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Complete Question

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Answer:

The value is [tex]P = 294594.3 \ W[/tex]

Explanation:

From the question we are told that

   The height is  [tex]h  =  41 \  m[/tex]

   The efficiency of the turbine is [tex]\eta =  0.77[/tex]

   The flow rate is [tex]\r  V  = 1 m^3 / s[/tex]

    The head loss is  g =  2 m

Generally the head gain by the turbine is mathematically represented as      

        [tex]H  =  h -  d[/tex]

=>     [tex]H  =  41 - 2[/tex]

=>     [tex]H  =  39 \  m [/tex]

Generally the actual power generated by the turbine is mathematically represented as

          [tex]P =  \eta  *\gamma *    \r V * H[/tex]

Here [tex]\gamma[/tex] is the specific density of water with value

        [tex]\gamma   =  9810 N/m^3 [/tex]

So

       [tex]P =0.77  *9810 *   1 *  39[/tex]

       [tex]P = 294594.3 \ W[/tex]

Answer:

The ratio of the displacement amplitudes of two sound waves is 1.16.

Explanation:

Given that,

Frequency = 5.0 kHz

Intensity level difference = 3.0 dB

We know that,

The sound intensity is inversely proportional to the square of distance.

[tex]I\propto\dfrac{1}{r^2}[/tex]

The sound intensity for first wave is

[tex]\beta_{1}=10\log\dfrac{I_{1}}{I_{0}}[/tex]...(I)

The sound intensity for second wave is

[tex]\beta_{2}=10\log\dfrac{I_{2}}{I_{0}}[/tex]...(II)

We need to calculate the ratio of intensity

From equation (I) and (II)

[tex]\beta_{2}-\beta_{1}=10\log\dfrac{I_{2}}{I_{0}}-10\log\dfrac{I_{1}}{I_{0}}[/tex]

[tex]\Delta \beta=10\log(\dfrac{I_{2}}{I_{1}})[/tex]

Put the value into the formula

[tex]3.0=10\log(\dfrac{I_{2}}{I_{1}})[/tex]

[tex]\dfrac{I_{2}}{I_{1}}=e^{\dfrac{3.0}{10}}[/tex]

[tex]\dfrac{I_{2}}{I_{1}}=1.34[/tex]

We need to calculate the ratio of the displacement

Using formula of displacement

[tex]\dfrac{r_{1}}{r_{2}}=\sqrt{\dfrac{I_{2}}{I_{1}}}[/tex]

Put the value into the formula

[tex]\dfrac{r_{1}}{r_{2}}=\sqrt{1.34}[/tex]

[tex]\dfrac{r_{1}}{r_{2}}=1.16[/tex]

Hence, The ratio of the displacement amplitudes of two sound waves is 1.16.

Answer:

The time is [tex]t_t =   3.7583 \ s [/tex]

Explanation:

From the question we are told that

   The angle is  [tex]\theta =  30^o[/tex]

    The horizontal  distance is  [tex]d =  120 \ ft[/tex]

Generally when the ball is at maximum height before descending the velocity is zero and this velocity can be mathematically represented as

      [tex]v =  u  +  at[/tex]

here a =  -g  the negative sign is because the direction of motion is against gravity

So

       [tex]v =  v_i   +  at[/tex]

Here[tex] v_i [/tex] is the vertical  component of the initial  velocity of the ball which is  mathematically

represented as

        [tex]v_i = usin(\theta )[/tex]

So      

=>     [tex]0  =  usin(\theta )  -9.8t[/tex]

Generally the total time taken to travel the 120 ft  is mathematically represented as

        [tex]t_t = \frac{120}{v_h}[/tex]

Here [tex]v_h[/tex] is the horizontal component of the initial velocity which is mathematically represented as

     [tex]v_h  =  u cos(\theta )[/tex]

So

       [tex]t_t = \frac{120}{ u cos(\theta )}[/tex]

