what would happen to the size of the shadow if the distance between the light and the hand is increased.

The weight of a person is 500N and his foot imprint area is 0.5m^2.Calculate the total pressure exerted by person when he stands

Ghella [55]

Answer:

Pressure on both feet will be $500N/m^2$

Explanation:

Weight of the person F = 500 N

Area of foot print $A=0.5m^2$

Area of both the foot $A=2\times 0.5=1m^2$

We have to find pressure on both the feet

Pressure is equal to ratio of force and area

So pressure $P=\frac{F}{A}$

$P=\frac{500}{1}=500N/m^2$

So the pressure on both feet will be $500N/m^2$ when person stands on both feet.

7 0

3 years ago

What is the special name for doctors who deal with cancer?

Sladkaya [172]

Oncology is the study of cancer, so the doctors are called oncologists

8 0

4 years ago

Read 2 more answers

Consider a container of oxygen gas at a temperature of 23°C that is 1.00 m tall. Compare the gravitational potential energy of a

Sergio039 [100]

Answer:

Yes, it is reasonable to neglect it.

Explanation:

Hello,

In this case, a single molecule of oxygen weights 32 g (diatomic oxygen) thus, the mass of kilograms is (consider Avogadro's number):

$m=1molec*\frac{1mol}{6.022x10^{23}molec} *\frac{32g}{1mol}*\frac{1kg}{1000g}=5.31x10^{-26}kg$

After that, we compute the potential energy 1.00 m above the reference point:

$U=mhg=5.31x10^{-26}kg*1.00m*9.8\frac{m}{s^2}=5.2x10^{-25}J$

Then, we compute the average kinetic energy at the specified temperature:

$K=\frac{3}{2}\frac{R}{Na}T$

Whereas $N_A$ stands for the Avogadro's number for which we have:

$K=\frac{3}{2} \frac{8.314\frac{J}{mol*K}}{6.022x10^{23}/mol}*(23+273)K\\ \\K=6.13x10^{-21}J$

In such a way, since the average kinetic energy energy is about 12000 times higher than the potential energy, it turns out reasonable to neglect the potential energy.

Regards.

8 0

4 years ago

A uniform solid disk rolls without slipping down an incline making an angle θ with the horizontal. What is its acceleration? (En

Maru [420]

Answer:

aCM = (2/3)*g*Sin θ

Explanation:

Consider a uniform solid disk having mass M, radius R and rotational inertia I about its center of mass, rolling without slipping down an inclined plane.

In order to get the linear acceleration of the object’s center of mass, aCM ,

down the incline, we analyze this as follows:

The force of gravity (W = Mg) acting straight down is resolved into components parallel and perpendicular to the incline.

Since the object rolls without slipping there is a force of friction (Ff) acting on the object, at it’s point of contact with the incline, in the direction up the incline.

Newton’s 2nd Law gives then for acceleration down the incline

∑Fx' = m*aCM ⇒ m*g*Sin θ - Ff = m*aCM

The force of friction also causes a torque around the center of mass

having lever arm R so we can also write

τ = R*Ff = I*α

Solving for the friction, Ff = I*α / R

This is used in the expression derived from the 2nd Law:

m*g*Sin θ - Ff = m*g*Sin θ - (I*α / R) = m*aCM

The objects angular acceleration is related to the linear acceleration of the edge that contacts the incline by

a = R*α

Since the object rolls without slipping this has the same magnitude as aCM so we have that

α = aCM / R

Using this in

m*g*Sin θ - (I*α / R) = m*g*Sin θ - (I*(aCM / R) / R) = m*aCM

⇒ aCM = (m*g*Sin θ*R²) / (I + m*R²)

if I = (1/2)*m*R² (for a uniform solid disk)

we get

aCM = (2/3)*g*Sin θ

6 0

3 years ago

A car covers a distace of 20km in 20 seconds. Calculate its speed in km/h m/s.

BartSMP [9]

Answer:

speed = 3600 kilometers per hour

= 3600 km/h

speed = 1000 meters per second

= 1000 m/s

Answer From Gauth Math

6 0

3 years ago