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zavuch27 [327]
3 years ago
7

Cal ulate a moment of force o. 50 meter distance and 10 newton force​

Physics
1 answer:
lina2011 [118]3 years ago
8 0

Answer:

500N/M

Explanation:

given that

force=10N

distance=50M

moment=force*distance

=10×50=500j

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A -3.00 nc point charge is at the origin, and a second -5.50 nc point charge is on the x-axis at x = 0.800 m. find the electric
Liula [17]

The electric field produced by a single-point charge is given by

E(r)=k\frac{q}{r^2}

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge


To find the electric field at x=0.200 m, we need to find the electric field produced by each charge at that point, and then find their resultant.


1) The first charge is q=-3.00 nC=-3.00 \cdot 10^{-9} C, and it is located at x=0, so its distance from the point x=0.200 m is

r=0.200 m-0=0.2 m

Therefore, the electric field is

E_1=(8.99 \cdot 10^9 Nm^2C^{-2})\frac{(3.0 \cdot 10^{-9} C)}{(0.2 m)^2}=675 N/C

And since the charge is negative, the direction of the field is toward the charge, so toward negative x direction.


2) The second charge is q=-5.50 nC=-5.5 \cdot 10^{-9}C and it is located at x=0.800 m, so its distance from the point is

r=0.800 m-0.200 m=0.6 m

Therefore, the electric field is

E_2 = (8.99 \cdot 10^9 Nm^2C^{-2})\frac{(5.5 \cdot 10^{-9} C)}{(0.6 m)^2}=137.5 N

And since the charge is negative, the direction of the field is toward the charge, so toward positive x-direction.


3) The total electric field at x=0.200 m will be given by the difference between the two fields (because they are in opposite directions). Taking the x-positive direction as positive direction, we have

E=E_2 -E_1 =137.5 N/C/C-675 N/C=-537.5 N/C

and the sign tells us that the field is directed toward negative x-direction.

7 0
3 years ago
An electric appliance is connected by wires to a 240-volt source of voltage. If the combined resistance of the appliance and wir
uranmaximum [27]
  • Voltage=V=240V
  • Resistance=R=12ohm
  • Current be I

Apply ohms law

  • V=IR
  • 240=12I
  • I=240/12
  • I=20A
6 0
3 years ago
A van has a weight of 4000 lb and center of gravity at Gv. It carries a fixed 900 lb load which has a center of gravity at Gl. I
natulia [17]

Answer:

 x = 25 / μ     [ ft]

Explanation:

To solve this exercise we can use Newton's second law.

Let's set a reference system where the x axis is parallel to the road

Y axis  

       N_B + N_A - W_van - W_load = 0

       N_B + N_A = W_van + W_load

X axis

     fr = ma

     a = fr / m

the total mass is

        m = (W_van + W_load) / g

the friction force has the expression

      fr = μ N_{total}

      fr = μy (W_van + W_load)

we substitute

      a = μ (W_van + W_load)    \frac{g}{W_van + W_load}

      a = μ g

taking the acceleration let's use the kinematic relations where the final velocity is zero

       v² = v₀² - 2 a x

       0 = v₀² -2a x

        x = \frac{v_o^2}{2a}

        x = \frac{v_o^2}{2 \mu g}

        x = \frac{40^2}{2 \ 32 \  \mu}

        x = 25 / μ     [ ft]

5 0
3 years ago
A disk of radius 10 cm speeds up from rest. it turns 60 radians reaching an angular velocity of 15 rad/s. what was the angular a
Marianna [84]

Answer:

α = 1.875  rad/s²

t = 8 s

Explanation:

α = ω²/2θ = 15²/(2(60)) = 1.875 rad/s²

t = ω/a = 15 / 1.875 = 8 s

6 0
3 years ago
What is the binding energy (in MeV per nucleon) for the ⁶₃Li nucleus?
Strike441 [17]

Answer:

Explanation:

⁶₃Li will have 3 protons and 3 neutrons .

mass of proton in amu = 1.00727 amu

mass of neutron in amu = 1.00866 amu

mass of lithium nucleus in amu = 6.01512 amu

mass defect = 3 (  1.00727  + 1.00866 ) - 6.01512 amu

= .03267 amu

Binding energy = mass defect in amu x 931 Mev

= 30.41 MeV

binding energy per nucleon

no of nucleon = 3 + 3 = 6

binding energy per nucleon = 30.41 / 6 Mev

= 5.068 MeV .

4 0
3 years ago
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