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Korvikt [17]
3 years ago
6

(Level 5-6) 3. An object accelerates 15.0 m/s2 when a force of 8.ON is applied to it. What is the mass of the

Physics
1 answer:
VMariaS [17]3 years ago
5 0

Answer:

0.53 kg

Explanation:

Mass = \frac{Force}{acceleration}

Force (also weight) = 8.0N

Acceleration = 15.0 m/s^{2}

= \frac{8.0}{15}

= 0.53 kg ( to 2 decimal place )

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3 years ago
.1 An 8-ft 3 tank contains air at an initial temperature of 808F and initial pressure of 100 lbf/in. 2 The tank develops a small
Alina [70]

Correct temperature is 80°F

Answer:

T_f = 38.83°F

Explanation:

We are given;

Volume; V = 8 ft³

Initial Pressure; P_i = 100 lbf/in² = 100 × 12² lbf/ft²

Initial temperature; T_i = 80°F = 539.67 °R

Time for outlet flow; t_o = 90 s

Mass flow rate at outlet; m'_o = 0.03 lb/s

Final pressure; P_f = 30 lbf/in² = 30 × 12² lbf/ft²

Now, from ideal gas equation,

Pv = RT

Where v is initial specific volume

R is ideal gas constant = 53.33 ft.lbf/°R

Thus;

v = RT/P

v_i = 53.33 × 539.67/(100 × 12²)

v_i = 2 ft³/lb

Formula for initial mass is;

m_i = V/v_i

m_i = 8/2

m_i = 4 lb

Now change in mass is given as;

Δm = m'_o × t_o

Δm = 0.03 × 90

Δm = 2.7 lb

Now,

m_f = m_i - Δm

Thus; m_f = 4 - 2.7

m_f = 1.3 lb

Similarly in above;

v_f = V/m_f

v_f = 8/1.3

v_f = 6.154 ft³/lb

Again;

Pv = RT

Thus;

T_f = P_f•v_f/R

T_f = (30 × 12² × 6.154)/53.33

T_f = 498.5°R

Converting to °F gives;

T_f = 38.83°F

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3 years ago
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