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Art
3 months ago
12

A 1100kg car pulls a boat on a trailer. (a) what total force resists the motion of the car, boat,and trailer, if the car exerts

a 1900n force on the road and produces an acceleration of 0.550 m/s2? the mass of the boat plus trailer trailer is 700kg. (b) what is the force in the hitch betweenthe car and the trailer if 80% of the resisting forces are experienced by the boat and trailer?

Physics
1 answer:
inna [3.1K]3 months ago
3 0
Refer to the diagram shown below.

m₁ = 1100 kg represents the mass of the car.
m₂ = 700 kg indicates the combined mass of the trailer and boat.
F = 1900 N is the driving force acting on the vehicle.
N₁ denotes m₁g, the normal force on the car.
N₂ corresponds to m₂g, the normal force on the trailer and boat.
Frictional forces are represented by μN₁ and μN₂, where μ is the coefficient of kinetic friction.
T signifies the force in the connection between the car and the trailer.

Part (a)
Let R₁ signify the total resistance acting against the motion of the car, boat, and trailer.
With the acceleration at 0.550 m/s², it follows that
(m₁ + m₂ kg)*(0.55 m/s²) = F
(1100 + 700 kg)*(0.55 m/s²) = (1900 - R₁) N.
This leads to the equation 990 = 1900 - R.
Therefore, R₁ = 910 N.

Answer: The total resistive force amounted to 910 N.

Part (b)
The trailer and boat experience 80% of the resisting forces.
Let R₂ denote this resistive force.
Thus,
R₂ = 0.8*R₁ = 728 N.
Assuming T is the tension in the hitch connecting the car and trailer, it follows:
T - R₂ = m₂(0.55 m/s²)
(T - 728 N) = (700 kg)*(0.55 m/s²).
This leads to T - 728 = 385.
Thus, T equals 1113 N.

Answer: The tension in the hitch is 1113 N.

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The amount of energy necessary to remove an electron from an atom is a quantity called the ionization energy, Ei. This energy ca
Keith_Richards [3271]

Answer:

The resulting value is E_i = 1.5596 *10^{-18} \ J.

Explanation:

The question specifies that

The wavelength is \lambda = 48.2 nm = 48.2 *10^{- 9 }\ m.

The velocity is v = 2.371*10^6 \ m/s.

The mass of the electron is m_e = 9.109*10^{-31} \ kg.

The energy of the incoming light is typically depicted mathematically as

E = \frac{h * c}{\lambda}.

Here, c represents the speed of light with the value c = 3.0 *10^{8} \ m/s.

h stands for Planck's constant with a value of h = 6.62607015 * 10^{-34 } J\cdot s.

Thus,

E = \frac{6.62607015 * 10^{-34 }* 3.0 *10^{8}}{48.2 *10^{- 9 }}

=> E = 4.12 *10^{-18} \ J.

Typically, kinetic energy is represented as

E_k = \frac{1}{2} * m_e * v^2

=> E_k = \frac{1}{2} * 9.109*10^{-31} * (2.371*10^6 )^2.

=> E_k = 2.56 *0^{-18} \ J.

The ionization energy is generally expressed mathematically as

E_i = 4.12 *10^{-18} - 2.56 *0^{-18}

=> E_i = 1.5596 *10^{-18} \ J.

8 0
3 months ago
Calculate the heat required when 2.50 mol of a reacts with excess b and a2b according to the reaction: 2a + b + a2b → 2ab + a2 g
Softa [3030]

Answer:

Q = 12.5 kJ

Explanation:

The formula used to compute heat is:

Q = H° * n

Where:

Q: heat (J or kJ)

H°: enthalpy of reaction (kJ/mol)

n: moles

Now, as noted in the comments, the question lacks completeness, here is the part that is missing:

Given:

2A + B  A2B (1)

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ΔH° = 35.0 kJ/mol

Using these two reactions, we can determine the heat change.

Using the above two reactions, we need to establish the overall reaction (the one presented in the question), so let’s combine (1) and (2):

2A + B -------> A2B   H°1 = -25 kJ/mol

2A2B --------> 2AB + A2   H°2 = 35 kJ/mol

When we add these equations, one A2B cancels out with one A2B from reaction 2, thus, we have:

2A + B + 2A2B -------> 2AB + A2

So, for the enthalpy, the values are summed:

H°3 = -25 + 35 = 10 kJ/mol

Now we can calculate the heat:

Q = 10 * 2.5 = 25 kJ

However, as we have 2A in the reaction, it does not maintain a 1:1 mole ratio, instead, it is 1:2, which requires us to adjust; thus:

Q = 25 / 2 = 12.5 kJ

3 0
2 months ago
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