: Which of the following are the requirements for successful animal reproduction?

A 80 kg skier grips a moving rope that is powered by an engine and is pulled at constant speed to the top of a 25 degrees hill.

zloy xaker [14]

Answer:

$P=28.085\,hp$

Explanation:

Given that:

mass of 1 skier, $m=80kg$

inclination of hill, $\theta=25^{\circ}$

length of inclined slope, $l=220m$

time taken to reach the top of hill, $t=2.3 min= 138 s$

coefficient of friction, $\mu=0.15$

Now, force normal to the inclined plane:

$F_N=m.g.cos\theta$

$F_N=80\times 9.8\times cos25^{\circ}$

$F_N=710.54\,N$

Frictional force:

$f=\mu.F_N$

$f=0.15\times 710.54$

$f=106.58\,N$

The component of weight along the inclined plane:

$W_l=m.g.sin\theta$

$W_l=80\times 9.8\times sin25^{\circ}$

$W_l=331.33\,N$

Now the total force required along the inclination to move at the top of hill:

$F=f+W_l$

$F=106.58+331.33$

$F=437.91\,N$

Hence the work done:

$W=F.l$

$W=437.91\times 220$

$W=96340.80\,J$

Now power:

$P=\frac{W}{t}$

$P=\frac{96340.80}{138}$

$P=698.12\,W$

So, power required for 30 such bodies:

$P=30\times 698.12$

$P=20943.65\,W$

$P=\frac{20943.65}{745.7}$

$P=28.085\,hp$

8 0

3 years ago

List the requirements of practical fuel?

rewona [7]

Answer:

Here we have some of the requirement of practical fuel are

1. It must contain large amount of stored energy. So that more amount of power output available to run the engines, motors etc.

2. It must occur in abundance in nature or be easy to produce.

3. The fuel must be made up of elements that combine easily with oxygen. Foe example if hydrogen molecules reacts with oxygen. Then the products are at the reaction of lower energy than the reactants, the result is the explosive release of energy and the product of water.

7 0

3 years ago

A jet aircraft is traveling at 260 m/s in horizontal flight. The engine takes in air at a rate of 53.3 kg/s and burns fuel at a

IrinaVladis [17]

Answer:

The thrust of the jet engine is 4188.81 N.

Explanation:

Given that,

Speed = 260 m/s

Rate in air= 53.3 kg/s

Rate of fuel = 3.63 kg/s

Relative speed = 317 m/s

We need to calculate the rate of mass change in the rocket

Using formula of rate of mass

$\dfrac{dM}{dt}=\dfrac{dM_{a}}{dt}+\dfrac{dM_{f}}{dt}$

Put the value into the formula

$\dfrac{dM}{dt}=53.3+3.63$

$\dfrac{dM}{dt}=56.93\ kg/s$

We need to calculate the thrust of the jet engine

Using formula of thrust

$T=\dfrac{dM}{dt}u-\dfrac{dM_{a}}{dt}v$

Put the value into the formula

$T=56.93\times317-53.3\times260$

$T=4188.81\ N$

Hence, The thrust of the jet engine is 4188.81 N.

7 0

4 years ago

How many significant digits should the answer to the following problem have? (2.49303 g) * (2.59 g) / (7.492 g) =

Feliz [49]

The number of significant digits to the answer of the following problem is four.

<h3>What are the significant digits?</h3>

The number of digits rounded to the approximate integer values are called the significant digits.

The following problem is

(2.49303 g) * (2.59 g) / (7.492 g) =

On solving we get

= 0.86184566204

The answer is approximated to 0.86185

Thus, the significant digits must be four.

Learn more about significant digits.

brainly.com/question/1658998

#SPJ1

6 0

2 years ago

A transformer has 1000 turns in the primary coil and 100 turns in the secondary coil. If the primary coil is connected to a 120

Marianna [84]

Answer:

The voltage on the secondary is 12 V while the current is 0.5 A.

Explanation:

A transformer works by changing the level of the voltage and current on a circuit using a magnetic field and two coils. The ratio by wich they are changed is dependant on the ratio of turns between the primary and secondary of the transformer. In this case we have a ratio for the voltage of:

ratio = (turns on the secondary)/(turns on the primary)

ratio = 100/1000 = 0.1

So in this case the voltage delivered to the primary will be multiplied by 0.1. We can now calculate the voltage on the secondary:

Voltage secondary = Voltage primary* ratio = 120*0.1 = 12 V

The transformer maintains roughly the same power output on both sides, since the power output on a electric circuit is given by the product of the voltage by the current on that circuit, to maintain the same power when the voltage has been droped the current must be raised by the same ratio. So we have:

Current secondary = Current primary*(1/ratio) =0.05*(1/0.1) = 0.5 A

6 0

3 years ago