Generally the time taken to reach the maximum height is  

      [tex]t = \frac{t_t}{2}[/tex]

=>    [tex]t = \frac{120}{ u cos(\theta )} * \frac{1}{2} [/tex]

=>    [tex]t = \frac{60}{ u cos(\theta )} [/tex]

So

      [tex]0  =  usin(\theta )  -9.8*   [\frac{60}{ u cos(\theta )}][/tex]

        [tex]  usin(\theta )   = 9.8*   [\frac{60}{ u cos(\theta )}][/tex]

       [tex]  usin(\theta ) *  u cos(\theta)  =60*  9.8   [/tex]

        [tex]  u^2 sin(\theta ) cos(\theta)  =60*  9.8   [/tex]

       [tex]  u^2 sin(30 ) cos(30)  =60*  9.8   [/tex]

        [tex]  u^2 * \frac{1}{2}* \frac{\sqrt{3}}{2}  =588.6  [/tex]

         [tex]  u^2 *\sqrt{3}  =2354.4 [/tex]

          [tex]  u^2 = 1359.31 [/tex]

        [tex]  u = 36.87 \ ft/s  [/tex]

Substituting this value into the equation for total time

     [tex]t_t = \frac{120}{36.87 cos(30 )}[/tex]

     [tex]t_t =   3.7583 \ s [/tex]

Answer:

6. 3.9 km SE

7. 17 km NE

8. 3496 seconds

9. 36 m

10. 13.27 km/m

Explanation:

6. and 7. added both numbers

8. multiplied both numbers

9. 60 x 60 = 3600

3600 x 0.01 = 36

10. 60 x 3 = 180

180 + 10 =190

190 / 14.32 = 13.27

Answer:

The energy is   [tex]U =  18.98 \  J [/tex]

Explanation:

From the question we are told that

   The inductor is  [tex]L  =  4.74 \ H[/tex]

    The resistance of the resistor is [tex]R =  9.33 \  \Omega[/tex]

    The voltage of the battery is [tex]V =  26.4 \  V[/tex]

Generally the current flowing in the circuit is mathematically represented as

      [tex]I =  \frac{V}{R}[/tex]

=>   [tex]I =  \frac{26.4}{9.33 }[/tex]

=>   [tex]I =  2.83 \ A[/tex]

Generally the corresponding energy stored in the circuit is  

       [tex]U =  \frac{1}{2} * L  *  I^2[/tex]

        [tex]U =  \frac{1}{2} *  4.74  *  2.83 ^2[/tex]

       [tex]U =  18.98 \  J [/tex]

Answer:

The rowboat will move away from sally more quickly because the rowboat because the sailboat is larger in mass

Explanation:

Gnerally the row boat will move away from her quicker than the sailboat this is because the mass of the sail boat is larger than the row boat , hence the frictional force that opposes motion will be greater in the sailboat than in the row boat.

Explanation:

[tex]kinetic \: energy \: is \: the \: energy \: of \: a \: body \: in \: motion. \\ that \: is \: energy \: of \: a \: body \: that \: is \: moving.\\ while \\ potential \: energy \: is \: the \: energy \: of \: a \: body \: by \: the \: virtue \: of \: its \: position \: \\ that \: is \: energy \: of \: a \: body \: that \: is \: not \: moving.[/tex]

♨Rage♨

Answer:

Explanation:

The speed of the cat can be found by using the formula

[tex]v = \frac{d}{t} \\ [/tex]

d is the distance

t is the time taken

From the question we have

[tex]v = \frac{9}{16} = \\ = 0.5625[/tex]

We have the final answer as

Hope this helps you

Answer:

Humans be considered carbon sinks. Not only do humans

have a lot of carbon in them, they also use a lot of carbon.

hope this helps

Answer:

Airplane X has more gravitational potential energy than Airplane W

Explanation:

Gravitational potential energy is defined as "the energy acquired by an object due to its positional change in presence of gravitational force."

