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2 years ago

it possible that the net kinetic energy for two objects be zero while the net momentum is zero? Explain.

1 answer:

4 0

Of course. That's what you have when both objects are at rest. I'm guessing that you left a word out of the question, and it actually says that the net kinetic energy is NOT zero. In that case, the answer is still 'yes', but you have to think about it for a second.

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A truck is traveling down a road with a 4-percent grade at a speed of 75 mi/h when its brakes are applied to slow it down to 22.

kvasek [131]

Answer:

3.964 s

Explanation:

Metric unit conversion:

1 miles = 1.6 km = 1600 m.

1 hour = 60 minutes = 3600 seconds

75 mph = 75 * 1600 / 3600 = 33.3 m/s

22.5 mph = 22.5 * 1600/3600 = 10 m/s

Let g = 9.81 m/s2

Friction is the product of coefficient and normal force, which equals to the gravity

$F_f = \mu N = \mu mg$

The deceleration caused by friction is friction divided by mass according to Newton 2nd law.

$a = F_f / m = \mu mg / m = \mu g = 0.6 *9.81 = 5.886 m/s^2$

So the time required to decelerate from 33.3 m/s to 10 m/s so the wheels don't slide, with the rate of 5.886 m/s2 is

$t = \frac{\Delta v}{a} = \frac{33.3 - 10}{5.886} = 3.964 s$

3 0

1 year ago

ou purchase a rectangular piece of metal that has dimen- sions 5.0 * 15.0 * 30.0 mm and mass 0.0158 kg. The seller tells you tha

Natalija [7]

Answer: 7022.2kg/m³, yes, I was cheated

Explanation:

Density of an object is defined as the ratio of the mass of the object to its volume. Mathematically;

Density = Mass/Volume

Note that the unit of both mass and volume must be standard unit.

Given mass = 0.0158kg

Dimension of the metal = 5mm×15mm×30mm

Note that 1mm = 0.001m

The volume of the metal will be

0.005×0.015×0.03

= 0.00000225m³

Density = 0.0158/0.00000225

Average density of the metal = 7022.2kg/m³

Since the standard density of Gold is 19,320kg/m³ and is higher than the density prescribed for me, it shows the I was cheated.

4 0

2 years ago

Three different planet-star systems, which are far apart from one another, are shown above. The masses of the planets are much l

alex41 [277]

a) 4F0

b) Speed of planet B is the same as speed of planet A

Speed of planet C is twice the speed of planet A

Explanation:

The magnitude of the gravitational force between two objects is given by the formula

$F=G\frac{m_1 m_2}{r^2}$

where

G is the gravitational constant

m1, m2 are the masses of the 2 objects

r is the separation between the objects

For the system planet A - Star A, we have:

$m_1=M_p\\m_2 = M_s\\r=R$

So the force is

$F_A=G\frac{M_p M_s}{R^2}=F_0$

For the system planet B - Star B, we have:

$m_1 = 4 M_p\\m_2 = M_s\\r=R$

So the force is

$F=G\frac{4M_p M_s}{R^2}=4F_0$

So, the magnitude of the gravitational force exerted on planet B by star B is 4F0.

For the system planet C - Star C, we have:

$m_1 = M_p\\m_2 = 4M_s\\r=R$

So the force is

$F=G\frac{M_p (4M_s)}{R^2}=4F_0$

So, the magnitude of the gravitational force exerted on planet C by star C is 4F0.

The gravitational force on the planet orbiting around the star is equal to the centripetal force, therefore we can write:

$G\frac{mM}{r^2}=m\frac{v^2}{r}$

where

m is the mass of the planet

M is the mass of the star

v is the tangential speed

We can re-arrange the equation solving for v, and we find an expression for the speed:

$v=\sqrt{\frac{GM}{r}}$

For System A,

$M=M_s\\r=R$

So the tangential speed is

$v_A=\sqrt{\frac{GM_s}{R}}$

For system B,

$M=M_s\\r=R$

So the tangential speed is

$v_B=\sqrt{\frac{GM_s}{R}}=v_A$

So, the speed of planet B is the same as planet A.

For system C,

$M=4M_s\\r=R$

So the tangential speed is

$v_C=\sqrt{\frac{G(4M_s)}{R}}=2(\sqrt{\frac{GM_s}{R}})=2v_A$

So, the speed of planet C is twice the speed of planet A.

3 0

2 years ago

Exercise 2.4.6: Suppose you wish to measure the friction a mass of 0.1 kg experiences as it slides along a floor (you wish to fi

JulijaS [17]

Answer:

b = 0.6487 kg / s

Explanation:

In an oscillatory motion, friction is proportional to speed,

fr = - b v

where b is the coefficient of friction

when solving the equation the angular velocity has the form

w² = k / m - (b / 2m)²

In this exercise we are given the angular velocity w = 1Hz, the mass of the body m = 0.1 kg, and the spring constant k = 5 N / m. Therefore we can disperse the coefficient of friction

let's call

w₀² = k / m

w² = w₀² - b² / 4m²

b² = (w₀² -w²) 4 m²

Let's find the angular velocities

w₀² = 5 / 0.1

w₀² = 50

w = 2π f

w = 2π 1

w = 6.2832 rad / s

we subtitute

b² = (50 - 6.2832²) 4 0.1²

b = √ 0.42086

b = 0.6487 kg / s

8 0

2 years ago

Car 1 goes around a level curve at a constant speed of 65 km/h . The curve is a circular arc with a radius of 95 m . Car 2 goes

Arte-miy333 [17]

Answer:

The radius of the curve that Car 2 travels on is 380 meters.

Explanation:

Speed of car 1, $v_1=65\ km/h$

Radius of the circular arc, $r_1=95\ m$

Car 2 has twice the speed of Car 1, $v_2=130\ km/h$

We need to find the radius of the curve that Car 2 travels on have to be in order for both cars to have the same centripetal acceleration. We know that the centripetal acceleration is given by :

$a=\dfrac{v^2}{r}$

According to given condition,

$\dfrac{v_1^2}{r_1}=\dfrac{v_2^2}{r_2}$

$\dfrac{65^2}{95}=\dfrac{130^2}{r_2}$

On solving we get :

$r_2=380\ m$

So, the radius of the curve that Car 2 travels on is 380 meters. Hence, this is the required solution.

4 0

2 years ago