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2 years ago

Q3. Recall the equation that links gravitational field strength, mass and weight

Physics

1 answer:

Ierofanga [76]2 years ago

7 0

Gravitational potential energy = mass x gravitational field strength (g) x height

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A cow wanders 30 m North, turns 22 degrees right of its original path, and wanders another 40 m. Find its total displacement.

scoundrel [369]

Answer:OB=58.3m

Explanation:

So here cow wanders 30m in north and turns 22 degrees in right side and moves 40m more, as shown in figure given.

now take the starting point as a origin such that cow moves in x-y co-ordinate axis.

As shown in figure length OA is the length when cow moves in north or y direction. Later she takes 22 degrees turn to right and moves 40m more.

So the final displacement is the length of cow from the origin that is length OB.

now co-ordinates of B are [40cos22°,40sin22°+30] i.e [37.084,44.984]

now displacement of cow= length of OB

= $\sqrt{[37.084]^{2}+[44.984]^{2} }$

= $\sqrt{3398.78}$

OB = $58.3$

7 0

3 months ago

A circular sign has a diameter of 40 cm and is subjected to normal winds up to 150 km/h at 10°C and 100 kPa. Determine the drag

Marat540 [252]

Answer:

147.7 N

221.55 Nm

Explanation:

P = Pressure = 100000 Pa

$R_s$ = Mass-specific gas constant = 287.015 J/kg k

T = Temperature = 10+273 = 283 K

C = Drag coefficient = 1.1

A = Area

r = Radius = 0.2 m

v = Speed of wind = $\frac{150}{3.6}\ m/s$

L = Length of pole

Density

$\rho=\frac{P}{R_sT}\\\Rightarrow \rho=\frac{100000}{287.058\times 283}\\\Rightarrow \rho=1.2309\ kg/m^3$

Drag force

$F=\frac{1}{2}\rho CAv^2\\\Rightarrow F=\frac{1}{2}\times 1.2309\times 1.1\times \left(\pi \times 0.2^2\right)\times \left(\frac{150}{3.6}\right)^2\\\Rightarrow F=147.7\ N$

Force on the circular sign is 147.7 N

$M=F\times L\\\Rightarrow M=147.7\times 1.5\\\Rightarrow M=221.55\ Nm$

Bending moment at the bottom of the pole is 221.55 Nm

6 0

2 years ago

Can anyone help me with this question please

JulsSmile [24]

Explanation:

V=u+at

where,

v=final speed

u=initial speed,(starting speed)

a=acceleration

t=time

v=u+at = 6=2+a*2

6=2+2a

2a=6-2

2a=4

a=4/2 = 2

a =2

2. to find time taken

v=u+at

25=5*2t

2t=25-5

2t=20

t=20/2

t=10sec

3. finding final speed

v=u+at

v=4+10*2

=4+20

v=24m/sec

5.v=u+at

=5+8*10

=5+80

V=85m/sev

6. v=u+at

8=u+4*2

8=u+8

U=8/8

u=1

these are your missing values

5 0

2 years ago

Find the period of the leg of a man who is 1.83 m in height with a mass of 67 kg. The moment of inertia of a cylinder rotating a

cupoosta [38]

Answer:

2.2 s

Explanation:

Using the equation for the period of a physical pendulum, T = 2π√(I/mgh) where I = moment of inertia of leg about perpendicular axis at one point = mL²/3 where m = mass of man = 67 kg and L = height of man = 1.83 m, g = acceleration due to gravity = 9.8 m/s² and h = distance of leg from center of gravity of man = L/2 (center of gravity of a cylinder)

So, T = 2π√(I/mgh)

T = 2π√(mL²/3 /mgL/2)

T = 2π√(2L/3g)

substituting the values of the variables into the equation, we have

T = 2π√(2L/3g)

T = 2π√(2 × 1.83 m/(3 × 9.8 m/s² ))

T = 2π√(3.66 m/(29.4 m/s² ))

T = 2π√(0.1245 s² ))

T = 2π(0.353 s)

T = 2.22 s

T ≅ 2.2 s

So, the period of the man's leg is 2.2 s

7 0

2 years ago

What is the length of 0.005 mm in kilometers?

ZanzabumX [31]

The answer would be $5*10^9
$ .

If we divide $\frac{5*10^9}{1*10^6}$ , it gives us 0.005!

Hope this helped! c:

4 0

2 years ago