That being said, gravitational potential energy depends on the height of an object above the ground. It also depends on the mass of the object and even further, the amount of gravitational force that is applied.

And if we take a look at the question again, we'd agree that the two airplanes are flying at different heights, this means their gravitational potential energy will be different. And as such, Airplane X has more gravitational potential energy than Airplane W

Answer:

False

Explanation:

The cryosphere is all the part of the earth where water is in solid form. Where wind interacts with rocks, it is an example of atmosphere - geosphere interaction.

Answer:

The force of forearm is 239.9 N

The force of elbow is 215.89 N

Explanation:

Suppose, When you lift an object by moving only your forearm, the main lifting muscle in your arm is the biceps. Suppose the mass of a forearm is 1.50 kg. If the biceps is connected to the forearm a distance [tex]d_{b}[/tex] 2.50 cm from the elbow, how much force [tex]F_{b}[/tex] must the biceps exert to hold a 950 g ball at the end of the forearm at distance dball J 36.0 cm from the elbow, with the forearm parallel to the floor? How much force [tex]F_{l}[/tex] must the elbow exert,

Given that,

Mass of forearm = 1.50 kg

Distance of forearm = 2.50 cm

Mass of ball = 950 g

Distance of ball = 36.0 cm

We need to calculate the force of forearm

Using balancing torque about elbow

[tex]F_{b}\times d_{b}=w_{f}\times\dfrac{d_{f}}{2}+w_{ball}\times d_{ball}[/tex]

Put the value into the formula

[tex]F_{b}\times 0.025= 1.50\times9.8\times\dfrac{0.36}{2}+0.95\times9.8\times0.36[/tex]

[tex]F_{b}=\dfrac{1.50\times9.8\times\dfrac{0.36}{2}+0.95\times9.8\times0.36}{0.025}[/tex]

[tex]F_{b}=239.9\ N[/tex]

We need to calculate the force of elbow

Using balancing force

[tex]F_{b}=F_{l}+w_{f}+w_{b}[/tex]

[tex]F_{l}=F_{b}-w_{f}-w_{b}[/tex]

Put the value into the formula

[tex]F_{l}=239.9-(1.50\times9.8)-(0.95\times9.8)[/tex]

[tex]F_{l}=215.89\ N[/tex]

Hence, The force of forearm is 239.9 N

The force of elbow is 215.89 N

Answer:

throw them away............

Answer:

2.0 m

Explanation:

Energy is conserved.

Initial KE = Final PE + Work done by friction

½ mv² = mgh + Fd

½ mv² = mgh + mgμd

½ v² = gh + gμd

½ v² − gh = gμd

d = (½ v² − gh) / (gμ)

d = (½ (7.5 m/s)² − (10 m/s²) (2.1 m)) / ((10 m/s²) (0.35))

d = 2.0 m

Answer:

4m/s^2 ( A)

Explanation:

The solution is in the attached file

Answer:

10) The distance between the archer and the tree is 50.074 meters.

11) The speed of the banana when it hits the water is approximately 13.554 meters per second.

Explanation:

10) The arrow experiments a parabolic motion, which is the combination of horizontal motion at constant velocity and vertical uniform accelerated motion. In this case we need to find the horizontal distance between the archer and the tree, calculated by the following kinematic equation:

[tex]x = x_{o} +v_{o}\cdot t \cdot \cos \theta[/tex] (Eq. 1)

Where:

[tex]x_{o}[/tex] - Initial position of the arrow, measured in meters.

[tex]x[/tex] - Final position of the arrow, measured in meters.

[tex]v_{o}[/tex] - Initial speed of the arrow, measured in meters per second.

[tex]t[/tex] - Time, measured in seconds.

[tex]\theta[/tex] - Launch angle, measured in sexagesimal degrees.

