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Mnenie [13.5K]
4 years ago
12

Please please i want the awnser directly

Physics
2 answers:
disa [49]4 years ago
6 0
Well to start Off It Will Move ==>  Because One student is useing more force but If they were useing the same force it will not move because there equal.

Hope this help Mark Brainliest Plz Need it to level up
Vlad1618 [11]4 years ago
6 0
IF the difference between the two forces is enough to overcome friction, then the box swill move toward the RIGHT.
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If light travels from oil to water at an angle, what happens to the direction of the light ray in water with respect to the normal, is it moves away from the normal.
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3 years ago
A 1100 kg car is traveling around a flat 82.3 m radius curve. The coefficient of static friction between the car tires and the r
Novay_Z [31]

Answer:

The maximum speed of car will be 20.5m/sec

Explanation:

We have given mass of car = 1100 kg

Radius of curve = 82.3 m

Static friction \mu _s=0.521

We have to find the maximum speed of car

We know that at maximum speed centripetal force will be equal to frictional force m\frac{v^2}{r}=\mu _srg

v=\sqrt{\mu _srg}=\sqrt{0.521\times 82.3\times 9.8}=20.5m/sec

So the maximum speed of car will be 20.5m/sec

8 0
3 years ago
Read 2 more answers
A particle charge of 2.7 µC is at the center of a Gaussian cube 55 cm on edge. What is the net electric flux through the surface
Alexxx [7]

Answer:

3.05×10⁵ Nm²C⁻¹

Explanation:

According to Gauss' law,

∅' = q/e₀............... Equation 1

Where ∅' = net flux through the surface, q = net charge, e₀ = electric permittivity of the space

From the question,

Given: q = 2.7 μC = 2.7×10⁻⁶ C,

Constant: e₀ = 8.85×10⁻¹² C²/N.m²

Substituting these values into equation 1

∅' = (2.7×10⁻⁶)/(8.85×10⁻¹²)

∅' = 3.05×10⁵ Nm²C⁻¹

3 0
3 years ago
A team of eight dogs pulls a sled with waxed wood runners on wet snow (mush!). The dogs have average masses of 19.0 kg, and the
aliina [53]

Answer:

a) a = 3.29 m/s²

b) a = 3.51 m/s²

Explanation:

mass of dogs = 19 kg

loaded sled mass = 210 Kg

a)writing all the forces in x direction

\sum F_x=8F_d-f_{s(max)}= (m_s+8m_d)a\\a=\dfrac{8F_d-\mu_sm_sg}{m_s+8m_d}\\a=\dfrac{8\times 185-(0.14)(210)(9.81)}{210+8(19)}

a = 3.29 m/s²

b) acceleration  when sled start to move the friction will now be acting will be kinetic friction.  

a=\dfrac{8F_d-\mu_sm_sg}{m_s+8m_d}\\a=\dfrac{8\times 185-(0.1)(210)(9.81)}{210+8(19)}

a = 3.51 m/s²

3 0
3 years ago
2 diferencias y 2 similitudes entre la actividad física y el ejercicio físico.
blondinia [14]

Answer:

En el deporte, la persona que lo práctica tiene que adecuarse a él, por el contrario cuando hablamos de actividad física el objetivo pasa por programar, dosificar y adecuar en entrenamiento a las características, necesidades y objetivos de cada persona.

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