Ask question

Ask question

All categories

3 years ago

9

The diagram shows two people pulling a wagon in the same direction. One person is pullig with a force of 150 Newtons and the oth

er person is pulling with a force of 200 Newtons.
The net force of the cart is

50 newtons east
200 newtons east
350 newtons east
37,500 newtons east

2 answers:

Svetach [21]3 years ago

6 0

I think it’s 350 Newton east

PS: Sorry if I’m wrong

tamaranim1 [39]3 years ago

4 0

Answer:

350 Newtons

Explanation:

You might be interested in

8. If a sample of chloroform is initially at 25°C, what is its final temperature if 150.0 g of chloroform

sashaice [31]

Answer:

The final temperature is 31.95° C.

Explanation:

Given that,

Initial temperature of a sample of chloroform, $T_i=25^{\circ} C$

Mass of chloroform, m = 150 g

It absorbs 1 kJ of heat, Q = 10³ J

The specific heat of chloroform, c = 00.96 J/gºC

We need to find the final temperature. The heat absorbed by an object in terms of specific heat is given by :

$Q=mc\Delta T\\\\Q=mc(T_f-T_i)\\\\\dfrac{Q}{mc}+T_i=T_f\\\\T_f=\dfrac{10^3}{150\times 0.96}+25\\\\=31.95^{\circ} C$

So, the final temperature is 31.95° C.

5 0

3 years ago

The atomic number of an atom is always equal to the total number of

Basile [38]

C is correct. Have a good day!

3 0

3 years ago

Read 2 more answers

A compound made of two elements, iridium (Ir) and oxygen (O), was produced in a lab by heating iridium while exposed to air. The

natima [27]

1) Compund Ir (x) O(y)

2) Mass of iridium = mass of crucible and iridium - mass of crucible = 39.52 g - 38.26 g = 1.26 g

3) Mass of iridium oxide = mass of crucible and iridium oxide - mass of crucible = 39.73g - 38.26g = 1.47g

4) Mass of oxygen = mass of iridum oxide - mass of iridium = 1.47g - 1.26g = 0.21g

5) Convert grams to moles

moles of iridium = mass of iridium / molar mass of iridium = 1.26 g / 192.17 g/mol = 0.00656 moles

moles of oxygen = mass of oxygen / molar mass of oxygen = 0.21 g / 15.999 g/mol = 0.0131

6) Find the proportion of moles

Divide by the least of the number of moles, i.e. 0.00656

Ir: 0.00656 / 0.00656 = 1

O: 0.0131 / 0.00656 = 2

=> Empirical formula = Ir O2 (where 2 is the superscript for O)

Answer: Ir O2

5 0

4 years ago

Read 2 more answers

A block of aluminum with a mass of 140 g is cooled from 98.4°C to 62.2°C with a release of 4817 J of heat. From these data, calc

disa [49]

Answer:

specific heat = 0.951 j/g·°C

Explanation:

Heat flow equation => q = m·c·ΔT

q = heat flow = 4817 joules

m = mass in grams = 140 grams Aluminum

c = specific heat = ?

ΔT = Temperature Change in °C = 98.4°C - 62.2°C = 36.2°C

q = m·c·ΔT => c = q/m·ΔT = 4817j/(140g)(36.2°C) = 0.951 j/g·°C

6 0

3 years ago

Air at 7S°F and14.6 though a cfm(cubic ft per minute) and the The flow rate is 48000 psia flows though a rectangular duct of Ix2

IgorLugansk [536]

Explanation:

The given data is as follows.

Air is at $75^{o}F$ and 14.6 psia.

$\varepsilon$ = 0.00015 ft, Flow rate, (Q) = 48000 $ft^{3}/m$

(a) Formula to calculate hydraulic radius $(r_{H})$ is as follows.

$r_{H} = \frac{\text{free flow area}}{\text{wet perimeter}}$

= $\frac{2 \times 1}{2(1) + 2(2)}$

= $\frac{1}{3}$ ft

Formula for equivalent diameter is as follows.

$D_{eq} = 4 \times r_{H}$

= $4 \times \frac{1}{3} ft$

= $\frac{4}{3}$ ft

(b) Formula for velocity floe is as follows.

Q = VA

V = $\frac{Q}{A}$

= $\frac{48000}{2 \times 1}$ ft/min

= 24000 ft/min

(c) Formula to calculate Reynold's number is as follows.

$R_{e}$ = $\frac{D \times V \times \rho}{\mu}$

= $\frac{\frac{4}{3} \times 24000 \times 0.0744}{0.0443}$ (as $\rho = 0.0744 lb/ft^{3}$ and $\mu$ = 0.0443 lb/ft. hr)

= 53742.66 hr/min

As 1 hr = 10 min. So, $53742.66 hr/min \times \frac{60 min}{1 hr}$

= 3224559.6

(d) Formula to calculate pressure drop $(\Delta P)$ is as follows.

$\frac{\Delta P}{L} = \frac{4f \rho V^{2}}{2Dg_{c}}$

Putting the given values into the above formula as follows.

$\frac{\Delta P}{L} = \frac{4f \rho V^{2}}{2Dg_{c}}$

= $\frac{4 \times 0.00015 \times 100 \times 0.0744 \times (24000)^{2}}{2 \times \frac{4}{3} \times {4}{3}}$

= 6.238 $lb/ft^{2}$

5 0

3 years ago