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2 years ago

What is the stretch when you pull with a force of 2.3 N on a spring with a spring constant of 19 N/m?

Physics

1 answer:

Neporo4naja [7]2 years ago

8 0

Answer:

x=0.12 m

Explanation:

Use the equation F=kx and isolate x because that's what you're solving for (x=F/k). Plug in your given values for F and k and solve for x [x=(2.3 N)/(19 N/m)] and you get x=0.12 m.

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A light horizontal spring has a spring constant of 138 N/m. A 3.35 kg block is pressed against one end of the spring, compressin

BARSIC [14]

Answer:

$U_k = 0.113$

Explanation:

using the law of the conservation of energy:

$E_i -E_f=W_f$

$\frac{1}{2}Kx^2=NU_kd$

where K is the spring constant, x is the spring compression, N is the normal force of the block, $U_k$ is the coefficiet of kinetic friction and d is the distance.

Also, by laws of newton, N is calculated by:

N = mg

N = 3.35 kg * 9.81 m/s

N = 32.8635

So, Replacing values on the first equation, we get:

$\frac{1}{2}(138)(0.123)^2= (32.8635)U_k(0.281m)$

solving for $U_k$ :

$U_k = 0.113$

8 0

3 years ago

which statement is a Hypothesis? A: Most of the earthworms moved to the shaded area during experiment. B: If an earthworm is giv

True [87]

Answer:

Explanation:

It's saying what you think

3 0

3 years ago

A spherical capacitor contains a charge of 3.00 nC when connected to a potential difference of 230 V. If its plates are separate

Assoli18 [71]

Answer:

Part(a): the capacitance is 0.013 nF.

Part(b): the radius of the inner sphere is 3.1 cm.

Part(c): the electric field just outside the surface of inner sphere is $\bf{2.81 \times 10^{4}~n~C^{-1}}$ .

Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and ' $\epsilon_{0}$ ' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

$C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Part(a):

Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.

So the capacitance (C) of the shell is

$C &=& \dfrac{Q}{V} = \dfrac{3 \times 10^{-90}~C}{230~V} = 1.3 \times 10^{-11}~F = 0.013~nF$

Part(b):

Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

$&& \dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{C/4 \pi \epsilon_{0}} = \dfrac{1}{0.043} + \dfrac{1}{1.3 \times 10^{-11} \times 9 \times 10^{9}} = 31.79\\&or,& a = \dfrac{1}{31.79}~m = 0.031~m = 3.1~cm$

Part(c):

If we apply Gauss' law of electrostatics, then

$&& E~4 \pi a^{2} = \dfrac{Q}{\epsilon_{0}}\\&or,& E = \dfrac{Q}{4 \pi \epsilon_{0}a^{2}}\\&or,& E = \dfrac{3 \times 10^{-9} \times 9 \times 10^{9}}{0.031^{2}}~N~C^{-1}\\&or,& E = 2.81 \times 10^{4}~N~C^{-1}$

3 0

3 years ago

According to Newton’s first law of motion, what will an object in motion do when no external force acts on it?

KonstantinChe [14]

By definition, we have to:

Newton's first law states that any object will remain in a state of rest or with a uniform rectilinear motion unless an external force acts on it.

Therefore, according to the first law of Newton, if the object is already in motion and has no force acting on it then, it will remain with a uniform rectilinear motion.

Answer:

The object will remain with a uniform rectilinear movement when the external force does not act on it.

4 0

3 years ago

Read 2 more answers

Assuming that Albertine's mass is 60.0 kg , for what value of μk, the coefficient of kinetic friction between the chair and the

makvit [3.9K]

Answer:

$\mu_k=0.101$

Explanation:

It is given that,

Mass of Albertine, m = 60 kg

It can be assumed, the spring constant of the spring, k = 95 N/m

Compression in the spring, x = 5 m

A glass sits 19.8 m from her outstretched foot, h = 19.8 m

When she just reach the glass without knocking it over, a force of friction will also act on it. Using the conservation of energy for the spring mass system such that,

$\dfrac{1}{2}kx^2=\mu_k mgh$

$\mu_k=\dfrac{kx^2}{2mgh}$

$\mu_k=\dfrac{95\times (5)^2}{2\times 60\times 9.8\times 19.8}$

$\mu_k=0.101$

So, the coefficient of kinetic friction between the chair and the waxed floor is 0.101. Hence, this is the required solution.

3 0

3 years ago