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3 years ago

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An iceberg of density 920kg/m^3 floats in seawater of density 1025kg/m^3 with volume of 10^6m^3.What is the total mass of the ic

eberg?

1 answer:

zhenek [66]3 years ago

5 0

Im pretty sure the answer is 1.95x10^9

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The heating of groundwater forms

KIM [24]

The appropriate answer is c. geysers. A geysers a hot water fountain that spouts intermittently with great force, frequently accompanied by a thunderous roar. The world famous Old Faithful is located in Yellowstone National Park. This geyser erupts every 65 min sending a jet of water almost 60 meters into the air. Sinkholes and caves are formed by the action of groundwater on carbonate rocks which causes them to colapse and former these structures.

6 0

3 years ago

Read 2 more answers

The average mass of a car in the US is 1.440 x 10^6 g. Express this mass in kg.

wolverine [178]

Answer:

Average mass of acar in the US (in kg) = 1440 kg

Explanation:

Average mass of a car in the US (in g) = 1.440 × 10⁶ g

Mass in kg:

$\rm 1 \: g = {10}^{ - 3} \: kg \\ \\ \rm 1.440 \times 10^6 \ g = 1.440 \times 10^6 \times {10}^{ - 3} \ kg \\ \\ \rm = 1.440 \times 10^{6 - 3} \ kg \\ \\ \rm = 1.440 \times 10^3 \ kg \\ \\ \rm = 1.440 \times 1000 \ kg \\ \\ \rm = 1440 \ kg$

5 0

3 years ago

What is the magnitude of the force that is exerted on a 20 kg mass to give it an acceleration of 10.0

ANEK [815]

Answer:Mass of the body = 20 kg.

Final Velocity = 5.8 m/s.

Initial velocity = 0

Time = 3 seconds.

Using the Formula,

Acceleration = (v - u)/ t

= (5.8 - 0)/ 3

= 1.6 m /s².

Now, Using the Formula,

Force = mass × acceleration

= 20 × 1.6

=

Explanation: I REALLY HOPE THIS HELPS I'M KINDA NEW AT THIS :] :]

8 0

3 years ago

A 100-W lightbulb is placed in a cylinder equipped with a moveable piston. The lightbulb is turned on for 0.010 hour, and the as

Taya2010 [7]

Answer:

$w = - 508.53$ joules

$q = - 3091.47$ joules

Explanation:

Let us convert the time in hours into seconds

$0.010* 3600\\= 36$

Change in internal energy

$\delta E = p * \delta t$

where E is the internal energy in Joules

p is the power in watts

and t is the time in seconds

$\delta E = - 100 * 36\\$

$\delta E = - 3600$ Joules

Amount of work done by the system

$w = - P * \delta V$

where P is the pressure and V is the volume

Substituting the given values in above equation, we get -

$w = - 1 * ( 5.92 -0.90)\\$

$w = -5.02$ liter-atmospheres

Work done in Joules

$- 5.02 * 101.3\\$ $= 508.53$ Joules

$q = \delta E - w\\$

Substituting the given values we get -

$q = - 3600 - (-508.53)\\q = - 3091.47$

Thus

$w = - 508.53$ joules

$q = - 3091.47$ joules

7 0

3 years ago

When a body of mass 0.25 kg is attached to a vertical massless spring, it is extended 5.0 cm from its unstretched length of 4.0

lora16 [44]

Answer:

$d=0.165m$

Explanation:

Given

$m=0.25kg$ , $x_{1}=5cm*\frac{1m}{100cm}=0.05m$ , $x_{2}=4cm*\frac{1m}{100cm}=0.04m$ , $v=2\frac{rev}{s}$

The tension of the spring is

$F_{k}=K*x_{1}=m*g$

$K=\frac{m*g}{x_{1}}$

$K=\frac{0.25kg*9.8m/s^2}{0.05m}=49N/m$

The force in the spring is equal to centripetal force so

$F_{c}=\frac{m*v^2}{r}$

$v=w*r=2\pi*r$

But Fc is also

Fc=KxΔr

$F_{c}=K*(r-x_{2})$

Replacing

$m*4\pi^2*r=K*(r-x_{2})$

$0.25kg*4\pi^2*r=49*(r-0.04m)$

$r=0.205m$

total distance is

$d=0.205-0.04=0.165m$

3 0

3 years ago