All categories

1 year ago

What is the value of acceleration due to gravity at the pole, equator and the centre of the earth

Physics

1 answer:

fgiga [73]1 year ago

7 0

Answer:

In combination, the equatorial bulge and the effects of the surface centrifugal force due to rotation mean that sea-level gravity increases from about 9.780 m/s2 at the Equator to about 9.832 m/s2 at the poles, so an object will weigh approximately 0.5% more at the poles than at the Equator.

You might be interested in

A 16 kg object cuts 4 km in 25 minutes, find the applied force on the object?

GenaCL600 [577]

Answer:

14min i think im not quite sure

Explanation:

7 0

1 year ago

Find the magnitude of the resultant force and the angle it makes with the positive x-axis. (Let |a| = 22 lb and |b| = 16 lb. Rou

SVEN [57.7K]

Incomplete question as the angle between the force is not given I assumed angle of 55°.The complete question is here

Two forces, a vertical force of 22 lb and another of 16 lb, act on the same object. The angle between these forces is 55°. Find the magnitude and direction angle from the positive x-axis of the resultant force that acts on the object. (Round to one decimal places.)

Answer:

Resultant Force=33.8 lb

Angle=67.2°

Explanation:

Given data

Fa=22 lb

Fb=16 lb

Θ=55⁰

To find

(i) Resultant Force F

(ii)Angle α

Solution

First we need to represent the forces in vector form

$\sqrt{x} F_{1}=22j\\ F_{2}=u+v\\F_{2}=16sin(55)i+16cos(55)j\\F_{2}=16(0.82)i+16(0.5735)j\\F_{2}=13.12i+9.176j$

Total Force

$F=F_{1}+F_{2}\\ F_{2}=22j+13.12i+9.176j\\F_{2}=13.12i+31.176j$

The Resultant Force is given as

$|F|=\sqrt{x^{2} +y^{2} }\\|F|=\sqrt{(13.12)^{2} +(31.176)^{2} }\\ |F|=33.8lb$

For(ii) angle

We can find the angle bu using tanα=y/x

$tan\alpha =\frac{31.176}{13.12}\\ \alpha =tan^{-1} (\frac{31.176}{13.12})\\\alpha =67.2^{o}$

7 0

3 years ago

The water flowing through a 2.0 cm (inside diameter) pipe flows out through three 1.3 cm pipes. (a) If the flow rates in the thr

Verdich [7]

Answer:

a)54L/min

b)0.845

Explanation:

a) A x V= $A_1V_1+ A_2V_2+A_3V_3$

where suffix 1,2,3 refers to the three pipes.

=27L/min+16L/min+11 L/min

=54L/min

b) A x V=54L/min => $\frac{\pi }{4} d^2$ x v

d= 2 cm

$\frac{\pi }{4} d^2$ x v = 54

v= $\frac{4}{\pi }$ x $\frac{54}{2^2}$

-> $A_1$ x $V_1$ =27L/min => $\frac{\pi }{4} d_1^2$ x $v_1$

$d_1$ = 1.3cm

$\frac{\pi }{4} d^2$ x $v_1$ = 27

$v_1$ = $\frac{4}{\pi }$ x $\frac{27}{1.3^2}$

Next is to find the ratio of speed i.e $\frac{v}{v_1}$

$\frac{4}{\pi }$ x $\frac{54}{2^2}$ / $\frac{4}{\pi }$ x $\frac{27}{1.3^2}$ => $\frac{54}{27}$ $\frac{1.3^2}{2^2}$

$\frac{v}{v_1}$ = 0.845

8 0

2 years ago

A straight line is drawn on the surface of a 14-cm-radius turntable from the center to the perimeter. A bug crawls along this li

sleet_krkn [62]

Answer:

v = $\left[\begin{array}{c}0.66&0\end{array}\right]$ m/s

Explanation:

The position vector r of the bug with linear velocity v and angular velocity ω in the laboratory frame is given by:

$\overrightarrow{r}=vtcos(\omega t)\hat{x}+vtsin(\omega t)\hat{y}$

The velocity vector v is the first derivative of the position vector r with respect to time:

$\overrightarrow{v}=[vcos(\omega t)-\omega vtsin(\omega t)]\hat{x}+[vsin(\omega t)+\omega vtcos(\omega t)]\hat{y}$

The given values are:

$t=\frac{x}{v}=\frac{14}{3.8}=3.7 s$

$\omega=\frac{45\times2\pi}{60s}=4.7\frac{1}{s}$

8 0

2 years ago

This exercise uses the radioactive decay model. After 3 days a sample of radon-222 has decayed to 58% of its original amount. (a

Vadim26 [7]

Answer: a) 3.85 days

b) 10.54 days

Explanation:-

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = ?

t = time taken for decomposition = 3 days

a = let initial amount of the reactant = 100 g

a - x = amount left after decay process = $\frac{58}{100}\times 100=58g$

First we have to calculate the rate constant, we use the formula :

Now put all the given values in above equation, we get

$k=\frac{2.303}{3}\log\frac{100}{58}$

$k=0.18days^{-1}$

a) Half-life of radon-222:

$t_{\frac{1}{2}}=\frac{0.693}{k}$

$t_{\frac{1}{2}}=\frac{0.693}{0.18}=3.85days$

Thus half-life of radon-222 is 3.85 days.

b) Time taken for the sample to decay to 15% of its original amount:

where,

k = rate constant = $0.18days^{-1}$

t = time taken for decomposition = ?

a = let initial amount of the reactant = 100 g

a - x = amount left after decay process = $\frac{15}{100}\times 100=15g$

$t=\frac{2.303}{0.18}\log\frac{100}{15}$

$t=10.54days$

Thus it will take 10.54 days for the sample to decay to 15% of its original amount.

3 0

2 years ago