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3 years ago

11 Design Imagine that a scientistdiscovered a way to make africtionless surface. What wouldbe some useful applications forthis

Physics

1 answer:

m_a_m_a [10]3 years ago

3 0

Answer:

You could move something across the Earth with a little push. It would make fuel really efficient on those pathways. You could make a floor that is impossible to walk on. Everybody would just fall without traction.

Explanation:

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A uniform magnetic field is perpendicular to the plane of a circular loop of diameter 13 cm formed from wire of diameter 2.6 mm

I am Lyosha [343]

Answer:

Rate of change of magnetic field is $3.466\times 10^3T/sec$

Explanation:

We have given diameter of the circular loop is 13 cm = 0.13 m

So radius of the circular loop $r=\frac{0.13}{2}=0.065m$

Length of the circular loop $L=2\pi r=2\times 3.14\times 0.065=0.4082m$

Wire is made up of diameter of 2.6 mm

So radius $r=\frac{2.6}{2}=1.3mm=0.0013m$

Cross sectional area of wire $A=\pi r^2=3.14\times0.0013^2=5.30\times 10^{-6}m^2$

Resistivity of wire $\rho =2.18\times 10^{-8}m$

Resistance of wire $R=\frac{\rho L}{A}=\frac{2.18\times 10^{-8}\times 0.4082}{5.30\times 10^{-6}}=1.67\times 10^{-3}ohm$

Current is given i = 11 A

So emf $e=11\times 1.67\times 10^{-3}=0.0183volt$

Emf induced in the coil is $e=-\frac{d\Phi }{dt}=-A\frac{dB}{dt}$

$0.0183=5.30\times 10^{-6}\times \frac{dB}{dt}$

$\frac{dB}{dt}=3.466\times 10^3=T/sec$

8 0

3 years ago

Pls help me I don’t get it :(

Ksju [112]

Answer:

2nd and 4th

Explanation:

4 0

2 years ago

The volume of 2.0 kg of helium in a piston-cylinder device is initially 7 m3. Now the helium is compressed to 5 m3 while its pre

IrinaVladis [17]

Answer:

A) T1 = 269.63 K

T2 = 192.59 K

B) W = -320 KJ

Explanation:

We are given;

Initial volume: V1 = 7 m³

Final Volume; V2 = 5 m³

Constant Pressure; P = 160 KPa

Mass; m = 2 kg

To find the initial and final temperatures, we will use the ideal gas formula;

T = PV/mR

Where R is gas constant of helium = R = 2.0769 kPa.m/kg

Thus;

Initial temperature; T1 = (160 × 7)/(2 × 2.0769) = 269.63 K

Final temperature; T2 = (160 × 5)/(2 × 2.0769) = 192.59 K

B) world one is given by the formula;

W = P(V2 - V1)

W = 160(5 - 7)

W = -320 KJ

6 0

3 years ago

Three balls are kicked from the ground level at some angles above horizontal with different initial speeds. All three balls reac

Charra [1.4K]

Answer:

Time of flight A is greatest

Explanation:

Let u₁ , u₂, u₃ be their initial velocity and θ₁ , θ₂ and θ₃ be their angle of projection. They all achieve a common highest height of H.

H = u₁² sin²θ₁ /2g

H = u₂² sin²θ₂ /2g

H = u₃² sin²θ₃ /2g

On the basis of these equation we can write

u₁ sinθ₁ =u₂ sinθ₂=u₃ sinθ₃

For maximum range we can write

D = u₁² sin2θ₁ /g

1.5 D = u₂² sin2θ₂ / g

2 D =u₃² sin2θ₃ / g

1.5 D / D = u₂² sin2θ₂ /u₁² sin2θ₁

1.5 = u₂ cosθ₂ /u₁ cosθ₁ ( since , u₁ sinθ₁ =u₂ sinθ₂ )

u₂ cosθ₂ >u₁ cosθ₁

u₂ sinθ₂ < u₁ sinθ₁

2u₂ sinθ₂ / g < 2u₁ sinθ₁ /g

Time of flight B < Time of flight A

Similarly we can prove

Time of flight C < Time of flight B

Hence Time of flight A is greatest .

8 0

3 years ago

Determine the acceleration due to gravity for low Earth orbit (LEO) given: MEarth = 6.00 x 1024 kg, rEarth = 6.40 x 106 m, G = 6

Nana76 [90]

Answer:

The answer to the question is as follows

The acceleration due to gravity for low for orbit is 9.231 m/s²

Explanation:

The gravitational force is given as

$F_{G}= \frac{Gm_{1} m_{2}}{r^{2} }$

Where $F_{G}$ = Gravitational force

G = Gravitational constant = 6.67×10⁻¹¹ $\frac{Nm^{2} }{kg^{2} }$

m₁ = mEarth = mass of Earth = 6×10²⁴ kg

m₂ = The other mass which is acted upon by $F_{G}$ and = 1 kg

rEarth = The distance between the two masses = 6.40 x 10⁶ m

therefore at a height of 400 km above the erth we have

r = 400 + rEarth = 400 + 6.40 x 10⁶ m = 6.80 x 10⁶ m

and $F_{G}$ = $\frac{6.67*10^{-11} *6.40*10^{24} *1}{(6.8*10^{6})^{2} }$ = 9.231 N

Therefore the acceleration due to gravity = $F_{G}$ /mass

9.231/1 or 9.231 m/s²

Therefore the acceleration due to gravity at 400 kn above the Earth's surface is 9.231 m/s²

4 0

3 years ago

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