Answer:
The electric field strength is 
Explanation:
Given that,
Magnetic field = 0.150 T
Speed 
We need to calculate the electric field strength
Using formula of velocity


Where, v = speed
B = magnetic field
Put the value into the formula



Hence, The electric field strength is 
Answer:
38 cm from q1(right)
Explanation:
Given, q1 = 3q2 , r = 60cm = 0.6 m
Let that point be situated at a distance of 'x' m from q1.
Electric field must be same from both sides to be in equilibrium(where EF is 0).
=> k q1/x² = k q2/(0.6 - x)²
=> q1(0.6 - x)² = q2(x)²
=> 3q2(0.6 - x)² = q2(x)²
=> 3(0.6 - x)² = x²
=> √3(0.6 - x) = ± x
=> 0.6√3 = x(1 + √3)
=> 1.03/2.73 = x
≈ 0.38 m = 38 cm = x
Answer:

It will float.
Explanation:
Hello.
In this case, given the width, length and height, we can compute the volume as follows:

Moreover, since the density is computed via the division of the mass by the volume:

We obtain:

In such a way, since the solid has a lower density than the water, we infer it will float.
Best regards.
Answer:
a) 70 N, b) b. Each initially applied a force bigger than static friction to get the box moving and accelerating, then when the desired final speed was achieved they reduced the force to make the net force zero.
Explanation:
a) A constant speed means that magnitude of friction force is equal to the magnitude of the external force. The friction force is directly proportional to the normal force, which is equal to the weight of the box. Therefore, the magnitude of the force is 70 N.
b) Alice used initially a greater force to accelerate the box up to needed speed and later reduced the external force to keep speed constant. The right choice is option b.
This seems like a calculus problem. I'm assuming you would use cos and sin. so here's the vertical component +10.0m/s multiplied by sin60 = 8.66 rounded to the hundreths place. Now for horizontal, that would be +10.0m/s multiplied by cos60 = 5. hope this helped.