What is the resultant force of 500g on abject accelerating at 5m/s2

Physics

1 answer:

max2010maxim [7]3 years ago

6 0

Answer: 2.5 N

Explanation:

m = 500g = 0.5Kg

a = 5m/s2

F = ma = 0.5 x 5 = 2.5 N

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The position of a particle moving along the x axis varies in time according to the expression x = 4t 2, where x is in meters and

Pie

Answer:

a) X = 17.64 m

b) X = 17.64 + 4∆t^2 + 16.8∆t

c) Velocity = lim(∆t→0)⁡〖∆X/∆t〗 = 16.8 m/s

Explanation:

a) The position at t = 2.10s is:

X = 4t^2

X = 4(2.10)^2

X = 17.64 m

b) The position at t = 2.10 + ∆t s will be:

X = 4(2.10 + ∆t)^2

X = 17.64 + 4∆t^2 + 16.8∆t m

c) ∆X is the difference between position at t = 2.10s and t = 2.10 + ∆t so,

∆X= 4∆t^2 + 16.8∆t

Divide by ∆t on both sides:

∆X/∆t = 4∆t + 16.8

Taking the limit as ∆t approaches to zero we get:

Velocity =lim(∆t→0)⁡〖∆X/∆t〗 = 4(0) + 16.8

Velocity = 16.8 m/s

8 0

3 years ago

Read 2 more answers

One litre of crude oil weighs 9.6N. Calculate its specific weight, density and specific gravity.

Zepler [3.9K]

Answer:

The answer is " $\bold{9600 \frac{N}{m^3}, 978.59 \frac{kg}{m^3}, and \ 0.978}$ "

Explanation:

Given:

$\to v=1\ liter= 10^{-3} \ m^3\\\\\to w= 9.6 \ N\\$

calculation:

$Specific \ weight =\frac{w}{v}=\frac{9.6}{10^{-3}}=9600 \frac{N}{m^3} \\\\w=mg\\\\m= \frac{w}{g}=\frac{9.6}{9.81}=0.9785\ kg\\\\\rho\ (density)=\frac{m}{v}=\frac{0.9785}{10^{-3}}=978.59 \frac{kg}{m^3}\\\\specific \ gravity = \frac{\prho \ obj}{\rho w}=\frac{978.54}{1000}=0.978$