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Mariulka [41]
3 years ago
15

If a car can go from 0 to 60 mi/hr in 8.0 seconds, what would be its final speed after 5.0 seconds if its starting speed were 50

mi/hr?
Physics
2 answers:
Harman [31]3 years ago
6 0
a= \frac{V}{t} \ \ \ \Rightarrow\ \ a= \frac{60\ [mi/hr]}{8\ [s]} \\ \\the\ final\ speed\ after\ 5.0\ seconds\\ \\V=50\ [mi/hr]+5\ [s]\ \cdot \frac{60\ [mi/hr]}{8\ [s]}=50\ [mi/hr] + \frac{5}{8} \cdot 60\ [mi/hr]=\\ \\=50\ [mi/hr]+37.5\ [mi/hr]=87.5\ [mi/hr]
Lady_Fox [76]3 years ago
5 0
v_1=60\ \frac{mi}{h}\\\\t_1=8.0\ s=\frac{8}{3600}\ h=\frac{1}{450}\ h\\--------------\\a=\frac{v}{t}\\\\a=\frac{60}{\frac{1}{450}}\ \frac{mi}{h^2}=27000\ \frac{mi}{h^2}\\--------------\\v_o=50\ \frac{mi}{h}\\\\t=5.0\ s=\frac{5}{3600}\ h=\frac{1}{720}\ h\\--------------

v=v_o+at\\\\\\v=50\ \frac{mi}{h}+27000\ \frac{mi}{h^2}\cdot\frac{1}{720}\ h=50\ \frac{mi}{h}+37.5\ \frac{mi}{h}=87.5\ \frac{mi}{h}\leftarrow Answer

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ddd [48]

Answer:

1977.696 J

Explanation:

Given;

Weight of the box = 28.0 kg

Force applied by the boy = 230 N

angle between the horizontal and the force = 35°

Therefore,

the horizontal component of the force = 230 × cosθ

= 230 × cos 35°

= 188.405 N

Coefficient of kinetic friction, μ = 0.24

Force by friction, f = μN

here,

N = Normal force = Mass × acceleration due to gravity

or

N = 28 × 9.81 = 274.68 N

therefore,

f = 0.24 × 274.68

or

f = 65.9232 N

Now,

work done by the boy, W₁ = 188.405 N × Displacement  

= 188.405 N × 30

= 5652.15 J

and,

the

work done by the friction, W₂ = - 65.9232 N × Displacement  

= - 65.9232 N × 30 m

= - 1977.696 J

[ since the friction force acts opposite to the direction of motion, therefore the workdone will be negative]

8 0
3 years ago
An ideal solenoid having a coil density of 5000 turns per meter is 10 cm long and carries a current of 4.0
Diano4ka-milaya [45]
The rule that is used to get the strength of magnetic field at the center of solenoid (B) is:
B = <span>µ x n x I where:
</span>µ is the permeability of the medium where the solenoid is based. In this problem, the medium is air which means that µ = <span>µ </span><span>o = 4 pi x 10^-7 Tm/A
</span>I is the current passing (I = 4 amperes)
n is the number of turns per unit length (5000 turns)

Substituting in the mentioned equation, we find that:
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Answer:

(e) no

(a) Rs = R' + R'' + R'''

(b) 1/Rp = 1/R' + 1/R'' + 1/R'''

Explanation:

11 e)

Practically it is not possible to give limitless charge to a conductor. It depends to the number of valence electrons.

(a) When the three resistances R'. R'' and R''' is in series combination.

Let they are connected to the voltage V and the current in each resistance is I.

According to Ohm's law

Voltage across R', V' = I R'

Voltage across R'', V'' = I R''

Voltage across R''', V''' = I R'''

So, let the equivalent resistance is Rs.

I Rs = I R' + I R'' + I R'''

Rs = R' + R'' + R'''

(b)

When the three resistances R'. R'' and R''' is in parallel combination.

Let they are connected to the voltage V and the current in each resistance is I', I''. I'''.

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Current in R''', I''' = V/R'''

The equivalent resistance is Rp.

V/Rp = V/R' + V/ R'' + V/R'''

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Answer:

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So, the liquid phase will be the one that exhibits this behavior, as the specific heat of liquid water (4.184 J/gºC) is the highest among the three phases.

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