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Igoryamba
3 years ago
7

An 800 kg fishing boat going south at 12 m/s runs into a stopped pontoon boat (1400 kg). The boats stick together and move south

. What is the velocity of the boats after the collision?
Physics
1 answer:
kykrilka [37]3 years ago
7 0

Answer:

the velocity of the boats after the collision is 4.36 m/s.

Explanation:

Given;

mass of fish, m₁ = 800 kg

mass of boat, m₂ = 1400 kg

initial velocity of the fish, u₁ = 12 m/s

initial velocity of the boat, u₂ = 0

let the final velocity of the fish-boat after collision = v

Apply the principle of conservation of linear momentum for inelastic collision;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

800 x 12    +   1400 x 0 = v(800 + 1400)

9600 = 2200v

v = 9600/2200

v = 4.36 m/s

Therefore, the velocity of the boats after the collision is 4.36 m/s.

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In a physics laboratory experiment, a coil with 200 turns enclosing an area of 11.8 cm^2 is rotated during the time interval 4.9
Zanzabum

Answer:

Explanation:

Flux through the coil = nBA , n is no of turns , B is magnetic flux and A a is area of the coli

= 200 x 5.6 x 10⁻⁵ x 11.8 x 10⁻⁴

=  13216 x 10⁻⁹ weber .

b ) When the coil becomes parallel to magnetic field  , flux through it will become zero.

c ) e m f induced = change in flux / time

= 13216 x 10⁻⁹ / 4.9 x 10⁻²

= 2697.14 x 10⁻⁷ V

= 269.7 x10⁻⁶

269.7 μV.

6 0
3 years ago
An airplane has a momentum of 8.55 x 107 kg.m/s[S] and a velocity of 900 km/h[S]. Determine the mass of the airplane.
kow [346]

Answer:

342,000kg

Explanation:

p=mv

8.55*10^7 kg*m/s=m(900 km/h)

85,500,000 kg*m/s=m(900 km/h)

(85,500,000 kg*m/s)/(900 km/h)=m

Get same units.... 900km/h = 250m/s

m/s cancel in the division, you are left with just kg!!

85,500,000/250=342,000kg! That's it!

6 0
2 years ago
Systems distribute heat energy by the circulation of air.
Ulleksa [173]
That seems like a statement more than a question. Where's the question?
6 0
3 years ago
Read 2 more answers
The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles
belka [17]

Answer:

F_a=5.67\times 10^{-5}\ N

<u />F_b=3.49\times 10^{-5}\ N

F_c=9.16\times 10^{-5}\ N

Explanation:

Given:

  • mass of particle A, m_a=363\ kg
  • mass of particle B, m_b=517\ kg
  • mass of particle C, m_c=154\ kg
  • All the three particles lie on a straight line.
  • Distance between particle A and B, x_{ab}=0.5\ m
  • Distance between particle B and C, x_{bc}=0.25\ m

Since the gravitational force is attractive in nature it will add up when enacted from the same direction.

<u>Force on particle A due to particles B & C:</u>

F_a=G. \frac{m_a.m_b}{x_{ab}^2} +G. \frac{m_a.m_c}{(x_{ab}+x_{bc})^2}

F_a=6.67\times 10^{-11}\times (\frac{363\times 517}{0.5^2}+\frac{363\times 154}{(0.5+0.25)^2})

F_a=5.67\times 10^{-5}\ N

<u>Force on particle C due to particles B & A:</u>

<u />F_c=G.\frac{m_c.m_b}{x_{bc}^2} +G.\frac{m_c.m_a}{(x_{ab}+x_{bc})^2}<u />

F_c=6.67\times 10^{-11}\times (\frac{154\times 517}{0.25^2}+\frac{154\times 363}{(0.25+0.5)^2} )

F_c=9.16\times 10^{-5}\ N

<u>Force on particle B due to particles C & A:</u>

<u />F_b=G.\frac{m_b.m_c}{x_{bc}^2} -G.\frac{m_b.m_a}{x_{ab}^2}<u />

<u />F_b=6.67\times 10^{-11}\times (\frac{517\times 154}{0.25^2}-\frac{517\times 363}{0.5^2}  )<u />

<u />F_b=3.49\times 10^{-5}\ N<u />

3 0
3 years ago
Consider a 150 turn square loop of wire 17.5 cm on a side that carries a 42 A current in a 1.7 T. a) What is the maximum torque
ankoles [38]

Answer:

(a) 328 Nm

(b) 79.35 Nm

Explanation:

N = =150, side = 17.5 cm = 0.175 m, i = 42 A, B = 1.7 T

A = side^2 = 0.175^2 = 0.030625 m^2

(a) Torque = N x i x A x B x Sinθ

For maximum torque, θ = 90 degree

Torque = 150 x 42 x 0.030625 x 1.7 x Sin 90

Torque = 328 Nm

(b) θ = 14 degree

Torque =  150 x 42 x 0.030625 x 1.7 x Sin 14

Torque = 79.35 Nm

7 0
3 years ago
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