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2 years ago

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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.

1 answer:

Radda [10]2 years 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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Jumping up before the elevator hits. After the cable snaps and the safety system fails, an elevator cab free-falls from a height

butalik [34]

Answer:

a) I = -2257.6 Kg*m/s

b) F = -451,520N

Explanation:

part a.

we know that:

I = $P_f-P_i$

where I is the impulse, $P_f$ the final momentum and $P_i$ the initial momentum.

so:

I = $MV_f-MV_i$

where M is the mass, $V_f$ the final velocity and $V_i$ the initial velocity.

Therefore, we have to find the initial velocity or the velocity of the passenger just before the collition.

now, we will use the law of the conservation of energy:

$E_i=E_f$

so:

mgh = $\frac{1}{2}MV_i^2$

where g is the gravity and h the altitude. So, replacing values, we get:

(85kg)(9.8m/s^2)(36m)= $\frac{1}{2}(85kg)V_i^2$

solving for $V_i$ :

$V_i = 26.56m/s$

Then, replacing in the initial equation:

I = $MV_f-MV_i$

I = $(85kg)(0m/s)-(85kg)(26.56m/s)$

I = -2257.6 Kg*m/s

Then, the impulse is -2257.6 Kg*m/s, it is negative because it is upwards.

part b.

we know that:

Ft = I

where F is the average force, t is the time and I is the impulse. So, replacing values, we get:

F(0,005s) = -2257.6 Kg*m/s

solving for F:

F = -451520N

Finally, the force is -451,520N, it is negative because it is upwards.

3 0

3 years ago

It's nighttime, and you've dropped your goggles into a 3.2-mm-deep swimming pool. If you hold a laser pointer 1.1 mm above the e

Savatey [412]

Answer: 5.30m

Explanation:

depth of pool = 3.2 m

i = 67.75°

Using snell's law, we have,

n₁ × sin(i) = n₂ × 2 × sin(r)

n₁ = 1, n₂ =1.33, r= 44.09°

Hence,

Distance of Google from edge if pool is:

2.2 + d×tan(r) = 2.2 + (3.2 × tan(44.09°) =5.30m

5 0

4 years ago

- Explain how the Earth, the sun, and time are connected,

Radda [10]

Answer:

the rotation of earth is determining what part of the sun faces what part of earth making time the part thats away from the sun would be night and the one facing the sun itself would be day

6 0

3 years ago

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Near the surface of Earth an electric field points radially downward and has a magnitude of approximately 100 N/C. What charge (

pentagon [3]

Answer:

$q=2.997\times 10^{-4}$ C

Sign-Negative

Explanation:

We are given that

Electric field = $E=100NC^{-1}$ (Radially downward)

Acceleration= $0.19 ms^{-2}$ (Upward)

Mass of charge=3 g= $3\times 10^{-3}$ kg

1kg=1000g

We have to find the magnitude and sign of charge would have to be placed on a penny .

By newton's second law

$\sum F_y=ma$

$\sum F_y=qE-mg$

Substitute the values then we get

$qE-mg=ma$

Substitute the values then we get

$q(100)-3\times 10^{-3}(9.8)=3\times 10^{-3}(0.19)$

$100q-29.4\times 10^{-3}=0.57\times 10^{-3}$

$100q=0.57\times 10^{-3}+29.4\times 10^{-3}=29.97\times 10^{-3}$

$q=\frac{29.97\times 10^{-3}}{100}$

$q=2.997\times 10^{-4}$ C

Sign of charge =Negative

Because electric force acting in opposite direction of electric field therefore,charge on penny will be negative.

4 0

3 years ago

My uncle wants to play twister with me and my little sister but he gets a lil too wild. What do I do ?

Lelu [443]

Get even wilder so he knows your'e not the one to joke with!

Hope This Helps Him Calm Down:)

8 0

3 years ago

Read 2 more answers