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lions [1.4K]
3 years ago
15

A 45 kg wagon is being pulled with a rope that makes an angle of 38 degrees with the horizontal. The applied force is 410 N and

the coefficient of kinetic friction between the wagon and the ground is 0.18.
Physics
1 answer:
mel-nik [20]3 years ago
8 0

Answer:

7.35m/s²

Explanation:

From the question we are not told what to find but we can as well find the acceleration of the wagon.

According to newton second law of motion

\sum F_x = ma_x\\Fm - Ff = ma_x\\Fm - \mu R = ma_x\\Fm- \mu mg = ma_x\\

Fm is the moving force = 410N

\mu is the coefficient of friction = 0.18

m is the mass = 45kg

g is the acceleration dur to gravity = 9.8m/s²

a is the acceleration of the wagon

Substitute the given data into the equation ang get ax

Fm- \mu mg = ma_x\\410 - (0.18)(9.8)(45) = 45a_x\\410 - 79.38 = 45a_x\\330.62 = 45a_x\\a_x = 330.62/45\\a_x = 7.35m/s^2\\

Hence the acceleration of the wagon is 7.35m/s²

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Answer:

a) t = 20 [s]

b) Can't land

Explanation:

To solve this problem we must use kinematics equations, it is of great importance to note that when the plane lands it slows down until it reaches rest, ie the final speed will be zero.

a)

v_{f}=v_{i}-(a*t)

where:

Vf = final velocity = 0

Vi = initial velocity = 100 [m/s]

a = desacceleration = 5 [m/s^2]

t = time [s]

Note: the negative sign of the equation means that the aircraft slows down as it stops.

0 = 100 - 5*t

5*t = 100

t = 20 [s]

b)

Now we can find the distance using the following kinematics equation.

x -x_{o}=(v_{o}*t)+\frac{1}{2}*a*t^{2}

x - xo = distance [m]

x -xo = (0*20) + (0.5*5*20^2)

x - xo =  1000 [m]

1000 [m] = 1 [km]

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4 0
3 years ago
Of the following transitions in the Bohr hydrogen atom, the ________ transition results in the absorption of the highest-energy
8_murik_8 [283]

Answer:

The <em><u>n = 2 → n = 3</u></em> transition results in the absorption of the highest-energy photon.

Explanation:

E_n=-13.6\times \frac{Z^2}{n^2}ev

Formula used for the radius of the n^{th} orbit will be,

where,

E_n = energy of n^{th} orbit

n = number of orbit

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Here: Z = 1 (hydrogen atom)

Energy of the first orbit in H atom .

E_1=-13.6\times \frac{Z^2}{1^2} eV=-13.6 eV

Energy of the second orbit in H atom .

E_2=-13.6\times \frac{Z^2}{(2)^2} eV=-3.40 eV

Energy of the third orbit in H atom .

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Energy of the fifth orbit in H atom .

E_5=-13.6\times \frac{Z^2}{(2)^2} eV=-0.544 eV

Energy of the sixth orbit in H atom .

E_6=-13.6\times \frac{Z^2}{(2)^2} eV=-0.378 eV

Energy of the seventh orbit in H atom .

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1) n = 2 → n = 3

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