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3 years ago

11

E)

1 answer:

guapka [62]3 years ago

7 0

Answer:

Pressure, P = 32666.66 Pa

Explanation:

It is given that,

Surface area of foot of Bimaba is 150 cm² or 0.015 m².

Her weight is 50 kg

We need to find the pressure does she exert on the ground, as she stands on her one foot. The force acting per unit area is called pressure. It can be given by :

$P=\dfrac{F}{A}\\\\P=\dfrac{mg}{A}\\\\P=\dfrac{50\times 9.8}{0.015}\\\\P=32666.66\ Pa$

So, the pressure is 32666.66 Pa.

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if 6.75g of oxide of lead was reduced to 6.12g of metal, calculate the empirical formula of the oxide

Solnce55 [7]

Mass lost = weight of oxygen = 6.75 - 6.12 = 0.63 g

0.63 g / 16.0 g/mole = 0.0394 moles O
6.12 g / 207.2 g/mole = 0.0295 moles Pb

Ratio being roughly 3:4, the empirical formula is Pb3O4 (which is one of the mixed oxides of lead).

6 0

4 years ago

Girls stay away from my questions....<br><br><br>What is rectilinear motion explain in detail .....

miv72 [106K]

Answer:

Linear motion in which the direction of the velocity remains constant and the path is a straight line.

4 0

2 years ago

Read 2 more answers

A 3.91 kg cart is moving at 5.7 m/s when it collides with a 4 kg cart which was at rest. They collide and stick together.

Nesterboy [21]

Answer:

<em>The velocity after the collision is 2.82 m/s</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

It states the total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is

P=mv.

If we have a system of two bodies, then the total momentum is the sum of the individual momentums:

$P=m_1v_1+m_2v_2$

If a collision occurs and the velocities change to v', the final momentum is:

$P'=m_1v'_1+m_2v'_2$

Since the total momentum is conserved, then:

P = P'

Or, equivalently:

$m_1v_1+m_2v_2=m_1v'_1+m_2v'_2$

If both masses stick together after the collision at a common speed v', then:

$m_1v_1+m_2v_2=(m_1+m_2)v'$

The common velocity after this situation is:

$\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}$

There is an m1=3.91 kg car moving at v1=5.7 m/s that collides with an m2=4 kg cart that was at rest v2=0.

After the collision, both cars stick together. Let's compute the common speed after that:

$\displaystyle v'=\frac{3.91*5.7+4*0}{3.91+4}$

$\displaystyle v'=\frac{22.287}{7.91}$

$\boxed{v' = 2.82\ m/s}$

The velocity after the collision is 2.82 m/s

6 0

3 years ago

A circular ride has a radius of 32 m If the time of one revolution of a rider is 0.98 s what is the speed of the rider?

11111nata11111 [884]

Given that,

The radius of a circular path, r = 32 m

The time of one revolution of a rider is 0.98 s.

To find,

The speed of the rider.

Solution,

Let v is the speed of the rider. Speed is equal to total distance divided by time taken.

$v=\dfrac{2\pi r}{t}\\\\v=\dfrac{2\pi \times 32}{0.98}\\\\v=205.16\ m/s$

So, the speed of the rider is 205.16 m/s.

6 0

3 years ago

Which is true concerning the acceleration due to Earth's gravity (ge) ? It decreases with increasing altitude. B. It is differe

klasskru [66]

Answer:

Option A decreases with increase in altitude

Explanation:

This can be explained as the value of gravitational acceleration, 'g' is not same everywhere.

It has its maximum value at poles of the Earth and minimum on its equator.

Thus a person will weigh more at poles than equator.

This variation is in accordance to:

$g = \frac{GM_{E}}{radius^{2}}$

Thus the gravitational acceleration changes as inverse square of the Radius of the Earth.

Thus as we move away from the Earth's center, gravitational acceleration, g decreases.

6 0

3 years ago