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zhannawk [14.2K]
2 years ago
8

If a bow holds 500J of potential energy as the arrow is pulled back, how much kinetic energy will the arrow have after it has be

en released?
Physics
1 answer:
ale4655 [162]2 years ago
4 0

Answer:

500J

Explanation:

The arrow will have an energy of 500J after it has been released from its state of rest.

This is compliance with the law of conservation of energy which states that "in every system, energy is neither created nor destroyed but transformed from one form to another".

  • The energy at rest which is the potential energy is 500J
  • This energy will be converted to kinetic energy in total after the arrow has been released.
  • This way, no energy is lost and we can account for the energy transformations occurring.
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Two charges are located in the xx–yy plane. If q1=−4.10 nCq1=−4.10 nC and is located at (x=0.00 m,y=1.080 m)(x=0.00 m,y=1.080 m)
Gala2k [10]

Answer:

Explanation:

Due to first charge , electric field at origin will be oriented towards - ve of y axis.

magnitude

Ey = -8.99 x 10⁹ x 4.1 x 10⁻⁹ / 1.08² j

= - 31.6 j N/C

Due to second charge electric field at origin

= 8.99 x 10⁹ x 3.6  x 10⁻⁹ / 1.2²+ .6²

= 8.99 x 10⁹ x 3.6  x 10⁻⁹ / 1.8

= 18 N/C

It is making angle θ where

Tanθ = .6 / 1.2

= 26.55°

this field in vector form

= - 18 cos 26.55 i - 18 sin26.55 j

= - 16.10 i - 8.04 j

Total field

= - 16.10 i - 8.04 j + ( - 31.6 j )

= -16.1 i - 39.64 j .

Ex = - 16.1 i

Ey = - 39.64 j .

8 0
3 years ago
The speed that a tsunami can travel is modeled by the equation , where s is the speed in kilometers per hour and d is the averag
CaHeK987 [17]

The speed of tsunami is a.0.32 km. 

Steps involved  :

The equation s = 356d models the maximum speed that a tsunami can move at. It reads as follows: s = 200 km/h d =?

Let's now change s to s in the equation to determine d: s = 356√d 200 = 356√d √d = 200 ÷ 356 √d = 0.562 Let's square the equation now by squaring both sides: (√d)² = (0.562) ² d = (0.562)² = 0.316 ≈ 0.32

As a result, 0.32 km is roughly the depth (d) of water for a tsunami moving at 200 km/h.

To learn more about tsunami refer : brainly.com/question/11687903

#SPJ4

6 0
2 years ago
A bicycle tire has a pressure of 7.00×105 N/m2 at a temperature of 18.0ºC and contains 2.00 L of gas. What will its pressure be
stepan [7]

Answer:

p_2 = 664081 N/m^{2}

Explanation:

from the ideal gas law we have

PV = mRT

P = \rho RT

\rho = \frac{P}{RT}

HERE  R is gas constant for dry air  =  287  J K^{-1} kg^{-1}

\rho = \frac{7.00 10^{5}}{287(18+273)}

\rho = 8.38 kg/m^{3}

We know by ideal gas law

\rho = \frac{m_1}{V_1}

m_1 = \rho V_1 = 8.38 *2*10^{-3}

m_1 = 0.0167 kg

for m_2

m_2 = \rho V_i - V_removed

m_2 = 8.38*(.002 - 10^{-4})

m_2 = 0.0159 kg

WE KNOW

PV = mRT

V, R and T are constant therefore we have

P is directly proportional to mass

\frac{p_2}{p_1}=\frac{m_2}{m_1}

p_2 = p_1 * \frac{m_2}{m_1}

p_2 =7*10^{5} * \frac {.0159}{0.0167}

p_2 = 664081 N/m^{2}

8 0
2 years ago
As the launch force increase the launch velocity will
FrozenT [24]

Answer:

As the launch force increase the launch velocity will

<em><u>Increase</u></em>

The reason for your answer to number six is because

<em><u>There is a direct relationship between force and acceleration.</u></em>

<em><u /></em>

Explanation:

<em>It is known all over the place that, there is a direct relationship between Force and acceleration of an object leading to an increase in force being directly proportional to the increase in the acceleration of the given object and vice versa.</em>

5 0
2 years ago
The force required to stretch a Hooke’s-law spring varies from 0 N to 63.5 N as we stretch the spring by moving one end 5.31 cm
Alika [10]

Answer:

Force constant will be 1195.85 N/m

Work done will be 1.6859 J

Explanation:

We have given the force,  F = 63.5 N

Spring is stretched by 5.31 cm

So x = 0.0531 m

Force is given , F = 63.5 N

We know that force is given by F=kx

So 63.5=k\times 0.0531

k = 1195.85 N/m

Now we have to find the work done

We know that work done is given by

W=\frac{1}{2}kx^2=\frac{1}{2}\times 1195.85\times 0.0531^2=1.6859J

8 0
2 years ago
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