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2 years ago

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A sharp edged orifice with a 50 mm diameter opening in the vertical side of a large tank discharges under a head of 5m. If the c

oefficient of contraction is 0.62 and the coefficientof velocity is 0.98, what is thedischarge?

1 answer:

Sever21 [200]2 years ago

3 0

Answer:

0.24

Explanation:

See attached file

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A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then

Elan Coil [88]

Answer:

A) $W_{ff} =-744.12J$

B) $F_f=-W_{ff}*sin\theta /hy = 112.75N$

C) $F_{f2}=207.58N$

Explanation:

This question is incomplete. The full question was:

<em>A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vf = 6.2 m/s. </em>

<em>Part (a) Write an expression for the work, Wf, done by the friction force between the ramp and the skateboarder in terms of the variables given in the problem statement. </em>

<em>Part (b) The ramp makes an angle θ with the ground, where θ = 30°. Write an expression for the magnitude of the friction force, fr, between the ramp and the skateboarder. </em>

<em>Part (c) When the skateboarder reaches the bottom of the ramp, he continues moving with the speed vf onto a flat surface covered with grass. The friction between the grass and the skateboarder brings him to a complete stop after 5.00 m. Calculate the magnitude of the friction force, Fgrass in newtons, between the skateboarder and the grass.</em>

For part A), we make a balance of energy to calculate the work done by the friction force:

$W_{ff}=\Delta E$

$W_{ff}=1/2*m*vf^2-m*g*hy$

$W_{ff}=-744.12J$

For part B), we use our previous value for the work:

$W_{ff}=-F_f*(hy/sin\theta)$ Solving for friction force:

$F_f=-W_{ff}*sin\theta /hy$

$F_f=112.75N$

For part C), we first calculate the acceleration by kinematics and then calculate the module of friction force by dynamics:

$Vf^2=Vo^2+2*a*d$

Solving for a:

$a=-3.844m/s^2$

Now, by dynamics:

$|F_f|=|m*a|$

$|F_f|=207.58N$

8 0

2 years ago

According to the saffir/simpson scale, what storm surge and type of damage was most likely experienced in homestead, florida whi

Naddika [18.5K]

Answer:

Explanation:

The Saffir-Simpson Hurricane Wind Scale gives 1 to 5 rating based on a hurricane's wind speed, storm surge and potential property damage. It rates according to categories. When placed at Hurricanes of Category 3 and above, they are termed major catastrophic hurricanes because of their potential for loss of life and damages of property. Ratings at Category 1 and 2 storms are dangerous but can be prevented by following directed measures.

Florida situated directly in the hurricane at east coast , had experiened Category 1- Category 5 ratings from hurricane and noticed in Land fall in Florida since 1894 with Winds ranging from 74-95 mph with some dangerous damages which can lead damages to Well-built homes, dedtroying roofs , Uprooting shallow rooted trees, and causing power outages due to destruction of poles

to 157 mph or higher causing Catastrophic damages making the area inhabitable due to loss of life and properties.

For example, Hurricane Dorian in September 2019 attacked Florida’s east coast resulting to a Category 2 storm producing tropical storm force winds,heavy rain and storm surge leading to damages of roof of building and disruption of power poles.

8 0

2 years ago

This sphere is now connected by a long, thin conducting wire to another sphere of radius R2 that is several meters from the firs

alekssr [168]

Here is the full question

A metal sphere with Radius R₁ has a charge Q₁. Take the electric potential to be zero at an infinite distance from the sphere

a) What are the electric field and electric potential at the surface of the sphere?

This sphere is now connected by a long, thin conducting wire to another sphere of radius R₂ that is several meters from the first sphere. Before the connection is made, this second sphere is uncharged. After electrostatic equilibrium has been reached:

b) what is the total charge on each sphere?

Assume that the amount of charge on the wire is much less than the charge on each sphere.

Answer:

a) The electric field (E) at the surface is the first sphere = $\frac{1}{4 \pi \epsilon _0}\frac{Q_1}{R_1^2}$

The electric potential (V) at the surface of the first sphere = $\frac{1}{4 \pi \epsilon _0}\frac{Q_1}{R_1}$

= $ER_1$

b)

The total charge of the first sphere $q_1 = \frac{Q_1R_1}{R_1+R_2}$

The total charge of the second sphere $q_2= \frac{Q_1R_2}{R_1+R_2}$

Explanation:

Given that;

the radius of the sphere = R

The radius of the first sphere = $R_1$

The radius of the second sphere = $R_2$

Charge on the first sphere = $Q_1$

a) The electric field (E) at the surface is the first sphere = $\frac{1}{4 \pi \epsilon _0}\frac{Q_1}{R_1^2}$

The electric potential (V) at the surface of the first sphere = $\frac{1}{4 \pi \epsilon _0}\frac{Q_1}{R_1}$

= $ER_1$

b) From the question. before the part b question; we learnt that the first sphere is now connected to another sphere;

Now that the two sphere are joined . Charges flows from one to another until their potentials are equal.

As Such; We use $q_1 \ and \ q_2$ to represent their charges respectively

The potential on the surface of the first sphere;

$V_1 = \frac{1}{4 \pi \epsilon _0}\frac{q_1}{R_1}$

The potential on the surface of the second sphere;

$V_2 = \frac{1}{4 \pi \epsilon _0}\frac{q_2}{R_2}$

$V_1=V_2$

∴

$\frac{1}{4 \pi \epsilon _0}\frac{q_1}{R_1}= \frac{1}{4 \pi \epsilon _0}\frac{q_2}{R_2}$

Thus, we can say :

$\frac{q_1}{q_2}= \frac{R_1}{R_2}$

and $Q_1 = q_1 + q_2$

As such ;

The total charge of the first sphere $q_1 = \frac{Q_1R_1}{R_1+R_2}$

The total charge of the second sphere $q_2= \frac{Q_1R_2}{R_1+R_2}$

3 0

2 years ago

What is the minimum work needed to push a 950-kg car 710 m up along a 9.0Â° incline? ignore friction?

kakasveta [241]

<span>1.0344645 MJ The minimum energy need is the potential energy of the car at the top of the ramp and is given by mass*gravity*height mass is known, gravity is assumed to be 9.81m/s^2 as it is on earth, and height must be calculated using trigonometry. height=sin(9 degrees)*710m=111meters so potential energy = 950kg*111m*9.81m/s^2=1.0344645 MJ Using the law of the conservation of energy we can assume that the energy expended to push the car up the incline was at least the potential energy gained by moving 111m against the pull of gravity.</span>

7 0

2 years ago

Read 2 more answers

A piece of copper of mass 100 g is being drilled through with a 1/2" electric drill. The drill operates at 40.0 W and takes 30.0

lesantik [10]

Answer:

correct option is c. 31.0°C

Explanation:

given data

copper of mass = 100 g

electric drill = 1/2"

power = 40.0 W

time = 30 s

C copper = 387 J/kgC

to find out

copper's increase in temperature

solution

we get here energy that is express a s

energy = 40 W × 30 s

energy = 1200 Watt seconds

and heat acquired by drill is here as

heat acquired = 100 × T × 387

here temperature rise in copper mass as

temperature rise in copper mass = $\frac{100}{1000}$ × T × 387

temperature rise in copper mass = 38.7 × T Watt second

we know that all the energy from the drill heats the copper

so we can say

38.7 × T = 1200

T = 31°C

so correct option is c. 31.0°C

4 0

2 years ago