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kolbaska11 [484]
3 years ago
11

An objects speed is 2.40 m/s and its momentum is 120.0 kg m/s. what is the mass of the object?

Physics
1 answer:
ziro4ka [17]3 years ago
5 0
Momentum = mass X Velocity
Using logic we can see that the velocity would be positive so 2.40 doesn't change.
120/2.40 = 50kg
Mass of object is 50 kg.
Hope that helps! :)
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The combustion of 1.631 g of sucrose, C12H22O11(s) , in a bomb calorimeter with a heat capacity of 5.30 kJ/°C results in an incr
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Answer:

= - 26.31 kJ

Explanation:

we know that number of moles is calculated as

Moles of C_{12}H_{22}O_{11} = \frac{mass}{molecular\ weight}

               = \frac{(1.631 g)}{(342.29 g/mol)}

                      = 0.00476 mol

Heat absorbed by calorimeter = heat\ capacity \times temperature\ rise

= 5.30 kJ/°C x (27.75 - 22.68)°C

= 26.87 kJ

Enthalpy of combustion

\Delta Hc = \frac{- 26.87}{0.00486}

= - 55290.12 kJ/mol

Negative sign shows that the heat is released

The balanced reaction

C_{12}H_{22} O_{11}(s) + 12 O_2(g) = 12 CO_2(g) + 11 H_2O(l)

ΔHc = ΔU + Δng (RT)

-55290.12 = ΔU + (12 - 12) *(RT)

\Delta U = - 55290.12 kJ/mol \times 0.00476 mol

= - 26.31 kJ

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Question 7 of 25
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Answer:

2PBr₃ + 3Cl₂ → 2PCl₃ + 3Br₂

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Explanation:

A balanced chemical equation is a chemical equation that have an equal number of elements of each type on both sides of the equation

Among the given chemical reactions, we have;

2PBr₃ + 3Cl₂ → 2PCl₃ + 3Br₂

In the above reaction;

The number of phosphorus, P, on either side of the equation = 2

The number of bromine atoms, Br, on either side of the equation = 6

The number of chlorine atoms, Cl, on either side of the equation = 6

Therefore, the number of elements in the reactant side and products side of the reaction are equal and the reaction is balanced

The second balanced chemical reaction is 2Na + MgCl₂ → 2NaCl + Mg

In the above reaction, there are two sodium atoms, Na,  one magnesium atom and two chlorine atoms on both sides of the reaction, therefore, the reaction is balanced

6 0
3 years ago
A 0.3-m-radius automobile tire rotates how many revolutions after starting from rest and accelerating at a constant 2.13 rad/s2
Aloiza [94]

Answer:

The automobile tire rotates 91 revolutions

Explanation:

Given;

angular acceleration of the automobile, α = 2.13 rad/s²

time interval, t = 23.2-s

To calculate the number of revolutions, we apply the first kinematic equation;

\theta = \omega_i  \ + \frac{1}{2} \alpha t^2

the initial angular velocity is zero,

\theta =0\ + \frac{1}{2} (2.13) (23.2)^2\\\\\theta = 573.2256 \ Rad

Find how many revolutions that are in 573.2256 Rad

N = \frac{\theta}{2 \pi} = \frac{573.2256}{2\pi} \\\\N = 91 \ revolutions

Therefore, the automobile tire rotates 91 revolutions

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4 years ago
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