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Blababa [14]
3 years ago
12

(1) find the density of a substance if the mass of the substance is 150kg and dimension 20mby10mby5m

Physics
1 answer:
Vikentia [17]3 years ago
3 0

Answer:

0.15kg/m³

Explanation:

Density = mass/ volume

Given that

Mass = 150kg

Note that volume = length x breadth x height

Volume = 20 x 10 x 5

Volume = 1000m³

Density = mass ➗ volume

Density = 150kg ➗ 1000m³

Density = 0.15kg/m³

I hope this was helpful, Please mark as brainliest  

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If the diameter of the space station is 730 m , how many revolutions per minute are needed for the "artificial gravity" accelera
nignag [31]
For circular motion we know that

<span>F=ma=v^2/r </span>

<span>Therefore: </span>

<span>v = sqrt (rma) </span>

<span>Also, for cicular motion: </span>

<span>rev/min. = 60v/(2r*pi) </span>

<span>So your equation is: </span>

<span>rev./min = 60sqrt(rma)/(2pi*r) </span>

<span>For the mass (m) we can just use 1 kg. 
</span>
rev./min = 60sqrt(730*1*9.8)/(2pi*730) =60sqrt(7154)/(4584.4) 
rev./min = 60sqrt(7154)/(4584.4) =1.11 rev/min
<span>
the answer is </span>1.11 rev/min<span>

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4 0
3 years ago
If Earth's obliquity was 157 degrees, would the seasons be more severe, less severe, or about the same?
ikadub [295]
The severity of the seasons on Earth is given not by the distance Earth-Sun but by the tilt of the Earth axis. This happens because that the sun rays are oblique in winter and perpendicular in summer (thus the same quantity of sun rays heats a bigger surface in winter - oblique rays). 
The present tild of the Earth axis is  23.5 degrees (from the vertical). If the axis were tilt at 157 degree this would be equivalent  to 180-157 =23 degree. Thus the severity of the seasons would be approximately the same but the seasons would be reversed (for example instead of winter we would have summer, instead of summer we would have winter). 
7 0
3 years ago
In the process of loading a ship, a shipping container gets dropped into the water and sinks to the bottom of the harbor. Salvag
jolli1 [7]

Answer:

57.885.8 kg   weight of the container

Explanation:

The volume of the balloon * density of water = buoyant force of balloon

 volume of a sphere = 4/3 pi r^3

                                   = 4/3 pi * (1.5)^3 = 14.14 m^3   <===balloon volume

Now,   find the buoyant force on the container ALONE ....

             5.8 * 2.6 * 2.8  * 1027  = 43 364  kg   <=====  buoyant force

Now add the buoyant force of the balloon to find the weight

             43 364  +   14.14 * 1027 = 57885.8   kg

3 0
2 years ago
11. Hans and Frans are two workers being considered for a job at the UPS loading dock. Hans boasts that he can lift a 100 kg box
Marta_Voda [28]

Answer:

Hans is more powerful

Explanation:

Power: This can be defined as the rate at which work is done or energy is used up.

The expression for power is given as,

P = E/t

P = mgh/t................. Equation 1

Where P = power, W = Work, t = time, m = mass, h = height, g = acceleration due to gravity.

Hans' power

P = mgh/t

Given: m = 100 kg, h = 2 m, g = 9.8 m/s², t = 3 s

Substitute into equation 1

P = 100(9.8)(2)/3

P = 653.33 W.

Frans' power

P' = mgh/t

Given; m = 200 kg, h = 5 m, t = 20 s.

P' = 200(5)(9.8)/20

P' = 9800/20

P' = 490 W

from the above,

since P>P'

Hence, Hans is more powerful

8 0
3 years ago
A student places blocks on a 100cm long see-saw as shown/
Hunter-Best [27]

Answer:

Part 1)

\tau_1 = 5 \times (0.50) = 2.5 N m

Part 2)

\tau_2 = 14 \times (0.30) = 4.2 N m

Part 3)

\tau_3 = 1.4 N m

Part 4)

Since torque on right side is more so here it will turn and slip over it

Explanation:

As we know that the block A is placed at distance

d = 50 cm from the hinge at 70 cm mark

So torque due to weight of A is given as

\tau_1 = 5 \times (0.50) = 2.5 N m

the block B is placed at distance

d = 30 cm from the hinge at 70 cm mark

So torque due to weight of B is given as

\tau_2 = 14 \times (0.30) = 4.2 N m

Now torque due to weight of the scale is given as

\tau_3 = 7(0.20)

\tau_3 = 1.4 N m

now torque on left side of scale is given as

\tau_{left} = \tau_1 + \tau_3

\tau_{left} = 2.5 + 1.4 = 3.9 N m

Torque on right Side is given as

\tau_{right} = \tau_2 = 4.2 Nm

Since torque on right side is more so here it will turn and slip over it

8 0
3 years ago
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