PLEASE HELP!!! PLEASE HELP!!!

Starting from rest, a solid sphere rolls without slipping down an incline plane. At the bottom of the incline, what does the ang

Marrrta [24]

Answer:

2/R*sqrt (g*s*sin(θ)) = w

Explanation:

Assume:

- The cylinder with mass m

- The radius of cylinder R

- Distance traveled down the slope is s

- The angular speed at bottom of slope w

- The slope of the plane θ

- Frictionless surface.

Solution:

- Using energy principle at top and bottom of the slope. The exchange of gravitational potential energy at height h, and kinetic energy at the bottom of slope.

ΔPE = ΔKE

- The change in gravitational potential energy is given as m*g*h.

- The kinetic energy of the cylinder at the bottom is given as rotational motion: 0.5*I*w^2

- Where I is the moment of inertia of the cylinder I = 0.5*m*R^2

We have:

m*g*s*sin(θ) = 0.25*m*R^2*w^2

2/R*sqrt (g*s*sin(θ)) = w

- The angular velocity depends on plane geometry θ , distance travelled down slope s, Radius of the cylinder R , and gravitational acceleration g

3 0

3 years ago

A block weighing 400 kg rests on a horizontal surface and supports on top of it ,another block of weight 100 kg which is attache

Paladinen [302]

Answer:

$F_a=1470\ N$

Explanation:

<u>Friction Force</u>

When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.

There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.

Please find the free body diagrams in the figure provided below.

The equilibrium condition for the mass 1 is

$\displaystyle F_a-F_{r1}-F_{r2}=m.a=0$

The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa

$\displaystyle F_a=F_{r1}+F_{r2}.....[1]$

The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is

$\displaystyle F_{r2}-T=0$

The friction forces are computed by

$\displaystyle F_{r2}=\mu \ N_2=\mu\ m_2\ g$

$\displaystyle F_{r1}=\mu \ N_1=\mu(m_1+m_2)g$

Recall N1 is the reaction of the surface on mass m1 which holds a total mass of m1+m2.

Replacing in [1]

$\displaystyle F_{a}=\mu \ m_2\ g\ +\mu(m_1+m_2)g$

Simplifying

$\displaystyle F_{a}=\mu \ g(m_1+2\ m_2)$

Plugging in the values

$\displaystyle F_{a}=0.25(9.8)[400+2(100)]$

$\boxed{F_a=1470\ N}$

8 0

2 years ago

The volume occupied by a sample of gas is 480 mL when the pressure is 115 kPa.What pressure must be applied to the gas to make i

balandron [24]

Answer:

The answer is

Explanation:

Using Boyle's law to find the final pressure

That's

$P_1V_1 = P_2V_2$

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

Since we are finding the final pressure

$P_2 = \frac{P_1V_1}{V_2}$

From the question

P1 = 115 kPa

V1 = 480 mL

V2 = 650 ml

So we have

$P_2 = \frac{115000 \times 480}{650} = \frac{55200000}{650} \\ = 84923.076923...$

We have the final answer as

Hope this helps you

7 0

3 years ago

Singly charged gas ions are accelerated from rest through a voltage of 10.3 V. At what temperature (in K) will the average kinet

Natasha_Volkova [10]

Answer:

Temperature of the gas molecules is 7.96 x 10⁴ K

Explanation:

Given :

Ions accelerated through voltage, V = 10.3 volts

The work done to change the position of singly charged gas ions is given by the relation :

W = q x V

Here q is charge of the ions and its value is 1.6 x 10⁻¹⁹ C.

Average kinetic energy of gas molecules is given by the relation:

K.E. = $\frac{3}{2}kT$

Here T is temperature and k is Boltzmann constant and its value is 1.38 x 10⁻²³ J/K.

According to the problem, the average kinetic energy of gas is equal to the work done to move the singly charged ions, i.e. ,

K.E. = W

$\frac{3}{2}kT = qV$

Rearrange the above equation in terms of T :

$T= \frac{2qV}{3k}$

Substitute the suitable values in the above equation.

$T=\frac{2\times1.6\times10^{-19}\times10.3 }{3\times1.38\times10^{-23} }$

T = 7.96 x 10⁴ K

5 0

3 years ago

What is (3.25 x 10^5) + (7.5 x 10^4)?

ivanzaharov [21]

Answer:

400000

Explanation:

So first solve one part:

(3.25 * 10^5)

(3.25 * 100,000)

= 325000

Then solve the next part:

(7.5 * 10^4)

(7.5 * 10000)

= 75000

Now lastly, add the two answers:

325000 + 75000 = 400000

Therefore,

(3.25 x 10^5) + (7.5 x 10^4) = 400000

5 0

2 years ago

Read 2 more answers