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2 years ago

What is the momentum of a 37-kg person riding south on an 18-kg bicycle at 1.2 m/s

Physics

1 answer:

Arlecino [84]2 years ago

3 0

The equation of momentum is:

$p=mv$

Where p is the momentum, m is the mass and v is the velocity. In this case the mass of the person and the mass of the bike so:

$m_{total}=mass_{person}+mass_{bike}=37kg+18kg=55kg$

And the velocity of the person riding the bike is 1.2 m/s and so:

$p=(55kg)(1.2m/s)=66 \frac{kgm}{s}$

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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

2 years ago

A solid sphere has a radius of 0.200 m and a mass of 150.0 kg. how much work is required to get the sphere rolling with an angul

Allisa [31]

Here in this case we can use work energy theorem

As per work energy theorem

Work done by all forces = Change in kinetic Energy of the object

Total kinetic energy of the solid sphere is ZERO initially as it is given at rest.

Final total kinetic energy is sum of rotational kinetic energy and translational kinetic energy

$KE = \frac{1}{2}Iw^2 +\frac{1}{2} mv^2$

also we know that

$I = \frac{2}{5}mR^2$

$w= \frac{v}{R}$

Now kinetic energy is given by

$KE = \frac{1}{2}(\frac{2}{5}mR^2)w^2 +\frac{1}{2} m(Rw)^2$

$KE = \frac{1}{5}mR^2w^2 +\frac{1}{2} mR^2w^2$

$KE = \frac{7}{10}mR^2w^2$

$KE = \frac{7}{10}*150*(0.200)^2(50)^2$

$KE = 10500 J$

Now by work energy theorem

Work done = 10500 - 0 = 10500 J

So in the above case work done on sphere is 10500 J

7 0

2 years ago

A 5.0 kg object moving at 5.0 m/s. KE = mv2 times 1/2

steposvetlana [31]

Answer: KE = 62.5J

Explanation:

Given that

Mass of object = 5kg

kinetic energy KE = ?

velocity of object = 5m/s

Since kinetic energy is the energy possessed by a moving object, and it depends on the mass (m) of the object and the velocity (v) by which it moves. Therefore, the object has kinetic energy.

i.e K.E = 1/2mv^2

KE = 1/2 x 5kg x (5m/s)^2

KE = 0.5 x 5 x 25

KE = 62.5J

Thus, the object has 62.5 joules of kinetic energy.

5 0

2 years ago

A 8.2-V battery is connected in series with a 38-mH inductor, a 150-Î© resistor, and an open switch.A 8.2-V battery is connected

tigry1 [53]

Answer:

(A). The current in the circuit is 19.25 mA.

(B). The store energy in the inductor is 7.04 μJ.

Explanation:

Given that,

Voltage = 8.2 V

Inductor = 38 mH

Resistance = 150 Ω

Time t = 0.110 ms

The battery has negligible internal resistance, so that the total resistance in the circuit is 150 ohms. Then use this equation for current at time t in terms of inductance

We need to calculate the current

Using formula of current

$I(t)=\dfrac{V}{R}\times(1-e^{-t\times\dfrac{R}{L}})$

Put the value into the formula

$I(t)=\dfrac{8.2}{150}\times(1-e^{-0.110\times10^{-3}\times\dfrac{150}{38\times10^{-3}}})$

$I(t)=0.01925\ A$

$I(t) = 19.25\ mA$

(B). We need to calculate the store energy in the inductor

Using formula of energy

$E=\dfrac{1}{2}LI^2$

Put the value into the formula

$E=\dfrac{1}{2}\times38\times10^{-3}\times(0.01925)^2$

$E=7.04\times10^{-6}\ J$

{tex]E=7.04\ \mu J[/tex]

Hence, (A). The current in the circuit is 19.25 mA.

(B). The store energy in the inductor is 7.04 μJ.

8 0

2 years ago

A simple pendulum consisting of a bob of mass m attached to a string of length L swings with a period T. If the pendulum is take

Juli2301 [7.4K]

To solve this problem we will use the definition of the period in a simple pendulum, which warns that it is dependent on its length and gravity as follows:

$T =2\pi \sqrt{\frac{L}{g}}$

Here,

L = Length

g = Acceleration due to gravity

We can realize that $2 \pi$ is a constant so it is proportional to the square root of its length over its gravity,

$T \propto \sqrt{\frac{L}{g}}$

Since the body is in constant free fall, that is, a point where gravity tends to be zero:

$g \rightarrow 0 \Rightarrow T \rightarrow \infty$

The value of the period will tend to infinity. This indicates that the pendulum will no longer oscillate because both the pendulum and the point to which it is attached are in free fall.

5 0

2 years ago