Ask question

Ask question

All categories

2 years ago

12

A 60.0 kg ballet dancer stands on her toes during a performance with four square inches (26.0 cm2) in contact with the floor. Wh

at is the pressure exerted by the floor over the area of contact in the following cases?
(a) if the dancer is stationary
(b) if the dancer is leaping upwards with an acceleration of 5.00 m/s2

1 answer:

Anit [1.1K]2 years ago

5 0

Answer:

Explanation:

Given

Mass of dancer $m=60\ kg$

Area of contact $A=26\ cm^2$

(a)If dancer is stationary

Pressure exerted by Floor is given by load per unit area of contact

$P=\frac{weight}{A}$

$P=\frac{mg}{A}$

$P=\frac{60\times 9.8}{26\times 10^{-4}}$

$P=226.153 kN$

(b)If dancer is leaping upwards with an acceleration of $a=5 m/s^2$

So apparent weight of dancer is $W'=mg+ma$

$W'=m(g+a)=60\cdot (9.8+5)$

$W'=888\ N$

$Pressure=\frac{W'}{A}$

$Pressure=\frac{888}{26\times 10^{-4}}$

$=341.53\ kN$

You might be interested in

At 0°C, frozen water (ice) changes to liquid water. When an ice cube is placed on something that is warmer than it heat will mov

kondaur [170]

Answer:

a common temperature is reached.

Explanation:

The principle of heat transfer is that heat flows from a substance of higher temperature to that of lower temperature until when a common temperature is reached.

For ice on getting in contact with a hot body; it changes to water at 0°c utilising the latent heat of fusion. After ice has been converted to water at 0°c, water starts increasing in temperature till a common temperature is reached.

8 0

2 years ago

A 6 kilogram block in outer space is moving at -100 m/s (to the left). It suddenly experiences three forces as shown below (in t

Blizzard [7]

Answer:

x = 1474.9 [m]

Explanation:

To solve this problem we must use Newton's second law, which tells us that the sum of forces must be equal to the product of mass by acceleration.

We must understand that when forces are applied on the body, they tend to slow the body down to stop it.

So as the body continues to move to the left, it is slowing down. Therefore we must calculate this deceleration value using Newton's second law. We must perform a sum of forces on the x-axis equal to the product of mass by acceleration. With leftward movement as negative and rightward forces as positive.

ΣF = m*a

$10 +12*sin(60)= - 6*a\\a = - 3.39[m/s^{2}]$

Now using the following equation of kinematics, we can calculate the distance of the block, before stopping completely. The initial speed must be 100 [m/s].

$v_{f}^{2} =v_{o}^{2}-2*a*x$

where:

Vf = final velocity = 0 (the block stops)

Vo = initial velocity = 100 [m/s]

a = - 3.39 [m/s²]

x = displacement [m]

$0 = 100^{2}-2*3.39*x\\x=\frac{10000}{2*3.39}\\x=1474.9[m]$

4 0

2 years ago

If an object is moving to the right and there is a net force acting on it to the left, what will happen to the object?

Ksju [112]

Hdjnenhdhdbbdbddbbdjjdjdjjejekrkrk

7 0

3 years ago

Please help me on this!

aalyn [17]

Answer:

erosion

Explanation:

grows into the eirth each year. the atmosphere and how the worldchanges every year

5 0

2 years ago

A disk of a radius 50 cm rotates at a constant rate of 100 rpm. What distance in meters will a point on the outside rim travel d

Iteru [2.4K]

Answer:

A point on the outside rim will travel 157.2 meters during 30 seconds of rotation.

Explanation:

We can find the distance with the following equation since the acceleration is cero (the disk rotates at a constant rate):

$d = v*t$

Where:

v: is the tangential speed of the disk

t: is the time = 30 s

The tangential speed can be found as follows:

$v = \omega*r$

Where:

ω: is the angular speed = 100 rpm

r: is the radius = 50 cm = 0.50 m

$v = \omega*r = 100 \frac{rev}{min}*\frac{2\pi rad}{1 rev}*\frac{1 min}{60 s}*0.50 m = 5.24 m/s$

Now, the distance traveled by the disk is:

$d = v*t = 5.24 m/s*30 s = 157.2 m$

Therefore, a point on the outside rim will travel 157.2 meters during 30 seconds of rotation.

I hope it helps you!

3 0

2 years ago