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vivado [14]
3 years ago
5

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

4 0
2 years ago
What human disease can result from exposure to ultraviolet radiation from the sun?
aev [14]
Melanoma skin cancer
8 0
3 years ago
Read 2 more answers
A small metal object is tied to the end of a string and whirled round a circular path of radius 30cm. The object makes 20 oscill
LUCKY_DIMON [66]

Answer:

(i) The angular speed of the small metal object is 25.133 rad/s

(ii) The linear speed of the small metal object is 7.54 m/s.

Explanation:

Given;

radius of the circular path, r = 30 cm = 0.3 m

number of revolutions, n = 20

time of motion, t = 5 s

(i) The angular speed of the small metal object is calculated as;

\omega = \frac{20 \ rev}{5 \ s} \times \frac{2 \pi \ rad}{1 \ rev} = \frac{40\pi \ rad}{5 \ s} = 8\pi \ rad/s = 25.133 \ rad/s

(ii) The linear speed of the small metal object is calculated as;

v = \omega r\\\\v = 25.133 \ rad/s \ \times \ 0.3 \ m\\\\v = 7.54 \ m/s

6 0
3 years ago
What is conclusion about potential difference (voltmeter readings) in a parallel electric circuit ?
Ainat [17]

Answer: the conclusion is that

Explanation:

5 0
3 years ago
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