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Montano1993 [528]
3 years ago
11

The resistance created by waves on a 120-m-long ship is tested in a channel using a model that is 4 m long Y Part A If the ship

travels at 70 km/h. what should be the speed of the model?
Physics
1 answer:
DanielleElmas [232]3 years ago
6 0

Answer:

V_m = 12.78 km/hr

Explanation:

given,

length of the ship = 120 m

length of model of the ship = 4 m

Speed at which the ship travels = 70 km/h

speed of model = ?

by using froude's law

  F_r = \dfrac{V}{\sqrt{L g}}

for dynamic similarities

  (\dfrac{V}{\sqrt{L g}})_P = (\dfrac{V}{\sqrt{L g}})_{model}

  (\dfrac{V_p}{\sqrt{L_p}}) = (\dfrac{V_m}{\sqrt{L_m}})

  (\dfrac{70}{\sqrt{120}}) = (\dfrac{V_m}{\sqrt{4}})

          V_m = 12.78 km/hr

hence, the velocity of model will be 12.78 km/h

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n alpha particle (q = +2e, m = 4.00 u) travels in a circular path of radius 5.94 cm in a uniform magnetic field with B = 1.10 T.
9966 [12]

Answer:

a). V = 3.13*10⁶ m/s

b). T = 1.19*10^-7s

c). K.E = 2.04*10⁵

d). V = 1.02*10⁵V

Explanation:

q = +2e

M = 4.0u

r = 5.94cm = 0.0594m

B = 1.10T

1u = 1.67 * 10^-27kg

M = 4.0 * 1.67*10^-27 = 6.68*10^-27kg

a). Centripetal force = magnetic force

Mv / r = qB

V = qBr / m

V = [(2 * 1.60*10^-19) * 1.10 * 0.0594] / 6.68*10^-27

V = 2.09088 * 10^-20 / 6.68 * 10^-27

V = 3.13*10⁶ m/s

b). Period of revolution.

T = 2Πr / v

T = (2*π*0.0594) / 3.13*10⁶

T = 1.19*10⁻⁷s

c). kinetic energy = ½mv²

K.E = ½ * 6.68*10^-27 * (3.13*10⁶)²

K.E = 3.27*10^-14J

1ev = 1.60*10^-19J

xeV = 3.27*10^-14J

X = 2.04*10⁵eV

K.E = 2.04*10⁵eV

d). K.E = qV

V = K / q

V = 2.04*10⁵ / (2eV).....2e-

V = 1.02*10⁵V

7 0
3 years ago
A wheel is rotating at 30.0 rpm. The wheel then accelerates uniformly to 50.0 rpm in 10.0 seconds. Determine the – a) rate of an
Mademuasel [1]

Answer:

The angular acceleration is 0.209\ rad/s^2

Explanation:

Given that,

Angular velocity, \omega_{i} = 30.0\ rpm

Angular velocity, \omega_{f} = 50.0\ rpm

Time t = 10.0 sec

We need to calculate the angular acceleration

Using formula of angular acceleration

\alpha=\dfrac{\Delta \omega}{\Delta t}

\alpha=\dfrac{\omega_{f}-\omega_{i}}{\Delta t}

\alpha=\dfrac{50.0-30.0}{10.0}

Now, we change the angular velocity in rad/s.

\omega=20\times\dfrac{2\pi}{60}

\omega=2.09\ rad/s

\alpha=\dfrac{2.09}{10.0}

\alpha=0.209\ rad/s^2

Hence, The angular acceleration is 0.209\ rad/s^2

5 0
3 years ago
Read 2 more answers
If an object has a mass of 1417g and it is moved 47 meters in 90 seconds, how much power was used?
astraxan [27]

Mass of the object is given as

m = 1417 g = 1.417 kg

now the speed of object is given as

v = \frac{d}{t}

here we know that

d = 47 m

t = 90 s

now we will have

v = \frac{47}{90} = 0.52 m/s

now we will have kinetic energy of the object as

KE = \frac{1}{2}mv^2

KE = \frac{1}{2}(1.417)(0.52)^2

KE = 0.19 J

now the power is defined as rate of energy

so here we can find power as

P = \frac{KE}{t}

P = \frac{0.19}{90} = 2.14\times 10^{-3} W

so above is the power used for the object

3 0
3 years ago
Symptoms of excessive stress include all of the following EXCEPT:
sweet-ann [11.9K]

Answer:

Symptoms of excessive stress include all of the following EXCEPT: increased energy level.

6 0
2 years ago
Psi communication refers to the transfer of information through a/an ________ process.
shtirl [24]

Answer:

unknown

Explanation:

Psi communication: The term "Psi communication" was initially originated in an article named 'Mass Communication and Para-Social Interaction' during 1956 by Donald Horton, and is also denoted as "parasocial interaction". It is described as a phenomenon that involves "one-sided interaction" between people, encompassing the viewer knowing a specific celebrity well whereas the viewer himself or herself is being completely unknown for the celebrity in return.

In the question above, the given statement represents that the Psi communication is responsible for transforming via an unknown process as one of its members are considered as unknown of the transfer, so the correct answer would be "unknown process".

4 0
3 years ago
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