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3 years ago

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A 2.00-kg block of ice is moving on a frictionless horizontal surface. At t = 0 the block is moving to the right with a velocity

of magnitude 3.00 m/s. (a) Calculate the velocity of the block (magnitude and direction) after a force of 5.00 N directed to the right has been applied for 4.00 s; (b) if instead a force of 7.00 N directed to the left is applied from t=0 to t = 4.00 s, what is the final velocity of the block?

1 answer:

enyata [817]3 years ago

4 0

Answer:

Explanation:

a )

We shall apply the concept of impulse .

Impulse = force x time = change in momentum

= 5 x 4 = 2 ( V - 3 ) , where V is final velocity of the object

20 = 2V - 6

V = 13 m /s

b )

Impulse applied = - 7 x 4 = - 28 kg m/s ( negative as direction of force is opposite motion )

If v be the final velocity

2 x 3 - 28 = 2 v ( initial momentum - change in momentum = final momentum )

- 22 = 2v

v = - 11 m /s

object will move with 11 m /s in opposite direction .

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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.

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Spiderman, whose mass is 74.0 kg, is dangling on the free end of a 11.0-m-long rope, the other end of which is fixed to a tree l

Anestetic [448]

Answer:

W = -1844.513 J

Explanation:

GIVEN DATA:

mass of spider man is m 74 kg

vertical displacement if spider is 11 m

final displacement = 11 cos 60.6 = - 6.753 m

change in displacement is = -6.753 - (-11) = 4.25 m

gravity force act on spiderman is f = mg = 74 × 9.8 = 725.2 N

work done by gravity is $W = F \delta r cos\theta$

$W = 725.2 \times 4.25 \times cos 180$

where 180 is the angle between spiderman weight and displacement

W = -1844.513 J

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3 years ago

A moving object must have which type of energy

xxMikexx [17]

Answer:

Kinetic Energy

Explanation:

Kinetic energy is the energy an object has because of its motion. If we want to accelerate an object, then we must apply a force. ... Kinetic energy can be transferred between objects and transformed into other kinds of energy. For example, a flying squirrel might collide with a stationary chipmunk.

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3 years ago

To develop muscle tone, a woman lifts a 2.50 kg weight held in her hand. She uses her biceps muscle to flex the lower arm throug

Romashka [77]

To solve this problem we will use the concepts related to Torque as a function of the Force in proportion to the radius to which it is applied. In turn, we will use the concepts of energy expressed as Work, and which is described as the Torque's rate of change in proportion to angular displacement:

$\tau = Fr$

Where,

F = Force

r = Radius

Replacing we have that,

$\tau = Fr$

$\tau = 21cm (\frac{1m}{100cm})* 550N$

$\tau = 11.55Nm$

The moment of inertia is given by 2.5kg of the weight in hand by the distance squared to the joint of the body of 24 cm, therefore

$I = 0.25Kg\cdot m^2 +(2.5kg)(0.24m)^2$

$I = 0.394kg\cdot m^2$

Finally, angular acceleration is a result of the expression of torque by inertia, therefore

$\tau = I\alpha \rightarrow \alpha = \frac{\tau}{I}$

$\alpha = \frac{11.55}{0.394}$

$\alpha = 29.3 rad/s^2$

PART B)

The work done is equivalent to the torque applied by the distance traveled by 60 °° in radians $(\pi / 3)$ , therefore

$W = \tau \theta$

$W = 11.5* \frac{\pi}{3}$

$W = 12.09J$

4 0

3 years ago

What would most likely happen as a result of the generator in a wind turbine breaking? The blades would not be turned. Less stea

Aleonysh [2.5K]

Answer:

What would most likely happen as a result of the generator in a wind turbine breaking?

The What would most likely happen as a result of the generator in a wind turbine breaking?

The blades would not be turned.

Less steam would be produced.

Electricity would not be generated.

Solar energy would not be absorbed.

The blades would not be turned.

Less steam would be produced.

Electricity would not be generated.

Solar energy would not be absorbed.

Explanation:

F

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3 years ago

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