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2 years ago

How can the direction of a tensional force be changed without diminishing the force?

1 answer:

4 0

Given what we know, we can confirm that the tensional force of a system can in theory be changed without diminishing its force through the use of an ideal pulley.

<h3>What is an ideal pulley?</h3>

A pulley is a small wheel through which a string or chain is run.
These are used in order to change the direction of a force.
An ideal pulley would be one in which there is no friction and the pulley itself would have no mass.
Therefore, the force would be able to change directions without giving part of its force to the pulley system.

Therefore, we can confirm that the only known way to change the direction of a force without diminishing its value would be through the use of a frictionless and massless pulley system otherwise known as an ideal pulley.

To learn more about Friction visit:

brainly.com/question/13357196?referrer=searchResults

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A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separation between the plates is 2.30 mm. If the separa

Elanso [62]

Answer:

Explanation:

plate separation = 2.3 x 10⁻³ m

capacity C₁ = ε A / d

= ε A / 2.3 x 10⁻³

C₂ = ε A / 1.15 x 10⁻³

$\frac{C_2}{C_1}$ = $\frac{2.3}{1.15}$

a ) when charge remains constant

energy = $\frac{q^2}{2C}$

q is charge and C is capacity

energy stored initially E₁= $\frac{q^2}{2C_1}$

energy stored finally E₂ = $\frac{q^2}{2C_2}$

$\frac{E_1}{E_2} = \frac{C_2}{C_1}$ = $\frac{2.3}{1.15}$

$E_2$ = $\frac{1.15}{2.3 } \times E_1$

= $\frac{1.15}{2.3 } \times 8.38$

= 4.19 J

b )

In this case potential diff remains constant

energy of capacitor = 1/2 C V²

energy is proportional to capacity as V is constant .

$\frac{E_2}{E_1} = \frac{C_2}{C_1}$

$\frac{E_2}{8.38} = \frac{2.3}{1.15}$

$E_2$ = 16.76 .

8 0

3 years ago

A 2 kg mass is free falling in the negative Y direction when a 10 N force is exerted in the minus X direction. What is the accel

lara31 [8.8K]

Answer:

The mass's acceleration is 5 m/s^2 in the minus X direction and 9,8 m/s^2 in the minus Y direction.

Explanation:

By applying the second Newton's law in the X and Y direction we found that in the minus X direction an external force of 10 N is exerted, while in the minus Y direction the gravity acceleration is acting:

X-direction balance force: $-10 [N] = m.ax$

Y-direction balance force: $-m*9,8 \frac{m}{s^2} = m.ay$

Where ax and ay are the components of the respective acceleration and m is the mass. By solving for each acceleration:

$ax=(-10 [N]) / m$

$ay=-m*9,8\frac{m}{s^2} / m$

Note that for the second equation above the mass is cancelled and, the Y direction acceleration is minus the gravity acceleration:

$ay=-9,8\frac{m}{s^2}$

For the x component aceleration we must replace the Newton unit:

$N =\frac{kg.m}{s^2}$

$ax= -10 \frac{kg.m}{s^2} / (2 kg)$

$ax= - 5 \frac{m}{s^2}$

6 0

3 years ago

A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizo

jeka57 [31]

Answer:

a. A = 0.735 m

b. T = 0.73 s

c. ΔE = 120 J decrease

d. The missing energy has turned into interned energy in the completely inelastic collision

Explanation:

4 kg * 10 m /s + 6 kg * 0 m/s = 10 kg* vmax

vmax = 4.0 m/s

¹/₂ * m * v²max = ¹/₂ * k * A²

m * v² = k * A² ⇒ 10 kg * 4 m/s = 100 N/m * A²

A = √1.6 m ² = 1.26 m

At = 2.0 m - 1.26 m = 0.735 m

T = 2π * √m / k ⇒ T = 2π * √4.0 kg / 100 N/m = 1.26 s

T = 2π *√ 10 / 100 *s² = 1.99 s

T = 1.99 s -1.26 s = 0.73 s

E = ¹/₂ * m * v²max =

E₁ = ¹/₂ * 4.0 kg * 10² m/s = 200 J

E₂ = ¹/₂ * 10 * 4² = 80 J

200 J - 80 J = 120 J decrease

The missing energy has turned into interned energy in the completely inelastic collision

3 0

3 years ago

How much work was done on a 45 kg weight when lifted 1.4 meters in the air?

BabaBlast [244]

Explanation:

w = f x d

45 x 1.4 = 630j

to get newton's do 45 x gravitation field strength

8 0

3 years ago

Bowling balls are roughly the same size, but come in a variety of weights. Given its official radius of roughly 0.110 m, calcula

velikii [3]

Answer:

6.1328 kg

60.16284 N

Explanation:

r = Radius of ball = 0.11 m

$\rho$ = Density of fluid = $1.1\times 10^3\ kg/m^3$ (Assumed)

g = Acceleration due to gravity = 9.81 m/s²

m = Mass of ball

V = Volume of ball = $\frac{4}{3}\pi r^3$

The weight of the bowling ball will balance the buouyant force

$W=F_b\\\Rightarrow mg=V\rho g\\\Rightarrow m=\frac{V\rho g}{g}\\\Rightarrow m=V\rho\\\Rightarrow m=\frac{4}{3}\pi 0.11^3\times 1.1\times 10^3\\\Rightarrow m=6.1328\ kg$

The mass of the bowling ball will be 6.1328 kg

Weight will be $6.1328\times 9.81=60.16284\ N$

5 0

3 years ago