A 25kg object is 3.0 meters up. How much potential energy does it have

Why do sugars molecules stay intact when dissolved by water?

koban [17]

I hope this helps you on your question

6 0

3 years ago

Outside of the mirror in your bathroom (or bedroom), how else are you using mirrors in your daily life?

Elan Coil [88]

Answer:

Explanation:

Mirrors are widely used in our day to day life for example

1.E.N.T. specialist used a concave mirror to see the clear and better image of the infected part.

2. A convex mirror is used in vehicles as a rear-view mirror. It covers a large area, so it is commonly used in vehicles.

3. Convex mirrors are generally utilized for making magnifying glasses

7 0

3 years ago

What formal regions are located in the western hemisphere

Tju [1.3M]

Answer: Part of Europe, part of Africa, part of Antarctic and the entire America.

Explanation: Western hemisphere or west hemisphere encompasses all regions located west of the longitude 0 °, or meridian of Greenwich.

The formal regions located in the western hemisphere is Europe, Africa, Antarctic and America.

4 0

2 years ago

A pulley system with a mechanical advantage of 15 is used to lift a 1750 N piano to a third floor balcony that is 7 m above the

devlian [24]

Answer:

The force is $E = 117 \ N$

Explanation:

From the question we are told that

The mechanical advantage is $ME = 15$

The load is $L = 1750 \ N$

The height is $h = 7 \ m$

Generally mechanical advantage is mathematically represented as

$ME = \frac{L }{E }$

Here E is the force that must be applied to move the piano

=> $E = \frac{L}{ ME }$

=> $E = \frac{1750 }{15 }$

=> $E = 117 \ N$

6 0

2 years ago

The terminals of a 0.70 Vwatch battery are connected by a 80.0-m-long gold wire with a diameter of 0.200 mm What is the current

Komok [63]

Answer:

$I=0.047A$

Explanation:

Let's use Ohm's law:

$V=IR$

or

$I=\frac{V}{R}$ (1)

Where:

$V=Voltage\\I=Current\\R=Electrical\hspace{2 mm}Resistance$

We know the value of the voltage V, so we need to find the value of R in order to find I. Fortunately there is a relation between the resistivity of a conductor and its electrical resistance given by:

$R=\rho*\frac{l}{A}$ (2)

Where:

$R=Electrical\hspace{2 mm}Resistance\\l=Length\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}conductor=80m\\A=Cross\hspace{2 mm}sectional\hspace{2 mm}area\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}conductor=1.256637061*10^{-7} \\\rho=Electrical\hspace{2 mm}resistivity\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}material=2.35*10^{-8}$

Keep in mind that the electrical resistivity of the gold is a known constant which is $\rho_g_o_l_d=2.35*10^{-8}$ and the cross sectional area of the conductor is calculated as:

$A=\pi *(r^{2})=\pi *(0.0002m)^{2} =1.256637061*10^{-7} m^{2}$

Because we have a wire in this case, so we assume a cylindrical geometry.

Now replacing our data in (2)

$R=(2.35*10^{-8})*\frac{80}{1.256637061*10^{-7} } =14.96056465\Omega$

Finally, we know R and V, so replacing these values in (1) we will be able to find the current:

$I=\frac{0.7}{14.96056465}\approx0.047A$

7 0

2 years ago