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Daniel [21]
3 years ago
15

Two shopping carts are pushed with the same force. One cart has a larger mass than the other. How does the mass of the shopping

carts affect their acceleration?
Question 9 options:

The cart with the smaller mass will accelerate faster.


The cart with the larger mass will accelerate faster.


The cart with the smaller mass will not accelerate as fast as that with the larger mass.


Both shopping carts will accelerate at the same rate since they are the same size.
Chemistry
2 answers:
OlgaM077 [116]3 years ago
8 0

Answer: The correct answer is the cart with the smaller mass will accelerate faster.

Explanation: According to Newtons Second Law, the net force is the product of the mass of the object and the acceleration of that object. The acceleration is in the same direction as that of the net force.

Mathematically,

F=ma

where,

F = net force

m = Mass of the object

a = acceleration of the object

In the question,

Force applied on two shopping carts is the same and mass of first cart is more than the mass of the second card.

F_1=F_2\\m_1>m_2

Acceleration is equal to \frac{F}{m}

As, acceleration is inversely proportional to the mass of the object, more the mass of the object less will be its acceleration and the object will move slowly.

Hence, the cart with smaller mass will accelerate faster than the cart having larger mass.

Harman [31]3 years ago
5 0
This would problem will be answered using the concept of momentum. Momentum is the mass times the velocity, where mass is inversely proportional to the velocity. This would mean that an object with greater mass will travel at a much lesser velocity, or vice versa.

Hence, the cart with the smaller mass will accelerate faster.
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calculate the freezing point of 3.46 gram of a compound X in 160 gram of benzene when a separate sample of X was vaporized it's
Temka [501]

Answer: The freezing point of 3.46 gram of a compound X in 160 gram of benzene is 4.38^0C

Explanation:

The relation of density and molar mass is:

d=\frac{PM}{RT}

where

d = density = 3.27 g/ L

P = pressure of the gas  = 773 torr = 1.02 atm   (760 torr = 1atm)

M = molar mass of the gas  = ?

T = temperature of the gas = 116^0C=(116+273)K=389K

R = gas constant  = 0.0821Latm/Kmol

M=\frac{dRT}{P}=\frac{3.27g/L\times 0.0821Latm/Kmol\times 389K}{1.02atm}=102.3g/mol

The relation of depression in freezing point with molality:

\Delta T_f=k_f\times m

\Delta T_f = depression in freezing point = T_f^0-T_f = 5.45-T_f

k_f = freezing point constant  = 5.1

m = molality = \frac{\text {moles of X}}{\text {weight of solvent in kg}}=\frac{3.46\times 1000}{102.3\times 160}=0.21

5.45-T_f=5.1\times 0.21

T_f=4.38^0C

Thus the freezing point of 3.46 gram of a compound X in 160 gram of benzene is 4.38^0C

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What is the wavelength of an X-ray that has a frequency of 7.8 x 1017 Hz
SVEN [57.7K]

Answer:

λ = 0.38 ×10⁻⁹ m

Explanation:

Given data:

Wavelength of xray = ?

Frequency of xray = 7.8 ×10¹⁷ Hz

Solution:

Formula:

Speed of light = wavelength × frequency

speed of light = 3×10⁸ m/s

Now we will put the values in formula.

3×10⁸ m/s = λ × 7.8 ×10¹⁷ Hz

λ = 3×10⁸ m/s / 7.8 ×10¹⁷ Hz

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λ = 3×10⁸ m/s / 7.8 ×10¹⁷s⁻¹

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