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3 years ago

What is weight in Newton’s, of a 50.-kg person on earth

Physics

1 answer:

Svetlanka [38]3 years ago

8 0

Answer:

It is about 490 Newtons. 490.3325 to be exact

Explanation:

Pls mark as the Brainliest answer. Thank You very much, enjoy your day

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A uniform disk has a moment of inertia that is (1/2)MR2. A uniform disk of mass 13 kg, thickness 0.3 m, and radius 0.2 m is loca

kicyunya [14]

The angular momentum of an object is equal to the product of its moment of inertia and angular velocity.
L = Iω
I = 1/2 MR²
I = 1/2 x 13 x (0.2)
I = 1.3

ω = 2π/t
ω = 2π/0.3
ω = 20.9

L = 1.3 x 20.9
= 27.2 kgm²/s

6 0

4 years ago

Read 2 more answers

A cart, which has a mass of m - 2.50 kg, is sitting at the top of an inclined plane which is 3.30 meters long and which meets th

SashulF [63]

Answer:

Part(i) the time taken for this cart to reach the bottom of the inclined plane is 1.457 s

Part(ii) the velocity of the cart when it reaches the bottom of the inclined plane is 4.531 m/s

Part(iii) the kinetic energy of the cart when it reaches the bottom of the inclined plane is 25.663 J

Explanation:

Given;

mass of the cart = 2.5 kg

angle of inclination, β = 18.5⁰

length of inclined plane = 3.3m

Part(i) the time taken for this cart to reach the bottom of the inclined plane

s = ut + ¹/₂×at²

initial vertical velocity, u = 0

s = 3.3 m

s = ¹/₂×at²

$t = \sqrt{\frac{2s}{a} }$

acceleration, of the cart, a = gsinβ

a = 9.8sin(18.5) = 3.11 m/s²

$t = \sqrt{\frac{2X3.3}{3.11 }}= 1.457 s$

Part(ii) the velocity of the cart when it reaches the bottom of the inclined plane

V = a×t

V = 3.11 × 1.457 = 4.531 m/s

Part(iii) the kinetic energy of the cart when it reaches the bottom of the inclined plane

KE = ¹/₂MV²

= ¹/₂ × 2.5× (4.531)²

= 25.663 J

6 0

4 years ago

You run 100 meters in 15 seconds. What is your speed in m/s? and show your work

muminat

6.67 m/s. This can be derived by taking the distance divided by time

4 0

4 years ago

Please help me promise would mark you brainiest

Triss [41]

B. Please mark me the brilliantest pls

3 0

3 years ago

A stone with a weight of 5.29 N is launched vertically from ground level with an initial speed of 26.0 m/s, and the air drag on

makkiz [27]

Answer:

a) 19.4 m/s

b) 19 m/s

Explanation:

a) In the given question,

the potential energy at the initial point = Ui = 0

the potential energy at the final point = Uf = mgh

the kinetic energy at the initial point = Ki = 1/2 mv₀².

the kinetic energy at the final point = Kf = 0

work done by air= Ea= fh = 0.262 N

Now, using the law of conservation of energy

initial energy= final energy

Ki +Ui = Kf + Uf +Ea

1/2 mv₀² + 0 = 0 + mgh + fh

1/2 mv₀² = mgh + fh

h = v₀²/ 2g (1 +f/w)

calculate m

m= w/g = 5.29 /9.8

= 0.54 kg

h = 20 ²/ (2 x9.80) x (1 0.265/5.29)

h = 19.4 m.

b) 1/2 mv² + 2fh = 1/2 mv₀²

Vg = 19 m/s

6 0

3 years ago