Physics: At the local grocery store, you push a 14.5-kg shopping cart. You stop for a moment to add a bag of dog food to your cart. With a force of 12.0 N you now accelerate the cart from rest through a distance of 2.29 m in 3.00 s. What was the mass of the dog food?

At the local grocery store, you push a 14.5-kg shopping

briibearr

23.07.2021

The mass of the dog food added is 9.03 kg

Explanation:

Given;

mass of the shopping cart, m₁ = 14.5 kg

let the mass of the bag added = m₂

the force applied, F = 12 N

initial velocity of the cart-bag system, u = 0

distance traveled by the system, d = 2.29 m

time of motion of the system, t = 3.0 s

The acceleration of the system is calculated as;

$d = ut + \frac{1}{2} at^2\\\\2.29 = 0 + (\frac{1}{2} \times 3^2)a\\\\2.29 = 4.5 a\\\\a = \frac{2.29}{4.5} \\\\a = 0.51 \ m/s^2$

The total mass of the system (M) is calculated as follows;

F = Ma

M = F/a

M = (12)/(0.51)

M = 23.53 kg

The mass of the dog food added is calculated as;

m₂ = M - m₁

m₂ = 23.53 kg - 14.5 kg

m₂ = 9.03 kg

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Eylul30

11.08.2020

magnification will be -0.025

Explanation:

We have given the radius of curvature = 12 cm

And object distance = 3 m

So focal length $f=\frac{R}{2}=\frac{12}{2}=6cm$

Now for mirror we know that $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

So $\frac{1}{0.06}=\frac{1}{3}+\frac{1}{v}$

$16.66-0.333=\frac{1}{v}$

v = 0.750 m

Now magnification of the mirror is $m=\frac{-v}{u}=\frac{-0.750}{3}=-0.025$

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