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attashe74 [19]
3 years ago
13

What is a Newton unit? kg-m/sec2 m/sec m/sec2 m2/sec

Physics
1 answer:
KiRa [710]3 years ago
7 0
Newton = kg . m/sec^2 
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An object of volume 0.0882 m3 is
vfiekz [6]

The density of the fluid is 776.3 m^{-3}

<u>Explanation:</u>

Buoyant force is the upward pushing force whenever an object is trying to get immersed in fluid. So this is the force given by the fluid on the object which is trying to get immersed. The buoyant force is found to be directly proportional to the product of density of the object, volume of the object. And here the acceleration due to gravity will be acting as proportionality constant.

      Buoyant force = Density \times Volume \times Acceleration

As, buoyant force is given as 671 N and volume is 0.0882 m^{3} and acceleration is known as 9.8 m/s^{2}. Then density is

   \text { Density }=\frac{\text { Buoyant force }}{\text {Volume } \times \text {Acceleration}}

Thus,

   \text { Density }=\frac{671}{0.0882 \times 9.8}=\frac{671}{0.86436}=776.296 \mathrm{kg} / \mathrm{m}^{3}

Density is 776.3 kg m^{-3}.

7 0
3 years ago
A 70mm long blockhas cross-section of 50mm by 10mm the block is subjected to forces 60KN (tension) on the 50mm by 10mm face and
sammy [17]

Answer:

970 kN

Explanation:

The length of the block = 70 mm

The cross section of the block = 50 mm by 10 mm

The tension force applies to the 50 mm by 10 mm face, F₁ = 60 kN

The compression force applied to the 70 mm by 10 mm face, F₂ = 110 kN

By volumetric stress, we have that for there to be no change in volume, the total pressure applied by the given applied forces should be equal to the pressure removed by the added applied force

The pressure due to the force F₁ = 60 kN/(50 mm × 10 mm) = 120 MPa

The pressure due to the force F₂ = 110 kN/(70 mm × 10 mm) = 157.142857 MPa

The total pressure applied to the block, P = 120 MPa + 157.142857 MPa = 277.142857 MPa

The required force, F₃ = 277.142857 MPa × (70 mm × 50 mm) = 970 kN

7 0
2 years ago
At a certain elevation, the pilot of a balloon has a mass of 120 lb and a weight of 119 lbf. What is the local acceleration of g
Strike441 [17]

Answer:

31.905 ft/s²

Explanation:

Given that

Mass of the pilot, m = 120 lb

Weight of the pilot, w = 119 lbf

Acceleration due to gravity, g = 32.05 ft/s²

Local acceleration of gravity of found by using the relation

Weight in lbf = Mass in lb * (local acceleration/32.174 lbft/s²)

119 = 120 * a/32. 174

119 * 32.174 = 120a

a = 3828.706 / 120

a = 31.905 ft/s²

Therefore, the local acceleration due to gravity at that elevation is 31.905 ft/s²

3 0
3 years ago
A roller of radius 12.5 cm turns at 14 revolutions per second. What is the linear velocity of the roller in meters per second?
Firdavs [7]

12.5 times 14 and convert to meters its 1.75 meters per second

7 0
3 years ago
Read 2 more answers
An engine absorbs 1749 J from a hot reservoir and expels 539 J to a cold reservoir in each cycle. a. What is the engine’s effici
Masja [62]

Answer:

a)η = 69.18 %

b)W= 1210 J

c)P=3967.21 W

Explanation:

Given that

Q₁ = 1749 J

Q₂ = 539  J

From first law of thermodynamics

Q₁   = Q₂ +W

W=Work out put

Q₂=Heat rejected to the cold reservoir

Q₁ =heat absorb by hot  reservoir

W= Q₁- Q₂

W= 1210 J

The efficiency given as

\eta=\dfrac{W}{Q_1}

\eta=\dfrac{1210}{1749}

\eta=0.6918

η = 69.18 %

We know that rate of work done is known as power

P=\dfrac{W}{t}

P=\dfrac{1210}{0.305}\ W

P=3967.21 W

8 0
3 years ago
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