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3 years ago

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You have to use 4 N of force to lift a 36 N object with a lever. The lever has an effort arm of 45 m. What is the load arm lengt

h?

1 answer:

Ad libitum [116K]3 years ago

3 0

Answer:

times the numbers

Explanation:

as that is the formula

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An unstrained horizontal spring has a length of 0.39 m and a spring constant of 350 N/m. Two small charged objects are attached

Vera_Pavlovna [14]

Answer:

A) The possible algebraic signs will either be both positive (+) or both negative (-) charged since the 2 objects are repelling each other to stretch the string.

B) Magnitude of charges = 1.206 × 10^(-6) C

Explanation:

We are given;

Spring constant;k = 350 N/m

Spring length;L = 0.39 m

Stretched length of spring;x = 0.022 m

A) The spring stretches by 0.022m. Therefore, the total force is (350 × 0.022) N = 7.7N. The charged objects will either be both positive (+) or both negative (-) charged since they are repelling each other to stretch the string.

B) Force (F) required to stretch spring is given by the formula;

F = kx

Thus:

F = (350 × 0.022)

F = 7.7 N

Now, if we assume point charges, then the distance (r) between them will be given as:

r = (0.39 + 0.022) = 0.412 m

Coulomb's Law has a formula:

F = k(q1×q2)/r²

where k is coulomb's constant = 8.99 × 10^(9) Nm²/C²

Making q1 × q2 the subject, we have;

(q1 × q2) = Fr²/k = 7.7 × 0.412²/(8.99 × 10^(9))

(q1 × q2) = 14.54 × 10^(-11) C

We are told that both charges are equal, thus; |q1| = |q2|

So;

q = √(14.54 × 10^(-11)) = 1.206 × 10^(-6) C

6 0

3 years ago

What should you do before inserting a wire into an outlet box?

Westkost [7]

Answer:

I think D it could maybe B

7 0

2 years ago

Read 2 more answers

A circular loop of wire has radius of 9.50 cmcm. A sinusoidal electromagnetic plane wave traveling in air passes through the loo

m_a_m_a [10]

Answer:

The induced emf is $\epsilon = 0.1041 \ V$

Explanation:

From the question we are told that

The radius of the circular loop is $r = 9.50 \ cm = 0.095 \ m$

The intensity of the wave is $I = 0.0215 \ W/m^2$

The wavelength is $\lambda = 6.90\ m$

Generally the intensity is mathematically represented as

$I = \frac{ c * B^2 }{ 2 * \mu_o }$

Here $\mu_o$ is the permeability of free space with value

$\mu_o = 4 \pi *10^{-7} N/A^2$

B is the magnetic field which can be mathematically represented from the equation as

$B = \sqrt{ \frac{ 2 * \mu_o * I }{ c} }$

substituting values

$B = \sqrt{ \frac{ 2 * 4\pi *10^{-7} * 0.0215 }{ 3.0*10^{8}} }$

$B = 1.342 *10^{-8} \ T$

The area is mathematically represented as

$A = \pi r^2$

substituting values

$A = 3.142 * (0.095)^2$

$A = 0.0284$

The angular velocity is mathematically represented as

$w = 2 * \pi * \frac{c}{\lambda }$

substituting values

$w = 2 * 3.142 * \frac{3.0*10^{8}}{ 6.90 }$

$w = 2.732 *10^{8} rad \ s^{-1}$

Generally the induced emf is mathematically represented as

$\epsilon = N * B * A * w * sin (wt )$

At maximum induced emf $sin (wt) = 1$

So

$\epsilon = N * B * A * w$

substituting values

$\epsilon = 1 * 1.342 *10^{-8} * 0.0284 *2.732 *10^{8}$

$\epsilon = 0.1041 \ V$

7 0

3 years ago

Spiderman, whose mass is 74.0 kg, is dangling on the free end of a 11.0-m-long rope, the other end of which is fixed to a tree l

Anestetic [448]

Answer:

W = -1844.513 J

Explanation:

GIVEN DATA:

mass of spider man is m 74 kg

vertical displacement if spider is 11 m

final displacement = 11 cos 60.6 = - 6.753 m

change in displacement is = -6.753 - (-11) = 4.25 m

gravity force act on spiderman is f = mg = 74 × 9.8 = 725.2 N

work done by gravity is $W = F \delta r cos\theta$

$W = 725.2 \times 4.25 \times cos 180$

where 180 is the angle between spiderman weight and displacement

W = -1844.513 J

7 0

3 years ago

How do you think overpumping groudwater is related to the formation of sinkholes?

vladimir2022 [97]

Ground water keept the ground at a stable level when it is gone the cavern it was in has no support and is at risk of callaps

3 0

3 years ago