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Leni [432]
3 years ago
9

if one paperclip has the mass of 1 gram and 1000 paperclips has a mass of 1 kilogram how many kilograms are 8000 paperclips ?

Physics
1 answer:
elixir [45]3 years ago
4 0
8 kilograms. Hope I've helped you
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ndicate whether the statement is true or false. Whenever we move, we alter the rate at which we move into the future. A. True B.
Ksju [112]
Whenever we move, we alter the rate at which we move into the future. This statement is true. 
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3 years ago
A boy is playing with a ball of mass 72g attached to a string. He is moving it at constant speed in a horizontal circle of radiu
irina [24]

Answer:

Explanation:

radius of circle r = 0.9 m.

(a ) In a motion on circular path , work done is zero because force ( centripetal force ) acts perpendicular to displacement .

( b )

Tension in string T = m ω²r

Putting the values

60 = .072 x ω² x 0.9

ω² = 926

ω = 30.4 rad /s

angle made in 20 revolutions θ = 20 x 2π = 126.6 rad

time taken = θ / ω

= 126.6 / 30.4

= 4.16 s .

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3 years ago
The acceleration due to gravity on the moon is about 1/6 of the acceleration due to gravity on the earth. A net force F acts hor
uysha [10]

Answer:

c.a_m

Explanation:

We are given that

Acceleration due to gravity on the moon=a_m

Acceleration due to gravity on the earth=a_e

g_m=\frac{1}{6}g_e

Net force due to am on an object on moon=F_{net}=ma_m

There is no friction and no drag force and there is no gravity involved

Then, the force acting on an object on earth=F=ma_e

F=F_{net}(given)

ma_m=ma_e

a_e=a_m

Hence, option c is true.

3 0
3 years ago
80
Dahasolnce [82]

Answer:

1 million hahahahahahahahhahah

7 0
3 years ago
A 49 kg bear slides, from rest, 11 m down a lodgepole pine tree, moving with a speed of 3.3 m/s just before hitting the ground.
hichkok12 [17]

Answer:

a) \Delta U_g=-5.3kJ

b) K=0.27kJ

c) F_f=0.45kN

Explanation:

the gravitational potential energy is given by:

U_g=m.g.h\\

\Delta U_g=m.g.h_f-m.g.h_i\\\Delta U_g=49kg*9.8m/s^2*(0m-11m)\\\Delta U_g=-5.3kJ

The kinetic energy is given by:

K=\frac{1}{2}m.v^2\\

the initial kinetic energy is zero because the motion started from rest, so:

K=\frac{1}{2}*49kg*(3.3m/s^2)^2\\K=0.27kJ

applying the conservation of energy theorem:

U_g-W_f=K_f\\W_f=-(\Delta K+\Delta U)\\W_F=5.3kJ-0.27kJ\\W_F=-5.0kJ

The work done by the friction force is given by:

W_f=F_f.h.cos(\theta)\\

the angle of the force is 180 degrees because it's against the movement:

F_f=\frac{W_f}{h.cos(\theta)}\\\\F_f=\frac{-5.0kJ}{11m.cos(180^o)}\\\\F_f=0.45kN

8 0
3 years ago
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