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spayn [35]
3 years ago
15

Frank swam kilometers against the current in the same amount of time it took him to swim kilometers with the current. The rate o

f the current was kilometer per hour. How fast would Frank swim if there were no current?
Physics
1 answer:
mojhsa [17]3 years ago
3 0

Answer:

If there was no current, Frank would swim at the same rate of kilometers per hour.

Explanation:

Given that Frank swam the same amount of kilometers against the current and with the current in the same time, then we realize that the current is a variable that does not affect Frank's swimming speed. So if there was no current, it would surely swim the same amount of kilometers at the same time.

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An object is acted upon by a constant force which gives it a constant acceleration a. At a certain time t1, having started from
Elenna [48]

Answer:

t_2=\sqrt{2}(t_1)

Explanation:

The kinetic energy of a body is that energy it possesses due to its motion. In classical mechanics, this energy depends only on its mass and speed, as follows:

K=\frac{mv^2}{2}

The speed in an uniformly accelerated motion is given by:

v=at

Replacing this expression in the formula for the kinetic energy, we have:

K=\frac{ma^2t^2}{2}\\

So, if we have K_2=2K_1:

K_1=\frac{ma^2t_1^2}{2}(1)\\K_2=\frac{ma^2t_2^2}{2}\\2K_1=\frac{ma^2t_2^2}{2}\\K_1=\frac{ma^2t_2^2}{4}(2)\\

Equaling (1) and (2) and solving for t_2:

\frac{ma^2t_1^2}{2}=\frac{ma^2t_2^2}{4}\\t_2=\frac{4t_1^2}{2}\\t_2=\sqrt{2t_1^2}\\t_2=\sqrt{2}(t_1)

7 0
4 years ago
Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, and (d) 2.00 s for a ball thrown straig
kupik [55]

Answer:

Explanation:

V = Deltax/Deltat

V = 15.0 m/s

Displacement:

(a) Vf = Vi + adeltat

Vf = 15.0m/s - 9.8m/s^2 x 0.500s = 10.1m/s

Displacement = (15.0m/s x 0.500s) - (0.5)(9.8m/s^2)(0.500s)^2 = 6.275m

(b) Vf = 15.0m/s - 9.8m/s^2 x 1.00s = 5.2m/s

Displacement = (15.0m/s x 1.00s) - (0.5)(9.8m/s^2)(1s)^2 = 10.1m

(c) Vf = 15.0m/s - 9.8m/s^2 x 1.50s = 14.7m/s

Displacement = (15.0m/s x 1.50s) - (0.5)(9.8m/s^2)(1.5s)^2 = 11.475m

(d) Vf = 15.0m/s - 9.8m/s^2 x 2.00s =  19.6m/s

Displacement = (15.0m/s x 2.00s) - (0.5)(9.8m/s^2)(2s)^2 = 10.4m

4 0
4 years ago
A cell of e.M.F 1.5v and internal resistance 2.5ohm is connected in series with an ammeter of resistance 0.5ohm. Calculate the c
Tju [1.3M]

Given :

A cell of e.m.f  1.5 V and internal resistance 2.5 ohm is connected in series with an ammeter of resistance 0.5 ohm.

To Find :

The current in the circuit.

Solution :

We know, resistance of the ammeter is in series with the circuit.

So, total resistance is :

R = 2.5 + 0.5 ohm

R = 3 ohm

Also, e.m.f applied is 1.5 V .

Now, by ohm's law :

I = \dfrac{V}{R}\\\\I = \dfrac{1.5}{3}\\\\I = 0.5 \ A

Therefore, the current in the circuit is 0.5 A.

8 0
3 years ago
3.
Greeley [361]

Answer:

90 m/s

8100 m

Explanation:

Given:

v₀ = 0 m/s

a = 0.5 m/s²

t = 3 min = 180 s

Find: v and Δx

v = at + v₀

v = (0.5 m/s²) (180 s) + 0 m/s

v = 90 m/s

Δx = v₀ t + ½ at²

Δx = (0 m/s) (180 s) + ½ (0.5 m/s²) (180 s)²

Δx = 8100 m

8 0
4 years ago
Discuss Joule-Thompson effect with relevant examples and formulae.
Delicious77 [7]

Answer:

\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp

Explanation:

Joule -Thompson effect

 Throttling phenomenon is called Joule -Thompson effect.We know that throttling is a process in which pressure energy will convert in to thermal energy.

Generally in throttling exit pressure is low as compare to inlet pressure but exit temperature maybe more or less or maybe remains constant depending upon flow or fluid flow through passes.

Now lets take Steady flow process  

Let

 P_1,T_1 Pressure and temperature at inlet and

 P_2,T_2 Pressure and temperature at exit

We know that Joule -Thompson coefficient given as

\mu _j=\left(\frac{\partial T}{\partial p}\right)_h

Now from T-ds equation

dh=Tds=vdp

So

Tds=C_pdt-\left [T\left(\frac{\partial v}{\partial T}\right)_p\right]dp

⇒dh=C_pdt-\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp

So Joule -Thompson coefficient

\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp

This is Joule -Thompson coefficient for all gas (real or ideal gas)

We know that for Ideal gas Pv=mRT

\dfrac{\partial v}{\partial T}=\dfrac{v}{T}

So by putting the values in

\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp

\mu _j=0 For ideal gas.

6 0
3 years ago
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