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Lera25 [3.4K]
3 years ago
6

Find the velocity of an airplane traveling north at 100 kilometers per hour. It is also flying through a wind that is blowing we

st at 30 kilometers per hour.
Physics
1 answer:
artcher [175]3 years ago
7 0

Answer:

Explanation:

Find the velocity of an airplane traveling north at 100 kilometers per hour. It is also flying through a wind that is blowing west at 30 kilometers per hour.

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A cart with mass 2.0 kg moving on a frictionless linear air track at an initial speed of 1.0 m/s undergoes an elastic collision
Katen [24]

Answer:

a) P=0.8 Kg*m/s b) K=0.6 N c)P/K=MV/(1/2*MV²) d) V2f=1.5 m/s e) M2=0.53 Kg

Explanation:

During an elastic collision between 2 bodies, the momentum P is the same before and after the collision

For this case:

Before the collision:

M₁= mass of first car= 2 Kg

V₁= initial speed of the first car = 1 m/s

M₂= mass of the second car

V₂= initial speed of the second car = 0 m/s (as it is stationary)

After the collision:

V₁f= final speed of the first car after the collision= 0.6 m/s

V₂f= final speed of the second car after the collision

As momentum is the same after and before:

M₁V₁ + M₂V₂ = M₁V₁f + M₂V₂f consider that term M₂V₂=0 as V₂=0

Then, momentum for car N° 2 after the collision is: P₂= M₂V₂f and replacing from the above equation: P₂= M₁V₁ – M₁V₁f = M₁(V₁ – V₁f) = 2 Kg*(1m/s – 0.6m/s) = 0.8 Kg*m/s

As the kinetic energy “K” is also conservative:

½*M₁V₁² + ½*M₂V₂² = ½*M₁V₂f² + ½*M₂V₂f² Where ½*M₂V₂²=0

Then: K₂= ½*M₂V₂f² = ½*M₁(V₁² – V₁f²) = 0.64 N

Finally, to obtain M₂ and V₂f:

P₂=M₂V₂f and K₂=1/2*M₂V₂f2²

P₂/K₂= (M₂V₂f)/(1/2*M₂V₂f) =2/V₂f  

V₂f= 2*K₂/P₂=1.5 m/s and M₂=P₂/V₂f=0.53 Kg

8 0
3 years ago
A solid sphere of diameter D = 26 cm has a charge of Q = 4 nano-coulombs uniformly distributed on it. Calculate the magnitude of
kozerog [31]

Answer:

E = 4.83  N/ C

Explanation:

If we have a uniform charge sphere we can use the following formulas to calculate the Electric field due to the charge of the sphere:

E= \frac{k*Q}{r^{2} } : Formula (1) To calculate the electric field in the region outside the sphere r ≥ a

Where:

K: coulomb constant (N*m²/C²)

a: sphere radius (m)

Q:  Total sphere charge (C)

r : Distance from the center of the sphere to the region where the electric field is calculated (m)

Equivalences

1nC=10⁻⁹C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

Q=4nC=4 *10⁻⁹C

D = 26 cm = 26*10⁻²m = 0.26m

a = D/2 = 0.13m

r= R+a = 2.6 m+ 0.13m = 2.73m

Problem development

Magnitude of the electric field at r = 2.73m from the center of the sphere  

r>a , We apply the Formula (1) :

E= \frac{k*Q}{r^{2} }

E= \frac{9*10^{9}*4*10^{-9} }{2.73^{2} }

E= 4.83 N/ C

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