All categories

3 years ago

1. A car has a mass of 4000 kg. What force is required to make the car accelerate at 20 mss

Chemistry

1 answer:

shtirl [24]3 years ago

8 0

Answer:

As Per Given Information

Mass of Car 4000 Kg

Acceleration of car 20m/s²

we have to find the force required to make the car car accelerate at given magnitude .

Using Formulae

Force = Mass × Acceleration

On Putting the value we obtain

↝ Force = 4000 × 20

↝ Force = 80000 N

↝ Force = 80KN

So, the force required to maintain the given acceleration is 80 KiloNewton .

You might be interested in

On hot days, you may have noticed that potato chip bags seem to “inflate”, even though they have not been opened. If I have a 25

Taya2010 [7]

The new volume of the bag : 747.43 ml

<h3>Further explanation</h3>

Charles's Law

When the gas pressure is kept constant, the gas volume is proportional to the temperature

$\tt \dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

V₁=250 ml

T₁=19+273=292 K

T₂=600+273=873 K

$\tt V_2=\dfrac{V_1.T_2}{T_1}\\\\V_2=\dfrac{250\times 873}{292}\\\\V_2=747.43~ml$

7 0

3 years ago

A reaction A(aq)+B(aq)↽−−⇀C(aq) has a standard free‑energy change of −4.20 kJ/mol at 25 °C. What are the concentrations of A, B,

Dafna1 [17]

Answer : The concentration of $A,B\text{ and }C$ at equilibrium are 0.132 M, 0.232 M and 0.168 M respectively.

Explanation :

The given chemical reaction is,

$A(aq)+B(aq)\rightleftharpoons C(aq)$

First we have to calculate the equilibrium constant for the reaction.

The relation between the equilibrium constant and standard free‑energy is:

$\Delta G^o=-RT \ln k$

where,

$\Delta G^o$ = standard free‑energy change = -4.20 kJ/mole

R = universal gas constant = 8.314 J/mole.K

k = equilibrium constant = ?

T = temperature = $25^oC=273+25=298K$

Now put all the given values in the above relation, we get:

$-4.20kJ/mole=-(8.314J/mole.K)\times (298K) \ln k$

$k=5.45$

Now we have to calculate the concentrations of A, B, and C at equilibrium.

The given equilibrium reaction is,

$A(aq)+B(aq)\rightleftharpoons C(aq)$

Initially 0.30 0.40 0

At equilibrium (0.30-x) (0.40-x) x

The expression of equilibrium constant will be,

$k=\frac{[C]}{[A][B]}$

$5.45=\frac{x}{(0.30-x)\times (0.40-x)}$

By solving the term x, we get

$x=0.168\text{ and }0.716$

From the values of 'x' we conclude that, x = 0.716 can not more than initial concentration. So, the value of 'x' which is equal to 0.716 is not consider.

The value of x will be, 0.168 M

The concentration of $A$ at equilibrium = (0.30-x) = 0.30 - 0.168 = 0.132 M

The concentration of $B$ at equilibrium = (0.40-x) = 0.40 - 0.168 = 0.232 M

The concentration of $C$ at equilibrium = x = 0.168 M

3 0

4 years ago

If the accepted value for the mass of an object is 20.0g and the student found that the mass of the object was 20.5g what is the

NikAS [45]

Answer:

Results

The percent error between 20 and 20.5 is 2.5%

Explanation:

Percent Error = | (20.5 − 20) / 20 | × 100 = | (0.5) / 20 | × 100 = | 0.025 | × 100 = 2.5% (three decimal places)Percent Error = 2.5%

3 0

3 years ago

Lithium (Li) and oxygen (O) are both in period 2. They both have _____.

AlladinOne [14]

They both have two electron shells

<h3>Further explanation</h3>

The period 2 element lies in the second row of the periodic system.

Consists of the elements: lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine, and neon

Lithium (Li)

atomic number : 3

electron configuration : [He] 2s¹

atomic number = number of proton=number of electron(in neutral atom)

So Li have 3 protons and 3 electrons

Because it fills the 2s orbital it has 2 shells

Oxygen (O)

atomic number : 8

electron configuration : [He] 2s²2p⁴

So O have 8 protons and 8 electrons

Because it fills the 2s and 2p orbital it has 2 shells

So Lithium (Li) and Oxygen (O) are both have two electron shells

6 0

3 years ago

Which 4 planets belong together? which 4 planets lumped together have the most similarities

andrey2020 [161]

Answer:

did you mean to add or attach a paper to this? We need more info to help

Explanation:

6 0

3 years ago

Read 2 more answers