A third class lever that is 5cm away from the fulcrum, and 20cm away from the fulcrum?

You have 2 minutes to get to PE from science class before you get a tardy. If PE is 100m away and you walk at a speed of 1.1m/s

mart [117]

Answer:

D

Explanation:

4 0

3 years ago

Read 2 more answers

The height of a typical playground slide is about 1.8 m and it rises at an angle of 30 ∘ above the horizontal.

Salsk061 [2.6K]

Answer:

$5.94\ \text{m/s}$

$1.7$

$0.577$

Explanation:

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

$\theta$ = Angle of slope = $30^{\circ}$

v = Velocity of child at the bottom of the slide

$\mu_k$ = Coefficient of kinetic friction

$\mu_s$ = Coefficient of static friction

h = Height of slope = 1.8 m

The energy balance of the system is given by

$mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 1.8}\\\Rightarrow v=5.94\ \text{m/s}$

The speed of the child at the bottom of the slide is $5.94\ \text{m/s}$

Length of the slide is given by

$l=h\sin\theta\\\Rightarrow l=1.8\sin30^{\circ}\\\Rightarrow l=0.9\ \text{m}$

$v=\dfrac{1}{2}\times5.94\\\Rightarrow v=2.97\ \text{m/s}$

The force energy balance of the system is given by

$mgh=\dfrac{1}{2}mv^2+\mu_kmg\cos\theta l\\\Rightarrow \mu_k=\dfrac{gh-\dfrac{1}{2}v^2}{gl\cos\theta}\\\Rightarrow \mu_k=\dfrac{9.81\times 1.8-\dfrac{1}{2}\times 2.97^2}{9.81\times 0.9\cos30^{\circ}}\\\Rightarrow \mu_k=1.73$

The coefficient of kinetic friction is $1.7$ .

For static friction

$\mu_s\geq\tan30^{\circ}\\\Rightarrow \mu_s\geq0.577$

So, the minimum possible value for the coefficient of static friction is $0.577$ .

8 0

3 years ago

Given a force of 100 N and acceleration of 5 m/s2, what is the mass

tatyana61 [14]

Answer:

20 kg

Explanation:

remember the equation f=ma.

100 N=force

5 m/s2= acceleration

so you need to divide force by acceleration: 100 N/ 5 m/s2= 20 kg, to get the mass.

8 0

2 years ago

A block of 7.80 kg kept on an inclined plane just begins to slide at an angle of inclination of 35.0°. Once it has been set into

Bas_tet [7]

Answer:

The answer is....i am sorry idk

8 0

4 years ago

What is the weight of a 8-kg substance in N, kN, kg·m/s2, kgf, lbm·ft/s2, and lbf?

kvv77 [185]

Answer:

$W = 78.48N\\W =0.0784kN\\W = 8kgf\\$

$W= 3.6lbf\\W= 115.2 lbm*ft/s2$

Explanation:

if

$m=8kg=3.6lb\\$

and g=9.81 m/s2=32.16 ft/s2

and

W=m*g

we can just replace de mass and gravity and we have

$W = 8kg * 9.81 \frac{m}{s^{2} } =78.48N\\W = \frac{78.48N}{1000} =0.0784kN\\W = 8kgf\\$

$W= 3.6lbf\\W= 3.6lbm *32.16 ft/s2 =115.2 lbm*ft/s2$

5 0

4 years ago