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erastovalidia [21]
2 years ago
5

An ant is crawling up a leaf that is hangin vertically. it lifts 0:007 kg crumb up 0:12 meters. to do this, the ant exerts 0.069

newtons of force. how much work dis the ant do on the crumb?
Physics
1 answer:
Afina-wow [57]2 years ago
5 0

Answer:

.000828 j

Explanation:

Work = F * d

          .0069 N * .12 m = .000828 j

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A) Charge q1 = +5.60 nC is on the x-axis at x = 0 and an unknown charge q2 is on the x-axis at x = -4.00 cm. The total electric
jeka94

Answer:

a) F₃₁ = 63.0 μN  

b) F₃₂ = - 14.0 μN

c) q₂ = - 5.0 nC

Explanation:

a)

  • Assuming that the three charges can be taken as point charges, the forces between them must obey Coulomb's Law, and can be found independent each other, applying the superposition principle.
  • So, we can find the force that q₁ exerts along the x-axis on q₃, as follows:

       F_{31} =\frac{k*q_{1}*q_{3} }{r_{13}^{2}} = \frac{9e9Nm2/C2*5.6e-9C*2.0e-9C}{(0.04m)^{2}}  = 63.0 \mu N   (1)

b)

  • Since total force exerted by q₁ and q₂ on q₃ is 49.0 μN, we can find the force exerted only by q₂ (which is along the x-axis only too) just by difference, as follows:

      F_{32} = F_{3} - F_{31}  = 49.0\mu N  - 63.0\mu N = -14.0 \mu N  (2)

c)

  • Finally, in order to find the value of q₂, as we know the value and sign of F₃₂, we can apply again the Coulomb's Law, solving for q₂, as follows:

      q_{2}  = \frac{F_{32} * r_{23}^{2} }{k*q_{3}} = \frac{(-14\mu N)*(0.08m)^{2}}{9e9Nm2/C2* 2 nC} = - 5 nC  (3)

6 0
2 years ago
a machine with a mechanical advantage of 2.5 requires an input force of 120 newtons. What output force is applied by this machin
creativ13 [48]
If your machine has a mechanical advantage of 2.5, then WHATEVER force you apply to the input, the force at the output will be 2.5 times as great.

If you apply 1 newton to the machine's input, the output force is

                 (2.5 x 1 newton)  =  2.5 newtons.

If you apply 120 newtons to the machine's input, the output force is

                 (2.5 x 120 newtons)  =  300 newtons.

4 0
3 years ago
A force of 20 N produces an acceleration of 10 m/s² in mass m1 and an acceleration of 5 m/s² in
Scrat [10]

Explanation:

F = 20N m= m1 a=10m/s²

m=m2 a=5m/s²

F = ma

<u>for the first one</u><u>:</u><u> </u>

f=m1 × a

20 = m1 ×10

20=10m1

m1=20/10

m1=2

<u>for</u><u> </u><u>the</u><u> </u><u>second</u><u> </u><u>one</u><u> </u><u>:</u>

f=m2×a

20=m2×5

m2= 20/5

m2= 4

since F=ma

F=(m1+m2) ×a

F =(4+2)×a

F =6×a

F=20(from the question above )

20=6×a

a=20/6

a=3.33

8 0
3 years ago
Read 2 more answers
A model rocket has an acceleration of 12 m/s2. A net force of 18 N is acting on the rocket. What is the mass of the rocket? its
larisa86 [58]

Answer:

C.) 1.5 kg

Explanation:

Start with the equation:

F_n_e_t=ma

Plug in what you know, and solve:

18=m(12)\\m=1.5kg

Find matching soluation:

C.) 1.5 kg

5 0
2 years ago
If you lift a five pound object 18 inches how many joules of energy did yo use?
kogti [31]
The work is 90 as 5 times 18

4 0
2 years ago
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