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wolverine [178]
3 years ago
7

Which statement describes the atomic number?​

Physics
1 answer:
Vadim26 [7]3 years ago
6 0

Answer:The term atomic number, conventionally denoted by the symbol Z, indicates number of protons present in the nucleus of an atom, which is also equal to the number of electrons in an uncharged atom. The number of neutrons is represented by the neutron number (N)

Explanation:

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a body accelerates uniformly from rest at 2m/s^2 for 5 seconds. Calculate its averege velocity in this time​
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3 years ago
Suppose that we are designing a cardiac pacemaker circuit. The circuit is required to deliver pulses of 1ms duration to the hear
olchik [2.2K]

Answer:

Minimum capacitance = 200 μF

Explanation:

From image B attached, we can calculate the current flowing through the capacitors.

Thus;

Since V=IR; I = V/R = 5/500 = 0.01 A

Maximum charge in voltage is from 5V to 4.9V. Thus, each capacitor will have 2.5V. Hence, change in voltage(Δv) for each capacitor will be ; Δv = 0.05 V

So minimum capacitance will be determined from;

i(t) = C(dv/dt)

So, C = i(t)(Δt/Δv) = 0.01[0.001/0.05]

C = 0.01 x 0.0002 = 200 x 10^(-6) F = 200 μF

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3 years ago
In amplitude modulation, which wave property remains constant?
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The carrier frequency remains constant during amplitude modulation. :)
8 0
3 years ago
A luggage handler pulls a suitcase of mass 19.6 kg up a ramp inclined at an angle 24.0 ∘ above the horizontal by a force F⃗ of m
Dvinal [7]

(a) 638.4 J

The work done by a force is given by

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement of the object

\theta is the angle between the direction of the force and the displacement

Here we want to calculate the work done by the force F, of magnitude

F = 152 N

The displacement of the suitcase is

d = 4.20 m along the ramp

And the force is parallel to the displacement, so \theta=0^{\circ}. Therefore, the work done by this force is

W_F=(152)(4.2)(cos 0)=638.4 J

b) -328.2 J

The magnitude of the gravitational force is

W = mg

where

m = 19.6 kg is the mass of the suitcase

g=9.8 m/s^2 is the acceleration of gravity

Substituting,

W=(19.6)(9.8)=192.1 N

Again, the displacement is

d = 4.20 m

The gravitational force acts vertically downward, so the angle between the displacement and the force is

\theta= 90^{\circ} - \alpha = 90+24=114^{\circ}

Where \alpha = 24^{\circ} is the angle between the incline and the horizontal.

Therefore, the work done by gravity is

W_g=(192.1)(4.20)(cos 114^{\circ})=-328.2 J

c) 0

The magnitude of the normal force is equal to the component of the weight perpendicular to the ramp, therefore:

R=mg cos \alpha

And substituting

m = 19.6 kg

g = 9.8 m/s^2

\alpha=24^{\circ}

We find

R=(19.6)(9.8)(cos 24)=175.5 N

Now: the angle between the direction of the normal force and the displacement of the suitcase is 90 degrees:

\theta=90^{\circ}

Therefore, the work done by the normal force is

W_R=R d cos \theta =(175.4)(4.20)(cos 90)=0

d) -194.5 J

The magnitude of the force of friction is

F_f = \mu R

where

\mu = 0.264 is the coefficient of kinetic friction

R = 175.5 N is the normal force

Substituting,

F_f = (0.264)(175.5)=46.3 N

The displacement is still

d = 4.20 m

And the friction force points down along the slope, so the angle between the friction and the displacement is

\theta=180^{\circ}

Therefore, the work done by friction is

W_f = F_f d cos \theta =(46.3)(4.20)(cos 180)=-194.5 J

e) 115.7 J

The total work done on the suitcase is simply equal to the sum of the work done by each force,therefore:

W=W_F + W_g + W_R +W_f = 638.4 +(-328.2)+0+(-194.5)=115.7 J

f) 3.3 m/s

First of all, we have to find the work done by each force on the suitcase while it has travelled a distance of

d = 3.80 m

Using the same procedure as in part a-d, we find:

W_F=(152)(3.80)(cos 0)=577.6 J

W_g=(192.1)(3.80)(cos 114^{\circ})=-296.9 J

W_R=(175.4)(3.80)(cos 90)=0

W_f =(46.3)(3.80)(cos 180)=-175.9 J

So the total work done is

W=577.6+(-296.9)+0+(-175.9)=104.8 J

Now we can use the work-energy theorem to find the final speed of the suitcase: in fact, the total work done is equal to the gain in kinetic energy of the suitcase, therefore

W=\Delta K = K_f - K_i\\W=\frac{1}{2}mv^2\\v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(104.8)}{19.6}}=3.3 m/s

6 0
3 years ago
1)why do we get electric shock while holding a live wire barefooted and not when wearing rubber shoes?
katrin2010 [14]
Electricity is always going to take the path of least resistance to ground. The rubber in your shoes is not a conductor of electricity, therefore you are not completing the circuit and you don't get shocked. Your bare feet, on the other hand ARE conductors of electricity, so when you hold the wire, you complete the circuit and become the path of least resistance to ground... ZAP!
5 0
3 years ago
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