Ask question

Ask question

All categories

Ostrovityanka [42]

2 years ago

8

As shown above an adhesive has been applied to contacting faces of two blocks so that the blocks interact with an adhesive force

that has a magnitude, Fadhesion A tension, T, is exerted by a wire attached to the upper blocks causing the blocks to remain at rest The two blocks have weights, Wbottom and Wtop. Which of the following must be true? (1 point) ♡ O Wtop=T OT = Wtop - Fadhesion O Wbottom Fadhesion O T = Wbottom + Wtop + Fadhesion

1 answer:

rusak2 [61]2 years ago

8 0

The magnitude of the adhesive force allows the top block to remain

attached to the bottom when the blocks and the wire are balanced.

The option that must be true is; $\mathbf{W_{bottom} = F_{adhesion}}$

Reason:

The tension exerted by the wire attached to the top block = T

Magnitude of the adhesive = $F_{adhesion}$

Weight of the top block = $W_{top}$

Weight of the bottom block = $W_{bottom}$

Given that with the exertion of the tension, the two blocks remain at rest, we have;

T = $W_{top}$ + $W_{bottom}$

The adhesive causes the bottom block to remain attached to the top block, we have;

Therefore, the magnitude of the adhesive force adds the bottom weight, to the top, weight, which gives;

The magnitude of the adhesive force = The weight of the bottom block

Therefore;

$W_{bottom}$ = $F_{adhesion}$

Learn more here:

brainly.com/question/18907970

You might be interested in

Why is it important to keep the muzzle of a firearm pointed in a safe direction, even though the firearm's safety is engaged?

Gnom [1K]

The correct answer is B. The safety only prevents you from pulling the trigger, but does not stop the pin from striking the primer. For example, if you drop the firearm, the pin may hit the primer and fire the firearm. It is always responsible to keep the firearm pointed in a safe direction so that if this happens, no consequences come out of it.

6 0

2 years ago

An organ pipe open at both ends has two successive harmonics with frequencies of 220 Hz and 240 Hz. What is the length of the pi

Harman [31]

Answer:

The length is $l = 8.6 \ m$

Explanation:

From the question we are told that

The frequencies of the two successive harmonics are $f_1 = 220 \ Hz$ , $f_2 = 240 \ Hz$

The speed of sound in the air is $v_s = 343 \ m/s$

Generally the frequency of a given harmonic is mathematically represented as

$f_n = \frac{n v }{2l}$

Here n defines the position of the harmonics

Now since the position of both harmonic is not know but we know that they successive then we can represented them mathematically as

$220 = \frac{n v}{2l}$

and

$240 = \frac{(n+1) v}{2l}$

So

$\frac{(n + 1 ) v}{2l} - \frac{n v}{2l} = 240-220$

=> $\frac{v}{2l} = 20$

=> $l = 8.6 \ m$

7 0

3 years ago

The maximum wavelength that an electromagnetic wave can have and still eject electrons from a metal surface is 459 nm. What is t

spin [16.1K]

Answer:

2.7067 eV

Explanation:

h = Planck's constant = $6.626\times 10^{-34}\ m^2kg/s$

c = Speed of light = $3\times 10^8\ m/s$

$\lambda_0$ = Threshold wavelength = 459 nm

Work function is given by

$W_0=\frac{hc}{\lambda_0}\\\Rightarrow W_0=\frac{6.626\times 10^{-34}\times 3\times 10^8}{459\times 10^{-9}}\\\Rightarrow W_0=4.33072\times 10^{-19}\ J$

Converting to eV

$1\ J=\frac{1}{1.6\times 10^{-19}}\ eV$

$4.33072\times 10^{-19}\ J=4.33072\times 10^{-19}\times \frac{1}{1.6\times 10^{-19}}\ eV=2.7067\ eV$

The work function W0 of this metal is 2.7067 eV

4 0

3 years ago

A 132 cm wire carries a current of 2.2 A. The wire is formed into a circular coil and placed in a B Field of intensity 1 T. a) F

EastWind [94]

Given Information:

Length of wire = 132 cm = 1.32 m

Magnetic field = B = 1 T

Current = 2.2 A

Required Information:

(a) Torque = τ = ?

(b) Number of turns = N = ?

Answer:

(a) Torque = 0.305 N.m

(b) Number of turns = 1

Explanation:

(a) The current carrying circular loop of wire will experience a torque given by

τ = NIABsin(θ) eq. 1

Where N is the number of turns, I is the current in circular loop, A is the area of circular loop, B is the magnetic field and θ is angle between B and circular loop.

We know that area of circular loop is given by

A = πr²

where radius can be written as

r = L/2πN

So the area becomes

A = π(L/2πN)²

A = πL²/4π²N²

A = L²/4πN²

Substitute A into eq. 1

τ = NI(L²/4πN²)Bsin(θ)

τ = IL²Bsin(θ)/4πN

The maximum toque occurs when θ is 90°

τ = IL²Bsin(90)/4πN

τ = IL²B/4πN

torque will be maximum for N = 1

τ = (2.2*1.32²*1)/4π*1

τ = 0.305 N.m

(b) The required number of turns for maximum torque is

N = IL²B/4πτ

N = 2.2*1.32²*1)/4π*0.305

N = 1 turn

8 0

3 years ago

A shot-putter accelerates a 7.2 kg shot from rest to 17 m/s . what work did the shot-putter do on the ball?

garri49 [273]

<span>1.0x10^3 Joules The kinetic energy a body has is expressed as the equation E = 0.5 M V^2 where E = Energy M = Mass V = Velocity Since the shot was at rest, the initial energy is 0. Let's calculate the energy that the shot has while in motion E = 0.5 * 7.2 kg * (17 m/s)^2 E = 3.6 kg * 289 m^2/s^2 E = 1040.4 kg*m^2/s^2 E = 1040.4 J So the work performed on the shot was 1040.4 Joules. Rounding the result to 2 significant figures gives 1.0x10^3 Joules</span>

6 0

3 years ago