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navik [9.2K]
3 years ago
14

If you pull on something to the right with a force of 50 N and I pull on the same object to the left with a force of 60 N, what

will happen?
The object will accelerate toward you
The object will accelerate toward me
The object will not accelerate at all
It is impossible to pull an object in two different directions
Physics
1 answer:
EleoNora [17]3 years ago
7 0
Nxxncnx. NMakalsllalakskdnd cxhcnmdnx
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Suppose that you are swimming in a river while a friend watches from the shore. In calm water, you swim at a speed of 1.25 m/s .
aliya0001 [1]

Answer: The observing friend will the swimmer moving at a speed of 0.25 m/s.

Explanation:

  • Let <em>S</em> be the speed of the swimmer, given as 1.25 m/s
  • Let S_{0} be the speed of the river's current given as 1.00 m/s.

  • Note that this speed is the magnitude of the velocity which is a vector quantity.
  • The direction of the swimmer is upstream.

Hence the resultant velocity is given as, S_{R} = S — S 0S_{0}

S_{R} = 1.25 — 1

S_{R} = 0.25 m/s.

Therefore, the observing friend will see the swimmer moving at a speed of 0.25 m/s due to resistance produced by the current of the river.

6 0
3 years ago
A cheetah can accelerate from rest at the rate of 4m /s
lesya [120]
Acceleration of cheetah (a) = 4m/s²
time = 10s
initial velocity(u) = 0
final velocity = v
distance travelled = s

v = u +at = 0 + 10×4 = 40m/s
s = (v²-u²)/2a = 40²/(2×4) = 1600/8 = 200m
   
6 0
3 years ago
WILL UPVOTE!!!Physics help please!!
liubo4ka [24]
Speed v = initial speed u + acceleration a x time t 
v=u+at = 2 + 4*3 = 14 m/s

8 0
3 years ago
Agri-Chem's contract with Enerco specified a maximum of 90,000 cu.ft. X 103 per day for its complexes. However, curtailments are
zvonat [6]

Answer:

Caustic soda would be least affected by a gas curtailment

Explanation:

Let,

X1= ammonia

X2= ammonium phosphate

X3= ammonium nitrate

X4 = Urea

X5= hydro-fluoric acid

X6= Chlorine

X7= Caustic soda

X8= Vinyl chloride monomer.

Agri chem’s current natural gas usage

= (1200 × 8 + 540 × 10 + 490 × 12 + …)                  

= 85,680,000 cu. ft. per day  

When, the curtailment is 20%, availability is

= 0.8 X  85,680

= 68,554,000 cu. ft. per day

Therefore, the gas constant

= 8X1 + 10X2 + 12X3 + 12X4 + 7X5 + 18X6 + 20X7 + 14X8 ≤68,544

When, the curtailment is 40% availability is

= 0.6 X 85,680

= 51,408,000 cu. ft. per day

Constant = 8X1 + 10X2 + 12X3 + 12X4 + 7X5 + 18X6 + 20X7 + 14X8≤ 51,408

                     

7 0
3 years ago
A 57.0 kg cheerleader uses an oil-filled hydraulic lift to hold four 120 kg football players at a height of 1.10 m. If her pisto
lapo4ka [179]

Answer:

The diameter of the piston of the players equals 55.136 cm.

Explanation:

from the principle of transmission of pressure in a hydraulic lift  we have

\frac{F_{1}}{A_{1}}=\frac{F_{2}}{A_{1}}

Since the force in the question is the weight of the individuals thus upon putting the values in the above equation we get

\frac{57.0\times 9.81}{\frac{\pi \times (19.0)^{2}}{4}}=\frac{4\times 120\times 9.81}{\frac{\pi \times D_{2}^{2}}{4}}

Solving for D_{2} we get

D_{2}^{2}=\frac{4\times 120}{57}\times 19^{2}\\\\\therefore D_{2}=\sqrt{\frac{4\times 120}{57}}\times 19\\\\D_{2}=55.136cm

5 0
3 years ago
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