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SpyIntel [72]
2 years ago
5

If you know how to SHiFT in sleep how do you do it

Physics
1 answer:
AleksandrR [38]2 years ago
8 0
You write the terms and conditions in your journal
You might be interested in
If a boat and its riders have a mass of 900 kg and the boat drifts in at 1.4 m/s how much work does sam do to stop it?
Rama09 [41]
The initial kinetic energy of the boat and its rider is
K_i =  \frac{1}{2} mv_i^2 =  \frac{1}{2}(900 kg)(1.4 m/s)^2=882 J

After Sam stops it, the final kinetic energy of the boat+rider is
K_f = 0 J
because its final velocity is zero.

For the law of conservation of energy, the work done by Sam is the variation of kinetic energy of the system:
W=K_f-K_i =0-882 J=-882 J
where the negative sign is due to the fact that the force Sam is applying goes against the direction of motion of the boat.
6 0
3 years ago
I need help with these questions
Feliz [49]
7. PE=0.5×700n/m×0.9m^2
0.9^2=0.81m
0.5×700×0.81= 283.5J

8. 2000=0.5×(x)×1.5m^2
1.5^2= 0.25
0.25×0.5=0.125
2000=0.125 (x)
2000/0.125=x
x=16000 n/m

9. 4000=0.5 (375 n/m)×(x)^2
0.5×187.5 (x)
4000/187.5=21.3333333333


6 0
3 years ago
Henry slides across an icy pond. The coefficient of kinetic friction betweenhis shoes and the ice is 0.09. If his mass is 115 kg
Kruka [31]

Answer:

101 N

Explanation:

4 0
2 years ago
A block-and-tackle pulley hoist is suspended in a warehouse by ropes of lengths 2 m and 3 m. the hoist weighs 430 n. the ropes,
ICE Princess25 [194]
Refer to the diagram shown below.

For horizontal equilibrium,
T₃ cos38 = T₂ cos 50
0.788 T₃ = 0.6428 T₂
T₃ = 0.8157 T₂                (1)

For vertical equilibrium,
T₂ sin 50 + T₃ sin 38 = 430
0.766 T₂ + 0.6157 T₃ = 430
1.2441 T₂ + T₃ = 698.392        (2)

Substitute (1) into (2).
(1.2441 + 0.8157) T₂ = 698.392
T₂ = 339.058 N
T₃ = 0.8157(399.058) = 276.571 N

Answer:
T₂ = 339.06 N
T₃ = 276.57 N

7 0
2 years ago
Determine the acceleration due to gravity for low Earth orbit (LEO) given: MEarth = 6.00 x 1024 kg, rEarth = 6.40 x 106 m, G = 6
Nana76 [90]

Answer:

The answer to the question is as follows

The  acceleration due to gravity for low for orbit is  9.231 m/s²

Explanation:

The gravitational force is given as

F_{G}= \frac{Gm_{1} m_{2}}{r^{2} }

Where F_{G} = Gravitational force

G = Gravitational constant = 6.67×10⁻¹¹\frac{Nm^{2} }{kg^{2} }

m₁ = mEarth = mass of Earth = 6×10²⁴ kg

m₂ = The other mass which is acted upon by  F_{G} and = 1 kg

rEarth = The distance between the two masses = 6.40 x 10⁶ m

therefore at a height of 400 km above the erth we have

r = 400 + rEarth = 400 + 6.40 x 10⁶ m = 6.80 x 10⁶ m

and  F_{G} = \frac{6.67*10^{-11} *6.40*10^{24} *1}{(6.8*10^{6})^{2} } = 9.231 N

Therefore the acceleration due to gravity =  F_{G} /mass  

9.231/1 or 9.231 m/s²

Therefore the acceleration due to gravity at 400 kn above the Earth's surface is  9.231 m/s²

4 0
3 years ago
Read 2 more answers
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