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Studentka2010 [4]
3 years ago
10

A floating ice block is pushed through a displacement d = (23 m) i - (9 m) j along a straight embankment by rushing water, which

exerts a force F = (200 N) i - (149 N) j on the block. How much work does the force do on the block during the displacement?
Physics
2 answers:
musickatia [10]3 years ago
4 0

Answer:

Work done, W = 5941 joules

Explanation:

It is given that,

Force exerted on the block, F=(200i-149j)\ N

Displacement, x=(23i-9j)\ m

Let W is the work done by the force do on the block during the displacement. Its formula is given by :

W=F.d

W=(200i-149j){\cdot} (23i-9j)                    

Since, i.i = j.j = k.k = 1

W=4600+1341

W = 5941 joules

So, the work done by the force do on the block during the displacement is 5941 joules. Hence, this is the required solution.

Sav [38]3 years ago
3 0

Answer:

Work = 5941 J

Explanation:

As we know that work done is given by the equation

W = F. d

here we know that

F  = (200 N)\hat i - (149 N) \hat j

also we have

d = (23m) \hat i - (9 m)\hat j

now from above formula we have

W = (200 N\hat i - 149 N \hat j).(23 m\hat i - 9m \hat j)

W = 5941 J

You might be interested in
A 44-cm-diameter water tank is filled with 35 cm of water. A 3.0-mm-diameter spigot at the very bottom of the tank is opened and
cricket20 [7]

Answer:

The frequency f = 521.59 Hz

The rate at which the frequency is changing = 186.9 Hz/s

Explanation:

Given that :

Diameter of the tank = 44 cm

Radius of the tank = \frac{d}{2} =\frac{44}{2} = 22 cm

Diameter of the spigot = 3.0 mm

Radius of the spigot = \frac{d}{2} =\frac{3.0}{2} = 1.5 mm

Diameter of the cylinder = 2.0 cm

Radius of the cylinder = \frac{d}{2} = \frac{2.0}{2} = 1.0 cm

Height of the cylinder = 40 cm = 0.40 m

The height of the water in the tank from the spigot = 35 cm = 0.35 m

Velocity at the top of the tank = 0 m/s

From the question given, we need to consider that  the question talks about movement of fluid through an open-closed pipe; as such it obeys Bernoulli's Equation and the constant discharge condition.

The expression for Bernoulli's Equation is as follows:

P_1+\frac{1}{2}pv_1^2+pgy_1=P_2+\frac{1}{2}pv^2_2+pgy_2

pgy_1=\frac{1}{2}pv^2_2 +pgy_2

v_2=\sqrt{2g(y_1-y_2)}

where;

P₁ and P₂ = initial and final pressure.

v₁ and v₂ = initial and final fluid velocity

y₁ and y₂ = initial and final height

p = density

g = acceleration due to gravity

So, from our given parameters; let's replace

v₁ = 0 m/s ; y₁ = 0.35 m ; y₂ = 0 m ; g = 9.8 m/s²

∴ we have:

v₂ = \sqrt{2*9.8*(0.35-0)}

v₂ = \sqrt {6.86}

v₂ = 2.61916

v₂ ≅ 2.62 m/s

Similarly, using the expression of the continuity for water flowing through the spigot into the cylinder; we have:

v₂A₂ = v₃A₃

v₂r₂² = v₃r₃²

where;

v₂r₂ = velocity of the fluid and radius at the spigot

v₃r₃ = velocity of the fluid and radius at the cylinder

v_3 = \frac{v_2r_2^2}{v_3^2}

where;

v₂ = 2.62 m/s

r₂ = 1.5 mm

r₃ = 1.0 cm

we have;

v₃ = (2.62  m/s)* (\frac{1.5mm^2}{1.0mm^2} )

v₃ = 0.0589 m/s

∴ velocity  of the fluid in the cylinder =  0.0589 m/s

So, in an open-closed system we are dealing with; the frequency can be calculated by using the expression;

f=\frac{v_s}{4(h-v_3t)}

where;

v_s = velocity of sound

h = height of the fluid

v₃ = velocity  of the fluid in the cylinder

f=\frac{343}{4(0.40-(0.0589)(0.4)}

f= \frac{343}{0.6576}

f = 521.59 Hz

∴ The frequency f = 521.59 Hz

b)

What are the rate at which the frequency is changing (Hz/s) when the cylinder has been filling for 4.0 s?

The rate at which the frequency is changing is related to the function of time (t) and as such:

\frac{df}{dt}= \frac{d}{dt}(\frac{v_s}{4}(h-v_3t)^{-1})

\frac{df}{dt}= -\frac{v_s}{4}(h-v_3t)^2(-v_3)

\frac{df}{dt}= \frac{v_sv_3}{4(h-v_3t)^2}

where;

v_s (velocity of sound) = 343 m/s

v₃ (velocity  of the fluid in the cylinder) = 0.0589 m/s

h (height of the cylinder) = 0.40 m

t (time) = 4.0 s

Substituting our values; we have ;

\frac{df}{dt}= \frac{343*0.0589}{4(0.4-(0.0589*4.0))^2}

= 186.873

≅ 186.9 Hz/s

∴ The rate at which the frequency is changing = 186.9 Hz/s  when the cylinder has been filling for 4.0 s.

8 0
3 years ago
One of the largest planes ever to fly, and the largest to fly frequently, is the Ukrainian-built Antonov An-124 Ruslan. The grav
Lerok [7]

Answer:

m = 236212 [kg]

Explanation:

The potential energy can be determined by means of the product of mass by gravity by height. In this way, we have the following equation.

P=m*g*h\\

where:

P = potential energy = 3360000000 [J]

m = mass [kg]

g = gravity acceleration = 9.81 [m/s²]

h = elevation = 1450 [m]

Now, we can clear the mass from the equation above:

3360000000=m*9.81*1450\\m = 236212 [kg]

4 0
3 years ago
What are particles of carbon called?
Ainat [17]
Co2 or greenhouse gas
8 0
3 years ago
BRAINLIEST AND MAY DOUBLE POINTS!!!
raketka [301]

Given that:

 Energy of bulb  (Work ) = 30 J,

 Time (t) = 3 sec

The power consumption =  ?

 We know that, Power can be defined as rate of doing work

              Power (P) = Work(Energy supplied) ÷ time

                               = 30 ÷ 3

                              = 10 Watts

<em> The power consumption is 10 W.</em>


3 0
3 years ago
The spring constant for the spring in above Table is 20 N/m.
satela [25.4K]
The answer is 0.025J.

W=1/2*k*x^2
W=1/2*20*0.050^2
W=0.025J
5 0
3 years ago
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