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3 years ago

Explan first netwons law in words

2 answers:

7 0

If something is at rest it will stay at rest aka stay still but if two things are moving at the same speed and same direction they are the same unless something happens to the other object. i hope this helps

inessss [21]3 years ago

3 0

An object either remains at rest until acted upon

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What constant acceleration (in ft/s2) is required to increase the speed of a car from 27 mi/h to 50 mi/h in 5 seconds

SIZIF [17.4K]

Answer:

6.746 ft/s^2

Explanation:

v(t)=50

v(0)=27

t=5/3600 = 1/720 hours

v(t)-v(0)= a(t-0)

50-27= a(1/720)

a= 23*720= 16560 mi/h^2

16560mi/h^2 * 5280/3600^2 (ft/s^2) =6.746 ft/s^2

4 0

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

7 0

3 years ago

Two groups of students were tested to compare their speed working math problems. Each group was given the same problems. One gro

oee [108]

Answer:

The people with caculators will probably answer faster due to thier ablitiy to use a device of technology

3 0

3 years ago

Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w

gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

$U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}$

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

$u(x) = \frac{-2}{x} + \frac{4}{x^2}$