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ad-work [718]
3 years ago
10

Using the standard convention "positive is up", what is the sign of the acceleration of a rock you throw up (a) as it is going u

p, and (b) as it is coming back down?
Physics
1 answer:
Sonbull [250]3 years ago
7 0

Answer:

(a) Negative

(b) Positive

Explanation:

Here is a question on sign.

(a) When the rock is going up, it is working against gravity. In this case, the acceleration due to gravity will take a negative value. This is because gravity works or acts downwards

(b) When the rock is coming down, it is actually working with gravity. So in this case, the sign will ne positive

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Electricity flows from positive to negative
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The forces in (Figure 1) are acting on a 1.0 kg object.What is ax , the x -component of the object's acceleration
a_sh-v [17]

The x -component of the object's acceleration is 2 m/s².

<h3>What's the resultant force along x- direction?</h3>
  • Forces along x axis direction are as follows
  1. 4N along +x axis, so it's taken as +4 N
  2. 2N along -x axis , so it's taken as -2N.
  • Resultant force along x direction = 4N - 2N = 2 N which is along + ve x direction.

<h3>What's the acceleration along x axis direction?</h3>
  • As per Newton's second law, Force = mass × acceleration of the object
  • Force along x axis= mass × acceleration along x axis= 2N
  • Acceleration = 2/ mass = 2/1 = 2 m/s²

Thus, we can conclude that the acceleration along x axis is 2 m/s².

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: The forces in (Figure 1) are acting on a 1.0 kg object. What is ax, the x-component of the object's acceleration?

Learn more about the acceleration here:

brainly.com/question/460763

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2 years ago
What is the net force acting upon this object? *
AnnZ [28]

2 N Right

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A modern compact fluorescent lamp contains 1.4 mg of mercury (Hg). If each mercury atom in the lamp were to emit a single photon
Reika [66]

Answer:

A. 1.64 J

Explanation:

First of all, we need to find how many moles correspond to 1.4 mg of mercury. We have:

n=\frac{m}{M_m}

where

n is the number of moles

m = 1.4 mg = 0.0014 g is the mass of mercury

Mm = 200.6 g/mol is the molar mass of mercury

Substituting, we find

n=\frac{0.0014 g}{200.6 g/mol}=7.0\cdot 10^{-6} mol

Now we have to find the number of atoms contained in this sample of mercury, which is given by:

N=n N_A

where

n is the number of moles

N_A=6.022\cdot 10^{23} mol^{-1} is the Avogadro number

Substituting,

N=(7.0\cdot 10^{-6} mol)(6.022\cdot 10^{23} mol^{-1})=4.22\cdot 10^{18} atoms

The energy emitted by each atom (the energy of one photon) is

E_1 = \frac{hc}{\lambda}

where

h is the Planck constant

c is the speed of light

\lambda=508 nm=5.08\cdot 10^{-7}nm is the wavelength

Substituting,

E_1 = \frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{5.08\cdot 10^{-7} m}=3.92\cdot 10^{-19} J

And so, the total energy emitted by the sample is

E=nE_1 = (4.22\cdot 10^{18} )(3.92\cdot 10^{-19}J)=1.64 J

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