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3 years ago

The thrust from a car is 590 N. The air resistance and friction together total 680 N. What happens to the speed of the car?

1 answer:

5 0

The net force in the direction the car is traveling (forward) is

(590 - 680) = -90N = 90 newtons backwards.

The car accelerates backwards at the rate of (90/its mass) m/sec-squared.

Another way to look at it: The car's forward speed decreases at that rate.

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If the atomic mass of Sodium-18 is 18.02597 u, what is the binding energy?

Klio2033 [76]

Answer:

<h3>The binding energy of sodium Na=<em>5.407791×10⁹J</em></h3>

Explanation:

<h3>Greetings !</h3>

Binding energy, amount of energy required to separate a particle from a system of particles or to disperse all the particles of the system. Binding energy is especially applicable to subatomic particles in atomic nuclei, to electrons bound to nuclei in atoms, and to atoms and ions bound together in crystals.

<h2>Formula : Eb=(Δm)c²</h2><h3>where:Eb= binding energy</h3><h3> .Δm= mass defect(kg)</h3><h3> c= speed of light 3.00×10⁸ms¯¹</h3><h2 /><h3><u>Given</u><u> </u><u>values</u></h3>

m= 18.02597
c=3.00×10⁸ms¯¹

<h3><u>required </u><u>value</u></h3>

Eb=?

<h3><u>Solution:</u></h3>

Eb=(Δm)c²
Eb=(18.02597)*(3.00*10⁸ms¯¹
Eb=5.407791*10⁹J

8 0

1 year ago

numerical question : what is the required heat to raise the temperature of 2 kg parrafin by 10 Celsius if 44000 joules is requir

kirill115 [55]

Answer:

The heat energy required is, E = 2200 J

Explanation:

Given,

The mass of paraffin, m = 2 Kg

The energy required to raise the temperature of the paraffin by 200° C = 44000 J

Then the heat energy required to raise the temperature of the paraffin by 10° C is given by,

Since 44000 J raises temperature by 200° C, then

E = 44000 J / 20

= 2200 J

Hence, the energy required to raise the temperature of the paraffin by 10° C is, E = 2200 J

8 0

3 years ago

A transformer consists of a 500 turn primary coil and a 2000-turnsecondary coil. If the current in the secondary is 3.0A, what i

Neporo4naja [7]

Answer:

The correct solution will be "12.0 A".

Explanation:

The given values are:

$N_p= 500 \ turn$

$N_s= 200 \ turn$

$I_s= 3.0 \ A$

By using the transformer formula, we get

⇒ $\frac{N_p}{N_s} =\frac{I_s}{I_p}$

⇒ $I_p = I_s\times \frac{N_s}{N_p}$

On substituting the given values, we get

⇒ $=3.0 \ A\times \frac{2000}{500}$

⇒ $=12.0 \ A$

8 0

3 years ago

A cat’s crinkle ball toy of mass 15g is thrown straight up with an initial speed of 3m/s . Assume in this problem that air drag

Readme [11.4K]

Answer:

(a)0.0675 J

(b)0.0675 J

(d)0.0675 J

(e)-0.0675 J

(f)0.459 m

Explanation:

15g = 0.015 kg

(a) Kinetic energy as it leaves the hand

$E_k = \frac{mv^2}{2} = \frac{0.015*3^2}{2} = 0.0675 J$

(b) By the law of energy conservation, the work done by gravitational energy as it rises to its peak is the same as the kinetic energy as the ball leave the hand, which is 0.0675 J

(d) The gravitational energy at peak point would also be the same as 0.0675J

(e) In this case as the reference point is reversed, we would have to negate the original potential energy. So the potential energy as the ball leaves hand is -0.0675J

(f) Since at the maximum height the ball has potential energy of 0.0675J. This means:

mgh = 0.0675

0.015*9.81h = 0.0675

h = 0.459 m

The ball would reach a maximum height of 0.459 m

4 0

3 years ago

Calculate a pendulum's frequency of oscillation (in Hz) if the pendulum completes one cycle in 0.5 s.

Marina86 [1]

Time taken to complete one oscillation for a pendulum is Time Period, T = 0.5 s
Frequency of the pendulum oscillation = 1 / Time Period => f = 1 / T = 1 / 0.5
Frequency f = 2 Hz

3 0

3 years ago