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Nata [24]
3 years ago
12

An object weighs 7.84 N when it is in air and 6.86 N when it is immersed in water of density 1000 kg/m3. What is the density of

the object?
Physics
1 answer:
kirill [66]3 years ago
3 0

Answer:

8000 kg/m^3

Explanation:

Weight in air = 7.84 n

Weight in water = 6.86 N

density of water = 1000 kg/m^3

Let d be the density of object

According to the Archimedes principle, when a body is immersed in a liquid partly or wholly, it experiences an upward force which is called buoyant force. The buoyant force is equal to the loss in weight of the body.

Loss in weight of the object = Weight of object in air - weight of object in water

Loss in weight = 7.84 - 6.86 = 0.98 N

Volume of body x density of water x g = 0.98

Let V be the volume of body

V x 1000 x 9.8 = 0.98

V = 10^-4 m^3

Weight in air = Volume of body x density of body x g

7.84 = 10^-4 x d x 9.8

d = 8000 kg/m^3

Thus, the density of body is 8000 kg/m^3.

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A 1350 kg uniform boom is supported by a cable. The length of the boom is l. The cable is connected 1/4 the
olchik [2.2K]

Answer:

Tension= 21,900N

Components of Normal force

Fnx= 17900N

Fny= 22700N

FN= 28900N

Explanation:

Tension in the cable is calculated by:

Etorque= -FBcostheta(1/2L)+FT(3/4L)-FWcostheta(L)= I&=0 static equilibrium

FTorque(3/4L)= FBcostheta(1/2L)+ FWcostheta(L)

Ftorque=(Fcostheta(1/2L)+FWcosL)/(3/4L)

Ftorque= 2/3FBcostheta+ 4/3FWcostheta

Ftorque=2/3(1350)(9.81)cos55° + 2/3(2250)(9.81)cos 55°

Ftorque= 21900N

b) components of Normal force

Efx=FNx-FTcos(90-theta)=0 static equilibrium

Fnx=21900cos(90-55)=17900N

Fy=FNy+ FTsin(90-theta)-FB-FW=0

FNy= -FTsin(90-55)+FB+FW

FNy= -21900sin(35)+(1350+2250)×9.81=22700N

The Normal force

FN=sqrt(17900^2+22700^2)

FN= 28.900N

4 0
3 years ago
A box is against a wall. A person pushes on the box, but the box does not move. Is this
Serhud [2]

Answer:

it is a force

Explanation:

a force is a push or a pull

7 0
2 years ago
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a hawk flies in a horizontal arc of radius 10.3 m at a constant speed of 4.8 m/s. find its centripetal acceleration. answer in u
n200080 [17]

The hawk’s centripetal acceleration is 2.23 m/s²

The magnitude of the acceleration under new conditions is 2.316 m/s²

radius of the horizontal arc = 10.3 m

the initial constant speed = 4.8 m/s

we know that the centripetal acceleration is given by

    a_{c}  = \frac{v^{2} }{r}

   a_{c}  = 23.04/10.3

    a_{c}  = 2.23 m/s²

It continues to fly but now with some tangential acceleration

a_{t} = 0.63 m/s²

therefore the net value of acceleration is given by the resultant of the centripetal acceleration and the tangential acceleration

so

a_{net}  =  \sqrt{a_{c} ^{2} +a_{t} ^{2}   }

a_{net}  =  \sqrt{4.97 + 0.396}

a_{net}  =  2.316 m/s²

So the magnitude of  net acceleration will become 2.316 m/s².

learn more about acceleration here :

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8 0
1 year ago
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Nonamiya [84]

\\ \bull\tt\longrightarrow P=\dfrac{V^2}{R}

  • P is power
  • R is resistance

\\ \bull\tt\longrightarrow R=\dfrac{V^2}{P}

Hence

\\ \bull\tt\longrightarrow R\propto V

\\ \bull\tt\longrightarrow R\propto \dfrac{1}{P}

  • Therefore if power is low then resistance will be high.

The first bulb has less power hence it has greater filament resistance.

5 0
3 years ago
Explain in details the concept of economic geography​
vova2212 [387]

Answer:

In the words of Hartshorn and Alexander: “Economic Geography is the study of the spatial variation on the earth’s surface of activities related to producing, exchanging and consuming goods and services. Whenever possible the goal is to develop generalizations and theories to account for these spatial variations.”

Explanation:

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3 years ago
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