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Ira Lisetskai [31]
3 years ago
9

there is a fish called an archer fish that shoots drops of water at insects resting on branches above the water. If the Archer f

ish is aiming at an insect with (x,y) coordinates of (0.5,0.25) and spits water at an angle of 40 degress relative to the surface of the water. At what speed v should an archerfish spit the water to shoot down a floating insect located at that distance?
Physics
1 answer:
Mariulka [41]3 years ago
4 0

Answer:

=3.5 m/s

Explanation:

y = x tanθ - 1/2 g x² / (u²cos²θ )

y = 0.25 , x = 0.5, θ = 40°

.25 = .50 tan40 - .5 x 9.8x x²/ u²cos²40

.25 = .42 - 2.0875/u²

u = 3.5 m / s.

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A total charge of 4.70 is distributed on two metal spheres. When the spheres are 10 cm apart, they each feel a repulsive force o
Anna35 [415]

Answer:0.114 C

Explanation:

Given

Total 4.7 C is distributed in two spheres

Let q_1 and q_2 be the charges such that

q_1+q_2=4.7

and Force between charge particles is given by

F=\frac{kq_1q_2}{r^2}

4.7\times 10^11=\frac{9\times 10^9\times q_1\cdot q_2}{0.1^2}

q_1\cdot q_2=0.522

put the value of q_1

q_2\left ( 4.7-q_2\right )=0.522

q_2^2-4.7q_2+0.522=0

q_2=\frac{4.7\pm \sqrt{4.7^2-4\times 1\times 0.522}}{2}

q_2=0.114 C

thus q_1=4.586 C

3 0
3 years ago
The ink drops have a mass m = 1.00×10^−11 kg each and leave the nozzle and travel horizontally toward the paper at velocity v =
luda_lava [24]

Answer:

9.98 × 10⁻⁹ C

Explanation:

mass, m = 1.00 × 10⁻¹¹ kg

Velocity, v = 23.0 m/s

Length of plates D₀ = 1.80 cm = 0.018 m

Magnitude of electric field, E = 8.20 × 10⁴ N/C

drop is to be deflected a distance d = 0.290 mm = 0.290 × 10⁻³ m

density of the ink drop = 1000 kg/m^3

Now,

Time = \frac{\textup{Distance}}{\textup{Velocity}}

or

Time = \frac{\textup{0.016}}{\textup{23}}

or

Time = 6.9 × 10⁻⁴ s

Now, force due to the electric field, F = q × E

where, q is the charge

Also, Force = Mass × acceleration

q × E = 1.00 × 10⁻¹¹ × a

or

a = \frac{q\times8.20\times10^4}{1\times10^{-11}}

Now from the Newton's equation of motion

d=ut+\frac{1}{2}at^2

where,  

d is the distance

u is the initial speed  

a is the acceleration

t is the time

or

0.290\times10^{-3}=0\times(6.9\times10^{-4})+\frac{1}{2}\times(\frac{q\times8.20\times10^4}{1\times10^{-11}})\times(6.9\times10^{-4})^2

or

q = 9.98 × 10⁻⁹ C

4 0
3 years ago
Write whether true or false...
elixir [45]

Answer:

a --> true, b --> false, c --> true, d -->false

Explanation:

a) since it stays floating the gravity force and the upqards push is the same

b) if it's balanced the rocket won't move from the ground, the force of the rocket, has to exceed the force of gravity

c) since it's going in a diretion the force of gravity is exceeding the force pushing it up

d) since that are speeding up at a rate, meaning growing, the force is unbalanced.

8 0
3 years ago
What is the current that reverses direction in a regular patter called
Oxana [17]

Answer:

Current that reverses direction in the regular pattern is called an alternating current, abbreviated as 'AC'.

Explanation:

hope this helps!

4 0
3 years ago
A car of mass 1000 kg is moving at 25 m/s. It collides with a car of mass 1200 kg moving at 30 m/s. When the cars collide, they
Alinara [238K]

Answer:

The total momentum of the cars before the collision is 61,000 kg.m/s

The total momentum of the cars after the collision is 61,000 kg.m/s

The velocity of the cars after the collision is 27.727 m/s

Explanation:

Given;

mass of the first car, m₁ = 1000 kg

initial velocity of the car, u₁ = 25 m/s

mass of the second car, m₂ = 1200 kg

initial velocity of the second car, u₂ = 30 m/s

The common velocity of the cars after collision = v

The total momentum of the cars before collision is calculated as;

P₁ = m₁u₁  +  m₂u₂

P₁ = (1000 x 25)  +  (1200 x 30)

P₁ = 61,000 kg.m/s

The total momentum of the cars after collision is calculated as;

P₂ = m₁v + m₂v

where;

v    is the common velocities of the cars after collision since they stick together.

P₂ = v(m₁ + m₂)

To determine "v" apply the principle of conservation of linear momentum for inelastic collision.

m₁u₁  +  m₂u₂  = v(m₁  + m₂)

(1000 x 25)  +  (1200 x 30) = v(1000 + 1200)

61,000 = 2,200v

v = 61,000/2,200

v = 27.727 m/s

The total momentum after collsion = v(m₁ + m₂)

                                                         = 27.727(1000 + 1200)

                                                          = 61,000 kg.m/s

Thus, momentum before and after collsion are equal.

8 0
3 years ago
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