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Nataly_w [17]
3 years ago
15

A gardener uses a water hose to fill a 30.0 L bucket. The gardener notes that it takes 1.00 min to fill the bucket. A nozzle wit

h an opening of cross sectional area 0.500 cm2 is then attached to the hose. The nozzle is held so that water is projected horizontally from a point 1.00 m above the ground. Over what horizontal distance can the water be projected
Physics
1 answer:
GalinKa [24]3 years ago
6 0

Answer:

The distance can the water be projected is 4.51 m

Explanation:

The speed of the water in the hose is equal to:

v1 = R/A1

If we solve the continuity for v2:

v2 = R/A2 (eq. 1)

The equation for the vertical position is:

yf = yi + vy*t - (1/2)gt²

yi = 0

vy = 0

Clearing t:

t=\sqrt{\frac{-2y_{f} }{g} } (eq. 2)

The equation for position is:

xf = xi + vxt = 0 + v2t = v2t (eq. 3)

Replacing equation 1 and 2 in equation 3:

x_{f} =\frac{R}{A_{2}  } \sqrt{\frac{-2y_{f} }{g} } =\frac{30}{0.5} \sqrt{\frac{(-2)*(-1)}{9.8} }*\frac{1000}{60}  =451.75cm=4.51m

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An electric current in a conductor varies with time according to the expression I(t) = 100 sin (120πt) , where I is in amperes a
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The  total charge passing a given point in the conductor from t = 0 to t = 1/240 s is 12000π Coulombs

<h3>Total charge through a conductor</h3>

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The expression that will be used to calculate the current is expressed s:

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While standing outdoors one evening, you are exposed to the following four types of electromagnetic radiation: yellow light from
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AM radio, FM radio, microwaves, sodium light.

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3 years ago
A 5.00-g bullet is fired horizontally into a 1.20-kg wooden block resting on a horizontal surface. The coefficient of kinetic fr
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Answer:

The initial velocity of the bullet is  u_b = 228.7 m/s

Explanation:

From the question we are told that

          The mass of the bullet is  m_b = 5.00g  =\frac{5}{1000} = 0.005kg

           The mass of the wooden block m_w = 1.20kg

          The coefficient of kinetic friction \mu_g = 0.20

          The  distance traveled by block and bullet d_{t } = 0.230m

let  F_ b denote the force on the bullet in the block and this mathematically denoted as

                            F_b = \mu_g mg

Where m is the mass of both bullet and block

  Substituting values

                             = (0.20)* (0.005 + 1.20)(9.8)

                             = 2.36N

Generally workdone is mathematically represented as

                         W = F *d_t

the  work in this cause in the kinetic energy used to move the block and bullet

                 i.e      W = \frac{1}{2}mv_f^2

Where  v_f is the final speed of the block and bullet

           Therefore  \frac{1}{2}mv_f^2  = F * d_t

making v_f the subject

                              v_f = \sqrt{\frac{2Fd_t}{m} }

Here m is the mass of both bullet and block

Now substituting values

                            v_f = \sqrt{\frac{2(2.36)(0.230)}{(0.005 + 1.20)} }

                                 = 0.949m/s

From the law of conservation of momentum this collision can be mathematically represented as

               

                             m_bu_b +m_wu_w = (m_w +m_b)v_f

Now from the question  m_wu_w =0  this because the block was at rest

                          Therefore u_b = \frac{(m_w +m_u) v_f}{m_b}

    Substituting values

                               u_b = \frac{(0.005 +1.2) * 0.949}{0.005}

                                    u_b = 228.7 m/s

   

3 0
3 years ago
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