All categories

3 years ago

D. What is one watt power? Calculate the power of

Physics

1 answer:

Tatiana [17]3 years ago

3 0

Answer:

437.5 watts.

Explanation:

One watt power (1-watt) is used to describe the situation of an electrical appliance where 1 joule of energy per seconds of time (1 J/s) is consumed or used.

According to the information in the question, a man can lift a load of 175N (Force) to a height of 25 m (distance) in 10 seconds (time).

Power = Workdone (J) ÷ time (s)

Workdone = Force (N) × distance (m)

Workdone = 175 × 25

Workdone (W) = 4375J

Power = W/t

P = 4375/10

P = 437.5 watts.

You might be interested in

I will do ALL of your school work!!

kakasveta [241]

Answer:

get it done for free on www.brainly.com

Explanation:

4 0

3 years ago

Read 2 more answers

A roller coaster has a vertical loop with radius 25.7 m. With what minimum speed should the roller-coaster car be moving at the

gogolik [260]

Answer:

15.88m/s

Explanation:

At the top of the roller coaster you will have three forces acting on the roller-coaster. See the image below. Fc is the centripetal force (for an object in circular motion), Fg is the gravitational force, and Fn is the normal force. To achieve the minimum speed we assume the roller-coaster is barely touching the vertical loop and so the normal force is zero. This leaves two acting forces.

$F_g = F_c\\mg = \frac{m\times v^2}{r}\\v = \sqrt{gr} = \sqrt{9.81 \times 25.7} = 15.88 m/s$

7 0

3 years ago

When reading a digital volt-ohmmeter (DVOM), you have a reading of 2168 mV, which is the same as:__________

levacccp [35]

Answer:

D, both A and B

Explanation:

2168 mV is the SI unit for potential difference and the Voltmeter.

The primary unit is Volt, represented as V. Due to the fact that there can be a much higher reading, or an even much more smaller one, comes the need for variants of the same unit.

10^-3 is called milli and represented as m

10^3 is called killo and represented as k

10^-6 is called micro and represented as µ

10^6 is called mega and represented as M

and even much higher variants of up to 10^12 and 10^-12

As we can see from the aforementioned example, 10^-3 is milli and represented as m

And our question gave us the unit in mV, which stands for millivolts.

Also, if we look at option B, it states, 2.168 volts. This 2.168 volts is also the same thing as A. Take a look at it this way, I said mV is 10^-3, right?

So, 2168*10^-3 is also 2168/100 which is 2.168. The only difference here is, once we make this conversion from mV, we have to drop the milli tag, because we have already made a conversion, and thus, leave it as V.

2168 mV = 2.168V

Hence why we picked option D, Both A & B as the right one

4 0

3 years ago

Phosphorous reacts with chlorine gas to produce phosphorous pentachloride. Calculate the mass of product produced when 25.0 g of

MariettaO [177]

Answer : The mass of product produced if the reaction occurred with a 70.5 percent yield will be, 20.67 grams.

Explanation : Given,

Mass of P = 25 g

Mass of $Cl_2$ = 25 g

Molar mass of P = 30.97 g/mole

Molar mass of $Cl_2$ = 71 g/mole

Molar mass of $PCl_5$ = 208.24 g/mole

First we have to calculate the moles of $P$ and $Cl_2$ .

$\text{Moles of }P=\frac{\text{Mass of }P}{\text{Molar mass of }P}=\frac{25g}{30.97g/mole}=0.807moles$

$\text{Moles of }Cl_2=\frac{\text{Mass of }Cl_2}{\text{Molar mass of }Cl_2}=\frac{25g}{71g/mole}=0.352moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2P+5Cl_2\rightarrow 2PCl_5$

From the balanced reaction we conclude that

As, 5 moles of $Cl_2$ react with 2 moles of $P$

So, 0.352 moles of $Cl_2$ react with $\frac{2}{5}\times 0.352=0.1408$ moles of $P$

That means, in the given balanced reaction, $Cl_2$ is a limiting reagent and it limits the formation of products and $P$ is an excess reagent because the given moles are more than the required moles.

Now we have to calculate the moles of $PCl_5$ .

As, 5 moles of $Cl_2$ react with 2 moles of $PCl_5$

So, 0.352 moles of $Cl_2$ react with $\frac{2}{5}\times 0.352=0.1408$ moles of $PCl_5$

Now we have to calculate the mass of $PCl_5$ .

$\text{Mass of }PCl_5=\text{Moles of }PCl_5\times \text{Molar mass of }PCl_5$

$\text{Mass of }PCl_5=(0.1408mole)\times (208.24g/mole)=29.32g$

Now we have to calculate the mass of product produced (actual yield).

$\%\text{ yield of }PCl_5=\frac{\text{Actual yield of }PCl_5}{\text{Theoretical yield of }PCl_5}\times 100$

$70.5=\frac{\text{Actual yield of }PCl_5}{29.32g}\times 100$

$\text{Actual yield of }PCl_5=20.67g$

Therefore, the mass of product produced if the reaction occurred with a 70.5 percent yield will be, 20.67 grams.

3 0

3 years ago

A 14.3-g bullet is fired into a 5.21 kg block of wood. The block is attached to a spring that has a spring force constant of 450

Natasha2012 [34]

Answer:

The initial speed of the bullet is $v_{o} = 889.199\,\frac{m}{s}$ .

Explanation:

The collision between bullet and block is inelastic and let suppose that motion occurs on a horizontal surface, so that changes in gravitational potential energy can be neglected. Initially, the intial speed of the bullet-block system can be determined with the help of the Work-Energy Theorem and the Principle of Energy Conservation:

$K = U_{k} + W_{loss}$

$\frac{1}{2}\cdot (5.224\,kg)\cdot v^{2} = \frac{1}{2}\cdot \left(450\,\frac{N}{m}\right)\cdot (0.22\,m)^{2} + (0.35)\cdot (5.224\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right) \cdot (0.22\,m)$

The initial speed of the bullet-block system is:

$v \approx 2.383\,\frac{m}{s}$

Now, the initial speed of the bullet is determined by applying the Principle of Momentum Conservation:

$(0.014\,kg)\cdot v_{o} = (5.224\,kg)\cdot \left(2.383\,\frac{m}{s} \right)$

$v_{o} = 889.199\,\frac{m}{s}$

6 0

3 years ago