What is true of all particles of gas enclosed within a container?

which has a higher acceleration:a 10kg object acted upon with a net force of 20N or an 18kg object acted on by a net force of 20

MA_775_DIABLO [31]

<span>Answer: The acceleration of 10 kg object is greater than that of 18 kg object.

Explanation:
According to Newton's Second law:
F = ma --- (A)

Let's find the acceleration for both 10 kg and 18 kg objects!
The net force on both of these masses = F = 20N

(1) Acceleration of 10 kg object
Mass = m = 10 kg
Plug in the values in equation (A):
20 = 10 * a
Acceleration = a = 2 m/s^2

(2) Acceleration of 18 kg object
Mass = m = 18 kg
Plug in the values in equation (A):

20 = 18 * a
Acceleration = a = 1.11 m/s^2

2 > 1.11; therefore, 10 kg object has the higher acceleration compared to the acceleration of the 18 kg object.</span>

7 0

4 years ago

Read 2 more answers

The volume of a cylindrical tin can with a top and a bottom is to be 16π cubic inches. If a minimum amount of tin is to be used

sergiy2304 [10]

Answer:

h = 4 in

Explanation:

GIVEN DATA:

volume of tin $= 16 \pi$

we know that

volume of cylinder is $v = \pi r^2 h$

so,

$16 \pi = \pi r^2 h$

$16 = r^2 h$

$r = \sqrt{\frac{16}{h}}$

construct formula for surface area

$S = 2\pi r^2 + 2\pi rh$

$S = \frac{2v}{h} + 2 \sqrt{v \pi h}$

minimize the function wrt h

$S' = \frac{2v}{h^2} + \sqrt{\frac{v \pi}{h} = 0$

solving for h we have

$h = [\frac{4 v}{\pi}]^{1/3}$

we kow $v = 16 \pi$ so

h = 4 in

8 0

3 years ago

The figure in Figure 1 shows two single-slit diffraction patterns. The distance between the slit and the viewing screen is the s

V125BC [204]

Answer:

"The wavelengths are the same for both. The width of slit 1 is larger than the width of slit 2."

Explanation:

The full question has not been provided, so I just copied this into the web and found this answer and explanation on quizlet:

"The wavelengths are the same for both. The width of slit 1 is larger than the width of slit 2.

D sin θ = m λ

if the wavelengths are the same, then if the angle is smaller, the slit width must be larger. The top photo shows a pattern that is more closely spaced. That means the angle is smaller. The slit width must be larger."

This answer/explanation should be correct, as we are looking at bright fringes and the formula being used corresponds to the parameters of the question.

Hope this helps!

8 0

2 years ago

A car is up on a hydraulic lift at a garage. The wheels are free to rotate, and the drive wheels are rotating with a constant an

masya89 [10]

Answer:

Explanation:

Given

Wheels are rotating with constant angular velocity let say $\omega$

Presence of constant angular velocity show that there is no angular acceleration thus there is no tangential acceleration.

But any particle on the rim will experience a constant acceleration towards center called centripetal acceleration.

(a) yes, there will be tangential velocity which is given by

$v=r\cdot \omega$

where r=radial distance from center

(b)tangential acceleration

there would be no tangential acceleration as velocity is constant

(c)centripetal acceleration

Yes, there will be centripetal acceleration given by

$a_c=\omega ^2\times r$

7 0

3 years ago

Two identical copper blocks are connected by a weightless, unstretchable cord through a frictionless pulley at the top of a thin

likoan [24]

Answer:

13.6 N

Explanation:

Since one side of the wedge is vertical and the wedge makes and angle of 33 with the horizontal, the angle between the weight of the copper block on the incline and the incline is thus 90 - 33 = 57.

Let M be the mass of the block that hangs, m be the mass of the block on the incline and T be the tension in the weightless unstretchable cord.

We assume the motion is downwards in the direction of the hanging block, M.

We now write equations of motion for each block.

So

Mg - T = Ma (1) and T - mgcos57 - F = ma where F is the frictional force on the block on the incline and a is their acceleration.

Now, since both blocks do not move, a = 0.

So, Mg - T = M(0) = 0 and T - mgcos57 - F = m(0) = 0

Mg - T = 0 (3) and T - mgcos57 - F = 0 (4)

From (3), T = Mg

Substituting T into (4), we have

T - mgcos57 - F = 0

Mg - mgcos57 - F = 0

So, Mg - mgcos57 = F

F = Mg - mgcos57

F = (M - mcos57)g

Since g = acceleration due to gravity = 9.8 m/s², and M = 2.94 kg and m = 2.85 kg.

We find F, thus

F = (2.94 kg - 2.85 kgcos57)9.8 m/s²

F = (2.94 kg - 2.85 kg × 0.5446)9.8 m/s²

F = (2.94 kg - 1.552 kg)9.8 m/s²

F = (1.388 kg)9.8 m/s²

F = 13.6024 kgm/s²

F ≅ 13.6 N

6 0

3 years ago