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lord [1]
2 years ago
5

How many times is larger than a centigram is a dekagram

Physics
1 answer:
S_A_V [24]2 years ago
8 0
A decagram is 1000 times bigger than a centigram
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Athlete’s foot is also called...
Aliun [14]

Answer:

Your answer is: D) Tinea Pedis

Explanation:

Hope this helped : )

6 0
3 years ago
How does the distance traveled by the coin compare to its displacement after ten flips?
Nataliya [291]

Answer:

The positive velocity occurs the instant the coin leaves our hand. It immediately begins slowing up until its upward velocity becomes zero at the maximum height.

Explanation:

hope helps ohjieun and jannatparia

3 0
2 years ago
Calculate the period of a satellite orbiting the Moon, 98 kmkm above the Moon's surface. Ignore effects of the Earth. The radius
beks73 [17]

Answer:

3.6*10^18s

Explanation:

To find the period of the satellite

We need to apply kephler's third law

Which is

MP² = (4π²/G) d³

d=semi-major axis which is the distance from center of moon = 98km+1740km = 1838km

where M= mass of the moon = 7.3x10^22kg

P=period

G=newtonian gravatational constant= 6.67x10^-11

To find the Period solve for P

P = √[(4π²/G M)xd³]

P=√(4 π²/6.67x10^-22*7.3x10^22kg) x (1.838x10^6m)³]

= 3.6*10^18s

6 0
3 years ago
What is the velocity (in m/s) of a 550 kg roller coaster cart at the bottom of the track if it started with 990,000 J of gravita
sesenic [268]

Answer:

60m/s

Explanation:

initial energy = final energy

g.p.e = k.e

k.e = 0.5 × mass × velocity²

g.p.e = 990000J as per Question

990000Nm = 0.5 × 550 × V²

V² = 3600

V = 60m/s

4 0
2 years ago
A slender, uniform metal rod of mass M and length lis pivoted without friction about an axis through its midpoint andperpendicul
lys-0071 [83]

Answer:

Explanation:

Moment of inertia of the metal rod pivoted in the middle

= M l² / 12

If the spring is compressed by small distance x twisting the rod by angle θ

restoring force by spring

= k x

moment of torque  about axis

= k x l /2

= k θ( l /2 )²     ( x / .5 l = θ )

=

moment of torque = moment of inertia of  rod  x angular acceleration

k θ( l /2 )²   = M l² / 12 d²θ/dt²

d²θ/dt² = 3 k/M  θ

acceleration =  ω² θ

ω² = 3 k/M

ω = √ 3 k / M

8 0
3 years ago
Read 2 more answers
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