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Kipish [7]
3 years ago
8

A man does 4,475 J of work in the process of pushing his 2.40 103 kg truck from rest to a speed of v, over a distance of 26.5 m.

Neglecting friction between truck and road, determine the following.
(a) the speed
(b) the horizontal force exerted on the truck
Physics
1 answer:
velikii [3]3 years ago
7 0

Answer:

1.9311 m/s

168.86792 N

Explanation:

v = Final velocity

m = Mass of truck = 2.4\times 10^3\ kg

s = Displacement = 26.5 m

Work done is given by

W=\dfrac{1}{2}m(v^2-u^2)\\\Rightarrow W=\dfrac{1}{2}m(v^2-0^2)\\\Rightarrow v=\sqrt{\dfrac{2W}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 4475}{2.4\times 10^3}}\\\Rightarrow v=1.9311\ m/s

The speed is 1.9311 m/s

Work done is given by

W=Fs\\\Rightarrow F=\dfrac{W}{s}\\\Rightarrow F=\dfrac{4475}{26.5}\\\Rightarrow F=168.86792\ N

The horizontal force is 168.86792 N

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Bezzdna [24]

Answer:

1200 Sm^2mol^-1

Explanation:

Given data :

conductivity of water ( kwater ) = 76 mS m^-1 = 0.076 Sm^-1

conductivity of kcl (aq)( Kkcl ) = 1.1639 Sm^-1

Kkcl = 1.1639 - 0.076 = 1.0879  Sm^-1

Resistance = 33.21 Ω

where conductivity can be expressed as = \frac{Cell constant}{Resistance }

hence cell constant = conductivity * Resistance

                                 = 1.0879 * 33.21 = 36.13m^-1

conductivity of  CH3COOH ( kCH3COOH ) =  36.13 / 300

                                                                       = 0.120 Sm^-1

<u>Determine the molar conductivity of acetic acid</u>

= ( kCH3COOH * 1000 ) / C

C = 0.1 mol dm

=  (0.120 * 1000) / 0.1  =  1200 Sm^2mol^-1

3 0
3 years ago
The ink drops have a mass m = 1.00×10^−11 kg each and leave the nozzle and travel horizontally toward the paper at velocity v =
luda_lava [24]

Answer:

9.98 × 10⁻⁹ C

Explanation:

mass, m = 1.00 × 10⁻¹¹ kg

Velocity, v = 23.0 m/s

Length of plates D₀ = 1.80 cm = 0.018 m

Magnitude of electric field, E = 8.20 × 10⁴ N/C

drop is to be deflected a distance d = 0.290 mm = 0.290 × 10⁻³ m

density of the ink drop = 1000 kg/m^3

Now,

Time = \frac{\textup{Distance}}{\textup{Velocity}}

or

Time = \frac{\textup{0.016}}{\textup{23}}

or

Time = 6.9 × 10⁻⁴ s

Now, force due to the electric field, F = q × E

where, q is the charge

Also, Force = Mass × acceleration

q × E = 1.00 × 10⁻¹¹ × a

or

a = \frac{q\times8.20\times10^4}{1\times10^{-11}}

Now from the Newton's equation of motion

d=ut+\frac{1}{2}at^2

where,  

d is the distance

u is the initial speed  

a is the acceleration

t is the time

or

0.290\times10^{-3}=0\times(6.9\times10^{-4})+\frac{1}{2}\times(\frac{q\times8.20\times10^4}{1\times10^{-11}})\times(6.9\times10^{-4})^2

or

q = 9.98 × 10⁻⁹ C

4 0
3 years ago
A thin, metallic spherical shell of radius 0.347 m0.347 m has a total charge of 7.53×10−6 C7.53×10−6 C placed on it. A point cha
USPshnik [31]

Answer:

E = 12640.78 N/C

Explanation:

In order to calculate the electric field you can use the Gaussian theorem.

Thus, you have:

\Phi_E=\frac{Q}{\epsilon_o}

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Q: net charge inside the Gaussian surface

εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:

\Phi_E=EA=E(4\pi r^2)=\frac{Q}{\epsilon_o}\\\\E=\frac{Q}{4\pi \epsilon_o r^2}

r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m

Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:

Q=7.53*10^{-6}C+3.65*10^{-6}C=1.115*10^{-5}C

Finally, you obtain for E:

E=\frac{1.118*10^{-5}C}{4\pi (8.85*10^{-12C^2/Nm^2})(0.795m)^2}=12640.78\frac{N}{C}

hence, the electric field at 0.795m from the center of the spherical shell is 12640.78 N/C

3 0
3 years ago
NEED HELP ASAPTsunamis are often called tidal waves because they are caused by tides.
garik1379 [7]

Answer:

false

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Answer:

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6 0
3 years ago
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