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ehidna [41]
1 year ago
15

A thin 1.5 mm coating of glycerine has been placed between two microscope slides of width 0.8 cm and length 3.9 cm . Find the fo

rce required to pull one of the microscope slides at a constant speed of 0.28 m/s relative to the other.
Physics
1 answer:
Radda [10]1 year ago
3 0

The  force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is zero.

<h3>Force required to pull one end at a constant speed</h3>

The force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is determined by applying Newton's second law of motion as shown below;

F = ma

where;

  • m is mass
  • a is acceleration

At a constant speed, the acceleration of the object will be zero.

F = m x 0

F = 0

Thus, the  force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is zero.

Learn more about constant speed here: brainly.com/question/2681210

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Now the elevator is moving downward with a velocity of v = -2.8 m/s but accelerating upward with an acceleration of a = 5.5 m/s2
borishaifa [10]

Answer:

160.75 N

Explanation:

The downward velocity has no effect on the force situation, it is only changes in velocity (plus, of course, gravity, which is always there) that require a force. At constant velocity, the bottom spring s_3 is supporting its mass m_3 to balance gravity.

As the elevator slows, though, it also ends up slowing down the spring arrangement, too. However, because the stretching takes time, it means that some damped harmonic motion will be set up in the spring chain.

When the motion has finally damped out, the net force the bottom spring s3 exerts on m3 has two components--that of gravity and of the deceleration of the elevator:

F_3net = m3 * (g + a) = 10.5×(9.81+5.5)= 10.5×15.31= 160.75 N

5 0
3 years ago
How fast (in rpm) must a centrifuge rotate if a particle 7.3 cm from the axis of rotation is to experience an acceleration of 1.
guajiro [1.7K]

The formula we use here is:

radial acceleration = ω^2 * R <span>

110,000 * 9.81 m/s^2 = ω^2 * 0.073 m 
<span>ω^2 = 110,000 * 9.81 / 0.073
ω = 3844.76 rad/s </span></span>

<span>and since: ω = 2pi*f  --> f = ω/(2pi)</span><span>
f = 3844.76 / (2pi) = 611.91 rps = 611.91 * 60 rpm 
<span>= 36,714.77 rpm </span></span>

5 0
3 years ago
A 12.0 cm object is 9.0 cm from a convex mirror that has a focal length of -4.5 cm. What is the distance of the image from the m
Rasek [7]

Answer:

- 3 cm

Explanation:

From the mirror formula;

1/f = 1/v + 1/u ; where f is the focal length, v is the image distance, and u is the object distance.

1/-4.5 = 1/9 + 1/v

1/v = -1/4.5 - 1/9

    = -1/3

Therefore;

v = -3 cm

Hence;

Image distance is - 3cm

5 0
2 years ago
n 38 g rifle bullet traveling at 410 m/s buries itself in a 4.2 kg pendulum hanging on a 2.8 m long string, which makes the pend
Kitty [74]

Answer:

68cm

Explanation:

You can solve this problem by using the momentum conservation and energy conservation. By using the conservation of the momentum you get

p_f=p_i\\mv_1+Mv_2=(m+M)v

m: mass of the bullet

M: mass of the pendulum

v1: velocity of the bullet = 410m/s

v2: velocity of the pendulum =0m/s

v: velocity of both bullet ad pendulum joint

By replacing you can find v:

(0.038kg)(410m/s)+0=(0.038kg+4.2kg)v\\\\v=3.67\frac{m}{s}

this value of v is used as the velocity of the total kinetic energy of the block of pendulum and bullet. This energy equals the potential energy for the maximum height reached by the block:

E_{fp}=E_{ki}\\\\(m+M)gh=\frac{1}{2}mv^2

g: 9.8/s^2

h: height

By doing h the subject of the equation and replacing you obtain:

(0.038kg+4.2kg)(9.8m/s^2)h=\frac{1}{2}(0.038kg+4.2kg)(3.67m/s)^2\\\\h=0.68m

hence, the heigth is 68cm

4 0
2 years ago
A highly volatile substance has an initial mass of 1200 g and its mass is reduced by 12% each second.
Softa [21]

Answer:

Explanation:

a) 1.00 - 0.12 = 0.88

m = 1200(0.88)^t

b) t = ln(m/1200) / ln(0.88)

c) m = 1200(0.88)^10 = 334.20 g

d) t = ln(10/1200) / ln(0.88) = 37.451... = 37 s

e) t = ln(1/1200) / ln(0.88) = 55.463... = 55 s

4 0
3 years ago
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