An object is said to move from a position of 10m East to a position of 5m west. Determine the object's distance travelled.

Consider a perfectly reflecting mirror oriented so that solar radiation of intensity i is incident upon, and perpendicular to, t

podryga [215]

The answer is Frad<span> = 2IA/c.</span>

4 0

3 years ago

Read 2 more answers

A long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so tha

Dvinal [7]

Answer:

0.087 m

Explanation:

Length of the rod, L = 1.5 m

Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.

time period, T = 3 s

the formula for the time period of the pendulum is given by

$T = 2\pi \sqrt{\frac{I}{mgd}}$ .... (1)

where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.

Moment of inertia of the rod about the centre of mass, Ic = mL²/12

By using the parallel axis theorem, the moment of inertia of the rod about the pivot is

I = Ic + md²

$I = \frac{mL^{2}}{12}+ md^{2}$

Substituting the values in equation (1)

$3 = 2 \pi \sqrt{\frac{\frac{mL^{2}}{12}+ md^{2}}{mgd}}$

$9=4\pi^{2}\times \left ( \frac{\frac{L^{2}}{12}+d^{2}}{gd} \right )$

12d² -26.84 d + 2.25 = 0

$d=\frac{26.84\pm \sqrt{26.84^{2}-4\times 12\times 2.25}}{24}$

$d=\frac{26.84\pm 24.75}{24}$

d = 2.15 m , 0.087 m

d cannot be more than L/2, so the value of d is 0.087 m.

Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.

3 0

2 years ago

A gas has an initial volume of 2.5 L at a temperature of 275 K and a pressure of 2.1 atm. The pressure of the gas increases to 2

olchik [2.2K]

Answer:

2.10L

Explanation:

Given data

V1= 2.5L

T1= 275K

P1= 2.1atm

P2= 2.7 atm

T2= 298K

V2= ???

Let us apply the gas equation

P1V1/T1= P2V2/T2

substitute into the expression we have

2.1*2.5/275= 2.7*V2/298

5.25/275= 2.7*V2/298

Cross multiply

275*2.7V2= 298*5.25

742.5V2= 1564.5

V2= 1564.5/742.5

V2= 2.10L

Hence the final volume is 2.10L

7 0

3 years ago

Can anyone help ????

user100 [1]

Explanation:

1. serie

2.Red

3.16 ohms

4.

I=V/R

6=V/16

V=16×6

V=96 volts

5 0

3 years ago

As shown in the diagram below, a rope attached to a 500.-kilogram crate is used to exert a force of 45 newtons at an angle of 65

kaheart [24]

Answer:

19.01 N

Explanation:

F = Force being applied to the crate = 45 N

$\theta$ = Angle at which the force is being applied = $65^{\circ}$

Horizontal component of force is given by

$F_x=F\cos\theta\\\Rightarrow F_x=45\times \cos65^{\circ}\\\Rightarrow F_x=19.01\ \text{N}$

The horizontal component of the force acting on the crate is 19.01 N.

4 0

2 years ago