelerates upward at 1.0 m/s2 for 10 m, starting from rest. a. How much work does gravity do on the elevator

cricket20 [7]

Answer:

a)= 98kJ

b)=108kJ

c) = 10kJ

Explanation:

a. The work that is done by gravity on the elevator is:

Work = force * distance

= mass * gravity * distance

= 1000 * 9.81 * 10

= 98,000 J

= 98kJ

b)The net force equation in the cable

T - mg = ma

T = m(g+a)

T = 1000(9.8 + 10)

T = 10800N

The work done by the cable is

W = T × d

= 10800N × 10

= 108000

=108kJ

c) PE at 10m = 1000 * 9.81 * 10 = 98,100 J

Work done by cable = PE +KE

108,100 J = KE + 98,100 J

KE = 10,000 J

= 10kJ

7 0

3 years ago

Read 2 more answers

a flag of mass 2.5 kg is supported by a single rope. A strong horizontal wind exerts a force of 12 N on the flag. Calculate the

tatuchka [14]

The free-body diagram of the forces acting on the flag is in the picture in attachment.

We have: the weight, downward, with magnitude
$W=mg = (2.5 kg)(9.81 m/s^2)=24.5 N$
the force of the wind F, acting horizontally, with intensity
$F=12 N$
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
$T \cos \alpha -F=0$
$T \sin \alpha -W=$
By dividing the second equation by the first one, we get
$\tan \alpha = \frac{W}{F}= \frac{24.5 N}{12 N}=2.04$
From which we find
$\alpha = 63.8 ^{\circ}$
which is the angle of the rope with respect to the horizontal.

By replacing this value into the first equation, we can also find the tension of the rope:
$T= \frac{F}{\cos \alpha}= \frac{12 N}{\cos 63.8^{\circ}}=27.2 N$

7 0

4 years ago