If we know that [tex]x_{o} = 0\,m[/tex], [tex]v_{o} = 65\,\frac{m}{s}[/tex], [tex]t = 0.85\,s[/tex] and [tex]\theta = 25^{\circ}[/tex], the horizontal distance between the archer and the tree is:

[tex]x = 0\,m + \left(65\,\frac{m}{s}\right)\cdot (0.85\,s)\cdot \cos 25^{\circ}[/tex]

[tex]x = 50.074\,m[/tex]

The distance between the archer and the tree is 50.074 meters.

11) The final speed of the banana ([tex]v[/tex]), measured in meters per second, just before hitting the water is determined by the Pythagorean Theorem:

[tex]v = \sqrt{v_{x}^{2}+v_{y}^{2}}[/tex] (Eq. 2)

Where:

[tex]v_{x}[/tex] - Horizontal speed of the banana, measured in meters per second.

[tex]v_{y}[/tex] - Vertical speed of the banana, measured in meters per second.

Each component of the speed are obtained by using these kinematic equations:

[tex]v_{x} = v_{o}\cdot \cos \theta[/tex] (Eq. 3)

[tex]v_{y} = v_{o}\cdot \sin \theta +g\cdot t[/tex] (Eq. 4)

Where [tex]g[/tex] is the gravitational acceleration, measured in meters per square second.

If we know that [tex]v_{o} = 7.6\,\frac{m}{s}[/tex], [tex]\theta = -40^{\circ}[/tex], [tex]g = -9.807\,\frac{m}{s^{2}}[/tex] and [tex]t = 0.75\,s[/tex], the components of final speed are, respectively:

[tex]v_{x} = \left(7.6\,\frac{m}{s} \right)\cdot \cos (-40^{\circ})[/tex]

[tex]v_{x} = 5.822\,\frac{m}{s}[/tex]

[tex]v_{y} = \left(7.6\,\frac{m}{s}\right)\cdot \sin (-40^{\circ})+\left(-9.807\,\frac{m}{s^{2}} \right) \cdot (0.75\,s)[/tex]

[tex]v_{y} = -12.240\,\frac{m}{s}[/tex]

And the speed of the banana right before hitting the water is:

[tex]v = \sqrt{\left(5.822\,\frac{m}{s} \right)^{2}+\left(-12.240\,\frac{m}{s} \right)^{2}}[/tex]

[tex]v \approx 13.554\,\frac{m}{s}[/tex]

The speed of the banana when it hits the water is approximately 13.554 meters per second.

I think it is D. Force

Answer:

Explanation:

Do you mean tire tracks?

If you do then the tracks show that there must have been a lot of heat involved. When the tires skid, the friction produced must be awfully hot to stop and 800 kg vehical (which is pretty small). The tire tracks that you see are bits of melted rubber taken from the tire.

Force is involved, but heat is the better answer.

gravity is necessary for friction to take place. If you were out in space, you could not get enough friction to leave tire tracks. (Force is awfully small).

Lubrication is there to give you a 4th choice. The exact opposite is what takes place. Lubrication reduces friction.

Answer:

61 m/s

Explanation:

        [tex]v_{f}^{2} - v_{o}^ {2} = 2* g* h (1)[/tex]

Answer:

Explanation:

Answer:

Fy = 14.3 [N]

Explanation:

To be able to solve this problem we must know that the force is a vector and has magnitude and direction, therefore it can be decomposed into the force in the X & y components:

When we have the components on the horizontal and vertical axes we must use the Pythagorean theorem.

[tex]F = \sqrt{F_{x}^{2} +F_{y}^{2} }[/tex]

where:

F = 15 [N]

Fx = horizontal component = 4.5 [N]

Fy = vertical component [N]

[tex]15=\sqrt{4.5^{2}+F_{y}^{2}}\\ 15^{2}= (\sqrt{4.5^{2}+F_{y}^{2}})^{2} \\225 = 4.5^{2}+F_{y} ^{2}\\ F_{y}^{2} =225 -4.5^{2}\\ F_{y}^{2}=204.75\\F_{y}=\sqrt{204.75}\\ F_{y}=14.3 [N][/tex]

