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N76 [4]
3 years ago
9

You find it takes 220 n of horizontal force to move an unloaded pickup truck along a level road at a speed of 2.4 m/s. you then

load up the pickup and pump up its tires so that its total weight increases by 42% while the coefficient of rolling friction decreases by 19%.
Physics
1 answer:
Vinvika [58]3 years ago
3 0
Kinetic friction for the truck is given by:

F = μ*N = μ*m*g

μ coefficient of friction
m mass of the truck
g gravitational constant = 9.81

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<
goblinko [34]

Answer:

Explanation:

initial velocity u = 32.7 m /s

final velocity v = 50.3 m /s

displacement s = 44500 m

acceleration a = ?

v² = u² + 2 a s

50.3² = 32.7² + 2 x a x 44500

2530.09 = 1069.29 + 89000a

a .016 m /s²

time taken t = ?

v = u + at

50.3 = 32.7 + .016 t

t = 1100 s

6 0
3 years ago
Which of the following solutions should be chosen to help restore an ecosystem naturally? A. reintroducing an animal to the ecos
jolli1 [7]

Answer:

A. reintroducing an animal to the ecosystem

Explanation:

As generally, all know that for restoring an ecosystem naturally, it requires reintroduction of an animal to the ecosystem. As though it helps in reimposing the ecosystem back, and also helps to improve our ecosystem in natural surroundings, natural terrain, and population density. Basically reintroducing an animal is also required for the balancing of the ecosystem. As everything requires a properly balanced nature.

7 0
3 years ago
What are tiny sacs at the end of the bronchioles filled with air called?
WARRIOR [948]

Answer:

Alveoli

The bronchioles end in tiny air sacs called alveoli, where oxygen is transferred from the inhaled air to the blood.

Hope this helps :)

7 0
4 years ago
A 4-79 permalloy solenoid coil needs to produce a minimum inductance of 1.1 . If the maximum allowed current is 4 , how many tur
daser333 [38]

The related concept to solve this exercise is given in the expressions that the magnetic field has both as a function of the number of loops, current and length, as well as inductance and permeability. The first expression could be given as,

The magnetic field H is given as,

H = \frac{nI}{l}

Here,

n = Number of turns of the coil

I = Current that flows in the coil

l = Length of the coil

From the above equation, the number of turns of the coil is,

n = \frac{Hl}{I}

The magnetic field is again given by,

H = \frac{B}{\mu_t}

Where the minimum inductance produced by the solenoid coil is B.

We have to obtain n, that

n = \dfrac{\frac{B}{\mu_t}l}{I}

Replacing with our values we have that,

n = \dfrac{\frac{1.1Wb/m^2 }{200000}(2m)}{4mA}

n = \dfrac{(\frac{1.1Wb/m^2 }{200000})(\frac{10^4 guass}{1Wb/m^2})(2m)}{4mA(\frac{10^{-3}A}{1mA})}

n = 27.5 \approx 28

Therefore the number of turn required is 28Truns

4 0
4 years ago
An electron emitted in the beta decay of bismuth-210 has a mean kinetic energy of 390 keV. (a) Find the de Broglie wavelength of
Sauron [17]

Explanation:

Given that,

The mean kinetic energy of the emitted electron, E=390\ keV=390\times 10^3\ eV

(a) The relation between the kinetic energy and the De Broglie wavelength is given by :

\lambda=\dfrac{h}{\sqrt{2meE}}

\lambda=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 390\times 10^3}}

\lambda=1.96\times 10^{-12}\ m

(b) According to Bragg's law,

n\lambda=2d\ sin\theta

n = 1

For nickel, d=0.092\times 10^{-9}\ m

\theta=sin^{-1}(\dfrac{\lambda}{2d})

\theta=sin^{-1}(\dfrac{1.96\times 10^{-12}}{2\times 0.092\times 10^{-9}})

\theta=0.010^{\circ}

As the angle made is very small, so such an electron is not useful in a Davisson-Germer type scattering experiment.

4 0
4 years ago
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