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Luden [163]
2 years ago
15

Suppose the truck that’s transporting the box In Example 6.10 (p. 150) is driving at a constant speed and then brakes and slows

at a constant acceleration. While coming to a stop, the driver looks in the rear-view mirror and notices that the box is not slipping. In what direction is the frictional force acting on the box?
Physics
1 answer:
Scorpion4ik [409]2 years ago
5 0

Answer:

Friction acts in the opposite direction to the motion of the truck and box.

Explanation:

Let's first review the problem.

A moving truck applies the brakes, and a box on it does not slip.

Now when the truck is applying brakes, only it itself is being slowed down. Since the box is slowing down with the truck, we can conclude that it is friction that slows it down.

The box in the question tries to maintains its velocity forward when the brakes are applied. We can think of this as the box exerting a positive force relative to the truck when the brakes are applied. When we imagine this, we can also figure out where the static friction will act to stop this positive force. Friction will act in the negative direction. Or in other words, friction will act in the opposite direction to the motion of the truck and box. This explains why the box slows down with the truck, as friction acts to stop its motion.

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The discovery and characterization of cathode rays was important in the development of the atomic theory because
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Answer:

All matter contained electrons

Explanation:

The discovery and characterization of cathode ray suggested that it was a subatomic particle and cathode ray ( electron) was the first discovered. It immensely became the strong explanatory tool for chemical bond. This can be attributed to  the the ease with which electron move from one atom to the other.  

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Karyotypes are done by matching up _____________________________ so that they are paired up. Question 11 options:
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Homologous Chromosomes

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What is the internal energy (to the nearest joule) of 10 moles of Oxygen at 100 K?
kkurt [141]

Answer:

U = 12,205.5 J

Explanation:

In order to calculate the internal energy of an ideal gas, you take into account the following formula:

U=\frac{3}{2}nRT        (1)

U: internal energy

R: ideal gas constant = 8.135 J(mol.K)

n: number of moles = 10 mol

T: temperature of the gas = 100K

You replace the values of the parameters in the equation (1):

U=\frac{3}{2}(10mol)(8.135\frac{J}{mol.K})(100K)=12,205.5J

The total internal energy of 10 mol of Oxygen at 100K is 12,205.5 J

6 0
2 years ago
Slick Willy is in traffic court (again) contesting a $50.00 ticket for running a red light. "You see, your Honor, as I was appro
Masteriza [31]

Answer:

61578948 m/s

Explanation:

λ_{actual} = λ_{observed} \frac{c+v_{o}}{c}

687 = 570 (\frac{3 * 10^{8} +v_{o} }{3 * 10^{8}} )

v_{o} = 61578948 m/s

So Slick Willy was travelling at a speed of 61578948 m/s to observe this.

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An object moving on the x axis with a constant acceleration increases its x coordinate by 82.9 m in a time of 2.51 s and has a v
Aneli [31]

We are given: Final velocity (v_f)=20 m/s .

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distance s = 82.9 m.

We know, equation of motion

v_f = v_i + at.

Let us plug values of final velocity, and time in above equation.

20=v_i+a(2.51)

20=v_i+2.51a

Subtracting 2.51a from both sides, we get

20-2.51a=v_i  -----------equation(1)

Using another equation of motion

v_f-v_i=2as

Plugging values of vi =20-2.51a, t=2.51 and distnace s=82.9 in this equation.

We get,

20-(20-2.51a)=2*a(82.90)

Now, we need to solve it for a.

20-20+2.51a=165.8a.

-163.29a=0

a=0.

So, the acceleration would be 0 m/s^2.


5 0
2 years ago
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