When an spoon is in a cup of water what is it? diffraction reflection refraction absorption bounce

A crate of eggs is located in the middle of the flatbed of a pickup truck as the truck negotiates a curve in the flat road. The

oksian1 [2.3K]

Answer:

v = 14.35 m/s

Explanation:

As we know that crate is placed on rough bed

so here when pickup will take a turn around a circle then in that case the friction force on the crate will provide the necessary centripetal force on the crate

So here we have

$\mu mg = \frac{mv^2}{R}$

here we have

$\mu g = \frac{v^2}{R}$

now we know that

$v = \sqrt{\mu Rg}$

here we have

$\mu = 0.600$

R = 35 m

g = 9.81 m/s/s

now plug in all values in above equation

$v = \sqrt{(0.600)(35)(9.81)}$

$v = 14.35 m/s$

4 0

3 years ago

A building that is 105.50 ft. feet tall is casting a shadow that is 37.50 ft. feet. If the building next to it, is casting a sha

Arlecino [84]

115.35 ft

Set the proportion up 37.50/105.50 = 41/x and solve for x

7 0

3 years ago

All of the following are examples of positive direction except:

charle [14.2K]

Answer:

right

Explanation:

think of a compass theres N.E.S.W north, east, south, west

8 0

3 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

3 years ago

If the pressure is 100,000 Pa, R is 8.314, the temperature is 300 K, and the number of moles of gas (n) is 100, what is the volu

djyliett [7]

We can solve the problem by using the ideal gas equation:
$pV=nRT$
where
p is the pressure of the gas
V is the volume of the gas
n is the number of moles of the gas
R is the gas constant
T is the absolute temperature of the gas

For the gas in our problem, we have:
$p=100000 Pa$
$R=8.314 J mol^{-1} K^{-1}$
$T=300 K$
$n=100$

If we rearrange the equation and we put these numbers into it, we find the volume of the gas:
$V= \frac{nRT}{p}= \frac{(100)(8.314)(300)}{100000}=2.49 m^3$

4 0

3 years ago