The work done to stretch the spring will be 112 J.
<h3>What is spring force?</h3>
The force required to extend or compress a spring by some distance scales linearly with respect to that distance is known as the spring force. Its formula is
F = kx
The given data in the problem is;
F is the spring force =?
K is the spring constant= 8.5 N/m
x is the length by which spring got stretched = 1.2m
The work is done to stretch the spring is;
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There is no need for tangential acceleration when moving in a circle at a constant speed.
<h3>What is centripetal acceleration?</h3>
centripetal acceleration refers to the speed at which a body moves through a circle. Due to the fact that velocity is a vector quantity (i.e., it has both a magnitude, the speed, and a direction), when a body travels in a circle, its direction is constantly changing, which causes a change in velocity, which results in an acceleration.
<h3>Which is an example of centripetal acceleration?</h3>
Centripetal acceleration occurs when you spin a ball on a string above your head. A car experiences centripetal acceleration when it is being driven in a circle. Additionally, a satellite in orbit around the Earth experiences centripetal acceleration.
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Answer:
There is absolutely No relationship between the weight of an object (which is constant) and the frictional force. If a block is sliding on a surface, that surface will be exerting a force on the block. That force can be resolved into a component parallel to the surface (which we call the frictional component), and a component perpendicular to the surface (called the normal component). For many situations, we find experimentally that the frictional component is approximately proportional to the normal component. The frictional component divided by the normal component is defined to be a quantity called the coefficient of kinetic or sliding friction. The coefficient of kinetic friction obviously depends on the nature of the surfaces involved. The normal component on an object can be decreased if you pull in the direction of the normal component (the weight does not change). However pulling this way on the object not only decreases the normal component, but it also decreases the frictional component since they are proportional. This is why it is easier to slide something if you pull up on it while you push it. If you push down, the normal and frictional components increase so it is harder to slide the object. The weight of an object is the downward force exerted by Earth’s gravity on that object, and it does not change no matter how you push or pull on the object.
Answer:
C) 7.35*10⁶ N/C radially outward
Explanation:
- If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.
- So, we can write the following equation:
- As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
- So, the +8 μC charge of the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
- So, on the outer surface of the shell there must be a charge that be the difference between them:
- Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:
- As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.
I believe the correct response would be B. It would decrease.
