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dimulka [17.4K]
3 years ago
11

A box is being pulled by two ropes. Eduardo pulls to the left with a force of 500 N, and Clara pulls to the right with a force o

f 200 N. The box moves because of the two forces applied to it. Leon records the forces and direction of the forces acting on the box in his lab notebook.
Physics
2 answers:
ZanzabumX [31]3 years ago
8 0
The box is moving in the direction of Eduardo with a force of 300N since he is pulling 300N stronger than Clara.
Ket [755]3 years ago
6 0

Answer:

C.Kinetict Friction

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A doubly ionized molecule (i.e., a molecule lacking two electrons) moving in a magnetic field experiences a magnetic force of 5.
Kobotan [32]

Explanation:

The given data is as follows.

           F = 5.75 \times 10^{-16} N

          q = 1.6 \times 10^{-19} C

          v = 385 m/s

       sin (63.9^{o}) = 0.876

Now, we will calculate the magnitude of magnetic field as follows.

              B = \frac{F}{qv sin (\theta)}

                  = \frac{5.75 \times 10^{-16} N}{1.6 \times 10^{-19} C \times 385 m/s \times 0.876}

                  = 0.01065 \times 10^{3} T

                  = 10.65 T

Thus, we can conclude that magnitude of the magnetic field is 10.65 T.

4 0
3 years ago
A horizontal uniform bar of mass 3 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to the
kirill115 [55]

Answer:

T₁ = 2.8125 N

Explanation:

The equilibrium equation of the moments at the point where string 2 is located on the bar is like this:

∑M₂ = 0

M₂ = F*d

Where:

∑M₂  : Algebraic sum of moments in the the point (2) of the bar

M₂ : moment in the point 2 ( N*m)

F  : Force ( N)

d  : Horizontal distance of the force to the point 2 ( N*m

Data

mb = 3 kg : mass of the  bar

mm = 1.5 kg :  mass of the  monkey

L = 3m : lengt of the bar

g = 9.8 m/s²: acceleration due to gravity

Forces acting on the bar

T₁ : Tension in string 1 (vertical upward)

T₂ : Tension in string 2 (vertical upward)

Wb :Weihgt of the bar (vertical downward)

Wm: Weihgt of the monkey  (vertical downward)

Calculation of the weight of the bar (Wb) and of the monkey(Wm)

Wb = m*g = 3 kg*9.8 m/s² = 29.4 N

Wm = m*g = 1.5 kg*9.8 m/s² = 14.7 N

Calculation of the distances  from forces the point 2

d₁₂ = (3-0.6) m = 2.4m  : Distance from T1 to the point 2

db₂ = (3÷2) m = 1.5 m : Distance from Wb to the point 2

dm₂ = (3÷2) m = 1.5 m : Distance from Wm to the point 2

Equilibrium  of moments at the point  2 on the bar

∑M₂ = 0

T₁(d₁₂) - Wb(db₂) - Wm(dm₂) = 0

T₁(2.4) -3*(1.5) - 1.5*(1.5) = 0

T₁(2.4) =3*(1.5) + 1.5*(1.5)

T₁(2.4) =6.75

T₁ = 6.75 / (2.4)

T₁ = 2.8125 N

5 0
2 years ago
Please help me it's URGENT (#10 btw and also how do I solve for time​
Afina-wow [57]

Answer:

t = 12s

Explanation:

Given:

v-initial = 0 m/s

x = 360 m

a = 5.0 m/s^2

Solve:

x = (v-initial)t + 1/2(a*t^2)

360 = 0t + 1/2 (5.0t^2)

360 = 2.5 t^2

144 = t^2

t = sqrt(144) = 12

Therefore, it takes 12 seconds.

5 0
3 years ago
a 1.05 kg water bottle is sitting on the teacher's desk it is .20 m tall and has a radius of .03 m find the force that the water
arsen [322]
The force applied would be 1.05*9.8 = 10.3 N
the pressure is equal to F/a
area will be πr^2 = 0.002826
thus pressure will be = 10.3/0.002826= 3644.72 N/m^2
4 0
3 years ago
About how many times greater is the density of a neutron star compared to a white dwarf?
irina1246 [14]

Answer:

Explanation:

over a million times

7 0
2 years ago
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