Study the diagram, which represents the general equation for photosynthesis.

What are the magnitude (size) and direction of the cumulative force acting on the car shown in the picture A. 30 N to the right

UNO [17]

Answer: The correct answer is option(A).

Explanation:

Total forces exerting on the car = F

$F_r=50 N$ = Force on car exerting in right direction

$F_l=20 N$ = Force on car exerting in left direction

$F_u=2500 N$ = Force on car exerting in upward direction

$F_d=2500 N$ = Force on car exerting in downward direction

$F=F_r+F_l+F_u+F_d$

$(-F_u)=F_d$ (negative sign shows the direction)

Since, upward force are equal in magnitude but opposite in direction by which they will balance out each other.so, the net force car will be due to two forces $F_r$ and $F_l$

$F=F_r+F_l=50 N+(-20 N)=30 N$ (negative shows the direction)

The magnitude (size) and direction of the cumulative force acting on the car will 30 N towards right direction.Hence, correct answer is option(A).

4 0

3 years ago

Read 2 more answers

This is due in five mintues please help

Fittoniya [83]

Answer:

5

Explanation:

Na, N, O, O, O

5 0

3 years ago

The vapor pressure of substance X is 100. mm Hg at 1080.°C. The vapor pressure of substance X increases to 600. mm Hg at 1220.°C

artcher [175]

Explanation:

The given data is as follows.

$P_{1}$ = 100 mm Hg or $\frac{100}{760}atm$ = 0.13157 atm

$T_{1}$ = $1080 ^{o}C$ = (1080 + 273) K = 1357 K

$T_{2}$ = $1220 ^{o}C$ = (1220 + 273) K = 1493 K

$P_{2}$ = 600 mm Hg or $\frac{600}{760}atm$ = 0.7895 atm

R = 8.314 J/K mol

According to Clasius-Clapeyron equation,

$log(\frac{P_{2}}{P_{1}}) = \frac{\Delta H_{vap}}{2.303R}[\frac{1}{T_{1}} - \frac{1}{T_{2}}$

$log(\frac{0.7895}{0.13157}) = \frac{\Delta H_{vap}}{2.303 \times 8.314 J/mol K}[\frac{1}{1357 K} - \frac{1}{1493 K}]$

$log (6) = \frac{\Delta H_{vap}}{19.147}[\frac{(1493 - 1357) K}{1493 K \times 1357 K}]$

0.77815 = $\frac{\Delta H_{vap}}{19.147J/K mol} \times 6.713 \times 10^{-5} K$

$\Delta H_{vap}$ = $2.219 \times 10^{5}$ J/mol

= $2.219 \times 10^{5}J/mol \times 10^{-3}\frac{kJ}{1 J}$

= 221.9 kJ/mol

Thus, we can conclude that molar heat of vaporization of substance X is 221.9 kJ/mol.

4 0

3 years ago

Which elements are more likely to lose electrons?<br> Ca, Li, F and Ne

Slav-nsk [51]

Ca, they only have two valence electrons, in order to become more stable, they would like to lose all of them

3 0

3 years ago

Suppose a solution is prepared by dissolving 15.0 g NaOH in 0.150 L of 0.250 M nitric acid. What is the final concentration of O

Kipish [7]

Answer:

2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.

Explanation:

Moles of NaOH = $\frac{15.0 g}{40 g/mol}=0.375 mol$

Molarity of the nitric acid solution = 0.250 M

Volume of the nitric solution = 0.150 L

Moles of nitric acid = n

$Molarity=\frac{Moles}{Volume(L)}$

$n=0.250 M\times 0.150 L=0.0375 mol$

$NaOH+HNO_3\rightarrow NaNO_3+H_2O$

According to reaction, 1 mole of nitric acid recats with 1 mole of NaOH, then 0.0375 moles of nitric acid will react with :

$\frac{1}{1}\times 0.0375 mol$ of NaOH

Moles of NaOH left unreacted in the solution =

= 0.375 mol - 0.0375 mol = 0.3375 mol

$NaOH(aq)\rightarrow Na^+(aq)+OH^-(aq)$

1 mole of sodium hydroxide gives 1 mol of sodium ions and 1 mole of hydroxide ions.

Then 0.3375 moles of NaOH will give :

$1\times 0.3375 moles=0.3375 mol$ of hydroxide ion

The molarity of hydroxide ion in solution ;

$=\frac{0.3375 mol}{0.150 L}=2.25 M$

2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.

3 0

3 years ago