Answer:

vₓ = 6.73 m/s

Explanation:

       [tex]v_{x} = \frac{\Delta x}{\Delta t} (1)[/tex]

       [tex]\Delta h = \frac{1}{2} * g * t^{2} (2)[/tex]

       [tex]t = \sqrt{\frac{2*\Delta h}{g} } =\sqrt{\frac{2*37.0m}{9.8m/s2}} = 2.75 s (3)[/tex]    

       [tex]v_{x} = \frac{\Delta x}{\Delta t} = \frac{x}{t} = \frac{18.5m}{2.75s} = 6.73 m/s[/tex]

Answer:

significant

Explanation:

The digits in a measurement that are considered significant are all of those digits that represent marked calibrations on the measuring device plus one additional digit to represent the estimated digit (tenths of the smallest calibration).

Answer:

(a) The net force is 80.394 N

    The acceleration of the crate is 0.804 m/s²

(b) the final velocity of the crate is 5.02 m/s

Explanation:

Given;

mass of the crate, m = 100 kg

applied force, F = 250 N

angle of inclination, θ = 45°

coefficient of friction, μ = 0.12

Applied force in y-direction, [tex]F_y = Fsin \theta = 250sin45 = 176.78 \ N[/tex]

Applied force in x-direction, [tex]F_x = Fcos \theta = 250cos45 = 176.78 \ N[/tex]

The normal force is calculated as;

N + Fy -W = 0

N = W - Fy

N = (100 x 9.8) - 176.78

N = 980 - 176.78 = 803.22 N

The frictional force is given by;

Fk = μN

Fk = 0.12 x 803.22

Fk = 96.386 N

(a) The net force is given by;

[tex]F_{net} = F_x - F_k\\\\F_{net} = 176.78-96.386\\\\F_{net} = 80.394 \ N[/tex]

Apply Newton's second law of  motion;

F = ma

[tex]a = \frac{F_{net}}{m}\\\\ a = \frac{80.394}{100}\\\\ a = 0.804 \ m/s^2[/tex]

(b) the velocity of the crate after 5.0 s

[tex]F = ma= \frac{m(v-u)}{t} \\\\Ft =m(v-u)\\\\v-u = \frac{Ft}{m}\\\\ v = \frac{Ft}{m} + u\\\\v = \frac{F_{net}*t}{m} + u\\\\v = \frac{80.394*5}{100} + 1\\\\v = 5.02 \ m/s[/tex]

Answer:

C

Explanation:

only if there is a net force of zero, the body will not move

some people may say B but that is wrong because maybe one force is greater than the other so the object would still move even though the forces are in opposite directions and parallel

Answer:

skeletal,cardiac,and smooth.

Explanation:

Answer:

Skeletal, smooth, and cardiac.

Explanation:

Skeletal Muscles:

Skeletal muscles are the most familiar type of muscles; they make up most of the muscle mass in the body. Flexible bands of connective tissue called tendons attach these muscles to the bones in the body. Skeletal muscles control voluntary movement in the body.

Smooth Muscles:

Smooth muscles are involuntary muscles that we don’t consciously control. They are found within the walls of many organs and control the movement of these organs. For example, they enable the movement of food through the digestive system.

Cardiac Muscles:

Cardiac muscles are a special type of involuntary muscle. Located in the heart, these muscles control the contractions of the heart.

can u give me brainliest???

Answer:

Force Exerted by Sedan on Truck = - F

Explanation:

This question can easily be solved by using Newton's Third Law of Motion. Newton's Third Law of Motion clearly states that for every action force there exists an equal in magnitude reaction force. However, the direction of the reaction force is opposite to the direction of the action force. Mathematically,

Action Force = - Reaction Force

Hence, we have here:

Force Exerted by Semi Truck on Sedan = F

So, from Newton's Third Law of motion:

Force Exerted by Sedan on Truck = - Force Exerted by Semi Truck on Sedan

Force Exerted by Sedan on Truck = - F