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2 years ago

Mechanical advantage allows you to apply a force over a what distance to what the distance an object moves

Physics

1 answer:

Sidana [21]2 years ago

7 0

Mechanical advantage allows you to apply a force over a short distance to increase the distance and object moves.

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Does A= 6.4 B=12 C=12.2 Form a RIGHT TRIANLGE?

Lina20 [59]

To answer this question, you must remember the equation:

a²+b²= c²

(6.4)² + (12)²= (12.2)²

<span>40.96 + 144 = 184.96
</span> (12.2)² = <span>148.84
</span>
184.96 ≠ 148.84

This cannot be a triangle

hope this helps

3 0

2 years ago

A 0.600 kg block is attached to a spring with spring constant 15 N/m. While the block is sitting at rest, a student hits it with

gulaghasi [49]

Answer:

8.8 cm

31.422 cm/s

Explanation:

m = Mass of block = 0.6 kg

k = Spring constant = 15 N/m

x = Compression of spring

v = Velocity of block

A = Amplitude

As the energy of the system is conserved we have

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{0.6\times 0.44^2}{15}}\\\Rightarrow A=0.088\ m\\\Rightarrow A=8.8\ cm$

Amplitude of the oscillations is 8.8 cm

At x = 0.7 A

Again, as the energy of the system is conserved we have

$\dfrac{1}{2}kA^2=\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2\\\Rightarrow v=\sqrt{\dfrac{k(A^2-x^2)}{m}}\\\Rightarrow v=\sqrt{\dfrac{15(0.088^2-(0.7\times 0.088)^2)}{0.6}}\\\Rightarrow v=0.31422\ m/s$

The block's speed is 31.422 cm/s

4 0

3 years ago

This is an object's change in motion per unit time in a specified direction.

timurjin [86]

Answer:

Velocity

Explanation:

Velocity is an object's change in motion per unit time in a specified direction

7 0

2 years ago

Read 2 more answers

A cube 6.0 cm on each side is made of a metal alloy. After you drill a cylindrical hole 2.0 cm in diameter all the way through a

Crank

To solve this problem it is necessary to apply the concepts related to Newton's second law, the definition of density and the geometric relationships that allow us to find the volume of the figures presented.

For the particular case of the Cube with equal sides its volume is determined by

$V_c = l^3$

$V_c = 6^3 = 216cm^3$

In the case of perforated material we have that its volume is given according to the cylindrical geometry, that is to say

$V_d = \pi r^2*l$

$V_d = \pi (\frac{2}{2})^2*6$

$V_d = 6\pi cm^3$

In this way the net volume would be

$\Delta V = V_c-V_d$

$\Delta V = 216cm^3-6\pi cm^3$

$\Delta V = 197.15cm^3 = 197.15*10^{-6}m^3$

We need to find the mass, but we have the Weight and Gravity so from Newton's second Law

$F= mg$

$m = \frac{F}{g}$

$m = \frac{6.6}{9.8}$

$m = 0.673kg$

PART A) From the relation of density as a unit of mass and volume we have to

$\rho = \frac{m}{V}$

$\rho = \frac{0.673}{197.15*10^{-6}}$

$\rho = 3413.64kg/m^3$

PART B) To find the weight of the cube then we apply the ratio of

$W = mg$

$W = V\rho g$

$W = (216*10^{-6})(3413.64)(9.8)$

$W = 7.22N$

3 0

3 years ago

Consider the interference pattern produced by two parallel slits of width a and separation d, in which d = 3a. The slits are ill

laila [671]

Answer:

a) m =1 θ = sin⁻¹ λ / d, m = 2 θ = sin⁻¹ ( λ / 2d) , c) m = 3

Explanation:

a) In the interference phenomenon the maxima are given by the expression

d sin θ = m λ

the maximum for m = 1 is at the angle

θ = sin⁻¹ λ / d

the second maximum m = 2

θ = sin⁻¹ ( λ / 2d)

the third maximum m = 3

θ = sin⁻¹ ( λ / 3d)

the fourth maximum m = 4

θ = sin⁻¹ ( λ / 4d)

b) If we take into account the effect of diffraction, the intensity of the maximums is modulated by the envelope of the diffraction of each slit.

I = I₀ cos² (Ф) (sin x / x)²

Ф = π d sin θ /λ

x = pi a sin θ /λ

where a is the width of the slits

with the values of part a are introduced in the expression and we can calculate intensity of each maximum

c) The interference phenomenon gives us maximums of equal intensity and is modulated by the diffraction phenomenon that presents a minimum, when the interference reaches this minimum and is no longer present

maximum interference d sin θ = m λ

first diffraction minimum a sin θ = λ

we divide the two expressions

d / a = m

In our case

3a / a = m

m = 3

order three is no longer visible

7 0

2 years ago