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anastassius [24]
4 years ago
15

Two lifeguards pull on ropes attached to a raft. If they pull in the same direction, the raft experiences a net external force o

f 300 N to the right. If they pull in opposite directions, the raft experiences a net external force of 115 N to the left.
Physics
1 answer:
makkiz [27]4 years ago
6 0

F_1=207.5\ N

F_2=92.5\ N    

Explanation:

Let F_1\ and\ F_2 is the two forces acting on the raft. If they pull in the same direction, the net force is given by :

F_1+F_2=300...............(1)

If they pull in opposite direction, net force is given by :

F_1-F_2=115...............(2)

It is required to find the magnitude of both forces. On solving equation (1) and (2) we get :

F_1+F_2=300

F_1-F_2=115

F_1=207.5\ N

F_2=92.5\ N          

So, the forces are 207.5 N and 92.5 N. Hence, this is the required solution.                  

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Stacey runs around a 400 meter track 2 times in 5 minutes. How does her displacement differ from her distance? *
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Answer:

Distance is 800 m and Displacement is 0 m

Explanation:

Total Distance

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= <u>800</u><u> </u><u>m</u>

Total Displacement

= <u>0</u><u> </u><u>m</u> since she returns to the same spot

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Unlike fireflies and glow worms, most objects don't make their own light. So how can you see what's around you?
VashaNatasha [74]
Light reflects off the cornea in the back of your eye thats what makes you see light 
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4 years ago
Hooke's law describes an ideal spring. Many real springs are better described by the restoring force (F Sp ) s =−kΔs−q(Δs) 3 (p)
Montano1993 [528]

Answer with Explanation:

We are given that

Restoring force,(FS_p)s=-k\Delta s-q(\Delta s)^3

k=350N/m

q=750 N/m^3

We have to find the work must you do to compress this spring 15 cm.

\Delta s=15 cm=0.15 m

Using 1 m=100 cm

Work done=\int_{0}^{0.15}-Fd(\Delta s)

W=-\int_{0}^{0.15}(-k\Delta s-q(\Delta s)^3))d(\Delta s)

W=k[\frac{(\Delta s)^2}{2}]^{0.15}_{0}+q[\frac{(\Delta s)^4}{4}]^{0.15}_{0}

W=0.01125k+0.000127q=0.01125\times 350+0.000127\times 750

W=4.033 J

Ideal spring work=0.5k(\Delta s)^2=0.5\times 350\times (0.15)^2=3.938 J

Percentage increase in work=\frac{4.033-3.938}{3.928}\times 100=2.4%

6 0
4 years ago
The current in a hair dryer measures 15 amps the resistance of the hair dryer is 8 ohms
suter [353]

The <u><em>voltage</em></u> at the input to the hair dryer is

V = (current) · (resistance)

V = (15 A) · (8 Ω)

<em>V = 120 volts</em>

<em>__________________________________</em>

The <em><u>power</u></em> dissipated by the hair dryer is

P = (current)²· (resistance)

P = (15 A)² · (8 Ω)

<em>P = 1800 watts </em>

<em />

3 0
3 years ago
What is the pressure exerted by a column of fresh water 10m high<br>(p=1000kg/m3 and g=10m/s)​
alexira [117]
  • Height (h) = 10 m
  • Density (ρ) = 1000 Kg/m^3
  • Acceleration due to gravity (g) = 10 m/s^2
  • We know, pressure in a fluid = hρg
  • Therefore, the pressure exerted by a column of fresh water
  • = hρg
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<u>Answer</u><u>:</u>

<u>1000</u><u>0</u><u>0</u><u> </u><u>Pa</u>

Hope you could understand.

If you have any query, feel free to ask.

8 0
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