I need help with these four problems.. anyone??

A speed skater moving across frictionless ice at 8.4 m/s hits a 5.7 m -wide patch of rough ice. She slows steadily, then continu

malfutka [58]

Answer:

Acceleration, $a=-2.48\ m/s^2$

Explanation:

Initial speed of the skater, u = 8.4 m/s

Final speed of the skater, v = 6.5 m/s

It hits a 5.7 m wide patch of rough ice, s = 5.7 m

We need to find the acceleration on the rough ice. The third equation of motion gives the relationship between the speed and the distance covered. Mathematically, it is given by :

$v^2-u^2=2as$

$a=\dfrac{v^2-u^2}{2s}$

$a=\dfrac{(6.5)^2-(8.4)^2}{2\times 5.7}$

$a=-2.48\ m/s^2$

So, the acceleration on the rough ice $-2.48\ m/s^2$ and negative sign shows deceleration.

8 0

3 years ago

To the nearest tenth, what is the area of the shaded segment when JA=8ft ??

Murrr4er [49]

$A=\dfrac{1}{6}\pi 8^2-\dfrac{8^2\sqrt{3}}{4}\approx 5.8\;[ft^2]$
Answer: <span>C. 5.8 ft squared</span>

7 0

3 years ago

Read 2 more answers

The platform height for Olympic divers is 10 m. A 60 kg diver steps off the platform to begin his dive.

azamat

Answer:

a) Ep = 5886[J]; b) v = 14[m/s]; c) W = 5886[J]; d) F = 1763.4[N]

Explanation:

a)

The potential energy can be found using the following expression, we will take the ground level as the reference point where the potential energy is equal to zero.

$E_{p} =m*g*h\\where:\\m = mass = 60[kg]\\g = gravity = 9.81[m/s^2]\\h = elevation = 10 [m]\\E_{p}=60*9.81*10\\E_{p}=5886[J]$

b)

Since energy is conserved, that is, potential energy is transformed into kinetic energy, the moment the harpsichord touches water, all potential energy is transformed into kinetic energy.

$E_{p} = E_{k} \\5886 =0.5*m*v^{2} \\v = \sqrt{\frac{5886}{0.5*60} }\\v = 14[m/s]$

c)

The work is equal to

W = 5886 [J]

d)

We need to use the following equation and find the deceleration of the diver at the moment when he stops his velocity is zero.

$v_{f} ^{2}= v_{o} ^{2}-2*a*d\\where:\\d = 2.5[m]\\v_{f}=0\\v_{o} =14[m/s]\\Therefore\\a = \frac{14^{2} }{2*2.5} \\a = 39.2[m/s^2]$

By performing a sum of forces equal to the product of mass by acceleration (newton's second law), we can find the force that acts to reduce the speed of the diver to zero.

m*g - F = m*a

F = m*a - m*g

F = (60*39.2) - (60*9.81)

F = 1763.4 [N]

3 0

3 years ago

A force of 6.00 N acts in the positive direction on a 3.00 kg object, originally traveling at +15.0 m/s, for 10.0 s. (a) What is

frozen [14]

Answer:

60 kg m/s

Explanation:

Let $a\;\; m/s^2$ be the acceleration of the object.