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wel
3 years ago
11

what must be the distance between source of sound and mountain to give an echo in 5 seconds ? ,(speed of sound at 0°c=33m\sec​

Physics
1 answer:
Tamiku [17]3 years ago
4 0

Answer:

distance must be = 330 × 5/2

= 330×2.5

=825m

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Uranium-238 eventually decays into
GalinKa [24]
Uranium-238 decays<span> by alpha emission </span>into<span> thorium-234, which itself </span>decays<span> by beta emission to protactinium-234, which </span>decays<span> by beta emission to </span>uranium<span>-234, and so on. The various </span>decay<span> products, (sometimes referred to as “progeny” or “daughters”) form a series starting at </span>uranium-238<span>.</span>
4 0
3 years ago
Water flows from the bottom of a storage tank at a rate of r(t) = 300 − 6t liters per minute, where 0 ≤ t ≤ 50. Find the amount
SSSSS [86.1K]

r(t) models the water flow rate, so the total amount of water that has flowed out of the tank can be calculated by integrating r(t) with respect to time t on the interval t = [0, 35]min

∫r(t)dt, t = [0, 35]

= ∫(300-6t)dt, t = [0, 35]

= 300t-3t², t = [0, 35]

= 300(35) - 3(35)² - 300(0) + 3(0)²

= 6825 liters

7 0
3 years ago
Runner A is initially 5.7 km west of a flagpole and is running with a constant velocity of 8.9 km/h due east. Runner B is initia
Oksanka [162]

Answer:

B meet A 0.01 km east of flagpole

Explanation:

given data

distance A = 5.7 km  west

velocity V1 = 8.9 km/h

distance B = 4.5 km  east

velocity V2 = 7 km/h

to find out

How far runners from the flagpole, when paths cross

solution

we know A and B are 5.7 + 4.5 = 10.2 km apart

and we consider here B will run distance x km for meet

so time will be for B is

time B = distance / velocity

time B = x / 7     ...................1

and

for A distance for meet = ( 10.2 - x ) km

so time A = distance / velocity

time A = ( 10.2 - x )  / 8.9      .............2

now equating equation 1 and 2

time A = time B

x / 7  =   ( 10.2 - x )  / 8.9

x = 4.490

so distance of B run for meet is 4.490 km

so distance from the flagpole when their paths cross is 4.5 - 4.490 = 0.01 km

so B meet A 0.01 km east of flagpole

4 0
3 years ago
A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 V. It is used to charge two storage batt
Natali [406]

Complete Question

A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 \Omega. It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300\Omega . If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?

Answer:

a

The additional resistance is R_z =  4.4 \Omega

b

The rate at which internal energy increase at the supply is Z_1 = 32 W

c

The rate at which internal energy increase in the battery  is  Z_1 = 32 W

d

The rate at which internal energy increase in the added series resistance is  Z_3 = 70.4 W

e

the increase rate of the chemically energy in the battery is C =  48 W

Explanation:

From the question we are told that

    The  open circuit voltage is  V =  40.0V

     The internal resistance is R = 2 \Omega

     The emf of each battery is e =  6.00 V

      The internal resistance of the battery is  r = 0.300V

      The  charging current is  I = 4.00 \ A

Let assume the the additional resistance to to added to the circuit is  R_z

 So this implies that

        The total resistance in the circuit is

                              R_T =  R + 2r +R_z

Substituting values

                             R_T = 2.6 +R_z

And  the difference in potential in the circuit is  

                         E = V -2e

                 =>   E =  40 - (2 * 6)

                        E =  28 V

Now according to ohm's law

            I = \frac{E}{R_T}

Substituting values

           4 = \frac{28}{R_z + 2.6}        

Making R_z the subject of the formula

So    R_z =  \frac{28 - 10.4}{4}

           R_z =  4.4 \Omega

The  increase rate of   internal energy at the supply is mathematically represented as

        Z_1  = I^2 R

Substituting values

     Z_1  = 4^2 * 2

     Z_1 = 32 W

The  increase rate of   internal energy at the batteries  is mathematically represented as

         Z_2 = I^2 r

Substituting values

         Z_2 = 4^2 * 2 * 0.3

         Z_2 = 9.6 \ W

The  increase rate of  internal energy at the added  series resistance  is mathematically represented as

        Z_3 = I^2 R_z

Substituting values

       Z_3 = 4^2 * 4.4

      Z_3 = 70.4 W

Generally the increase rate of the chemically energy in the battery is  mathematically represented as

         C = 2 * e * I

Substituting values

       C =  2 * 6  * 4

      C =  48 W

6 0
3 years ago
The spring to launch a pinball in a pinball machine is compressed 25 cm and has a spring constant of 140 N/m.
mote1985 [20]

Answer:

I think it is 5.6. This is my answer

8 0
3 years ago
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