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3 years ago

13

A 0.54 kg air hockey puck is initially at rest. What will it's kinetic energy energy be after a net force of 0.56 N acts on it f

or a distance of 0.84m?

1 answer:

Rufina [12.5K]3 years ago

8 0

Answer:

Kf = 470 mJ

Explanation:

According the work-energy theorem, the change in the kinetic energy of one object, is equal to the net work done on it.
Since the puck is initially at rest, the change is kinetic energy is just the final kinetic energy of the puck.
Assuming that the net force is horizontal, and causes a horizontal displacement also, we can find the net work on the puck as follows:

$W_{net} = F_{net} * \Delta X = 0.56 N * 0.84 m = 0.47 J = 470 mJ (1)$

As we have already said, (1) is equal to the final kinetic energy of the puck:
⇒ Kf = 470 mJ (2)

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How much mass should be attached to a vertical ideal spring having a spring constant (force constant) of 39.5 n/m so that it wil

mrs_skeptik [129]

The frequency of a simple harmonic oscillator such as a spring-mass system is given by
$f= \frac{1}{2 \pi} \sqrt{ \frac{k}{m} }$
where
k is the spring constant
m is the mass attached to the spring.

Re-arranging the formula, we get:
$m= \frac{k}{4 \pi^2 f^2}$
and since we know the constant of the spring:
$k=39.5 N/m$
and the frequency of oscillation:
f=1.00 Hz
we can find the value of the mass attached to it:
$m= \frac{39.5 Hz}{4 \pi^2 (1.00 Hz)^2} = 1.00 kg$

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3 years ago

Two workers are sliding 300 kg crate across the floor. One worker pushes forward on the crate with a force of 400 N while the ot

almond37 [142]

Answer:

The kinetic coefficient of friction of the crate is 0.235.

Explanation:

As a first step, we need to construct a free body diagram for the crate, which is included below as attachment. Let supposed that forces exerted on the crate by both workers are in the positive direction. According to the Newton's First Law, a body is unable to change its state of motion when it is at rest or moves uniformly (at constant velocity). In consequence, magnitud of friction force must be equal to the sum of the two external forces. The equations of equilibrium of the crate are:

$\Sigma F_{x} = P+T-\mu_{k}\cdot N = 0$ (Ec. 1)

$\Sigma F_{y} = N - W = 0$ (Ec. 2)

Where:

$P$ - Pushing force, measured in newtons.

$T$ - Tension, measured in newtons.

$\mu_{k}$ - Coefficient of kinetic friction, dimensionless.

$N$ - Normal force, measured in newtons.

$W$ - Weight of the crate, measured in newtons.

The system of equations is now reduced by algebraic means:

$P+T -\mu_{k}\cdot W = 0$

And we finally clear the coefficient of kinetic friction and apply the definition of weight:

$\mu_{k} =\frac{P+T}{m\cdot g}$

If we know that $P = 400\,N$ , $T = 290\,N$ , $m = 300\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , then:

$\mu_{k} = \frac{400\,N+290\,N}{(300\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$\mu_{k} = 0.235$

The kinetic coefficient of friction of the crate is 0.235.

5 0

3 years ago

Calculate the magnitude of the electric field inside the solid at a distance of 9.50 cm from the center of the cavity. Express y

WITCHER [35]

Question:

A point charge of -2.14uC is located in the center of a spherical cavity of radius 6.55cm inside an insulating spherical charged solid. The charge density in the solid is 7.35×10−4 C/m^3.

a) Calculate the magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity.

Express your answer using two significant figures.

Answer:

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity $3.65\times 10^5N/C$

Explanation:

A point charge ,q = $-2.14\times 10^{-6} C$ is located in the center of a spherical cavity of radius , $r =6.55\times 10^{-2}$ m inside an insulating spherical charged solid.

The charge density in the solid , d = $7.35 \times 10^{-4}C/m^3.$

Distance from the center of the cavity,R = $9.5\times 10^{-2 }m$

Volume of shell of charge= V = $(\frac{4\pi}{3})[ R^3 - r^3 ]$

Charge on the shell ,Q = $V \times d'$

$Q =(\frac{4\pi}{3})[ R^3 - r^3 ] \times d$

$Q = 4.1888*\times 10^{-4 }[8.57375 - 2.81011 ]\times 7.35\times 10^{-4}$

$Q = 4.1888\times 10^{-4} [5.76364 ] \times 7.35 \times 10^{-4}$

$Q =2.4143 \times 10^{-4} \times 7.35 \times 10^ { -4}$

$Q =1.7745 \times 10^{-6 }C$

Electric field at $9.5\times 10^{-2}$ m due to shell $E1 = \frac{k Q}{R^2}$

E1 = $\frac{ 9 \times 10^9\times 1.7745\times 10^{-6 }}{ 90.25\times 10^{-4}}$

$E1 =1.769\times 10^6 N/C$

Electric field at $9.5\times 10^{-2}$ due to 'q' at center $E2 = \frac{kq}{R^2}$

E2 = $\frac{ - 9 \times 10^9\times 2.14\times 10^{-6 }}{ 90.25\times 10^{-4}}$

$E2 =2.134\times 10^6 N/C$

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity

= E2- E1

$=[ 2.134 - 1.769 ]\times 10^6$

$= 3.65\times 10^5 N/C$

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