Answer:
False
whenever the string breaks, the ball will follow the straight line tangential path
Explanation:
No, the ball will not follow a curved path after the string breaks. Since, the the direction of velocity is tangential to each point of the circular motion. Therefore, it changes at every point. This produces an acceleration in the circle called centripetal acceleration. There is also a tangential component of acceleration acting on the ball during this motion.
So, whenever the string breaks, the ball will follow the straight line tangential path. Hence, the given statement is false.
Answer:
Explanation:
From the question we are told that:
Mass
Speed
Acceleration Time
Constant speed Time
Deceleration time
Generally the equation for Acceleration is mathematically given by
Therefore acceleration for the first 0.80 sec is
Therefore
Spring Reading=Normal force -Reaction
Answer:
His trip took 5.78 seconds
Explanation:
23.7m divided by 4.1m/s = 5.78048780488
B.meter
C.Rate
D.Speed
E.velocity
F.slope
G.refrence point
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Speed can be calculated if you know the distance that an object travels in one unit of time, therefore the correct answer is option D.
The total distance covered by any object per unit of time is known as speed. It depends only on the magnitude of the moving object.
The unit of speed is a meter/second. The generally considered unit for speed is a meter per second.
Thus, Speed can be calculated if you know the distance that an object travels in one unit of time, therefore the correct answer is option D.
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Answer:
D.Speed
Explanation:
The speed of an object is the distance the object travels in one unit of time.
Answer:
Explanation:
Given
density of cylinder is
Length of first cylinder is
radius
For cylinder 2
and are the height above water
E
as object is floating so its weight must be balanced with buoyant force
For 2nd cylinder
Dividing 1 and 2 we get
Answer:
first of all since the value of mass of the object and force F(horizontal) or the relation between them is not mentioned , so there can be multiple effects and let us study them in cases.
Let us consider that mass of the object is m and the coefficient of friction is the coefficient of static friction of the surface{since friction can be static as well as dynamic} whose value(μ) as is given in the question is 0.2. Since the surface and the force applied are both horizontal so there will be no vertical component of the force applied which means that the normal force is N=weight of the object=mg where g is the acceleration due to gravity(9.8 m/s^2 on average on surface of earth).
Case 1:
When F<=μN. This means that when the value of F is less than or equal to μN then the static friction is equal to F. So the body remains stationary.
Case 2:
When F>μN. This means that when the value of F is greater than μN then the static friction is less than F. So the body starts moving and if the force applied F is constant then the body will start accelerating because coefficient of kinetic friction is less than that of static friction.
Explanation:
When an object rests on a horizontal floor. the coefficient of static friction is 0.4 and the acceleration of gravity is 9.8m/s, then the maximum force of friction would be 1.962M and the minimum force of friction would be zero when no external force is applied, where M represents the mass of the object
Friction is a type of force that resists or prevents the relative motion of two physical objects when their surfaces come in contact.
When an object is beginning to move as a result of an external force, limiting friction is involved. This frictional force balances the external force until the object is in a condition where it remains stationary.
maximum Frictional force= μN
where μ is the coefficient of static friction
N is the normal reaction force having a value of mg
Thus, The maximum force of friction would be 1.962M, and the minimum force would be zero
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Answer:
E = 9.4 10⁶ N / C, The field goes from the inner cylinder to the outside
Explanation:
The best way to work this problem is with Gauss's law
Ф = E. dA = qint / ε₀
We must define a Gaussian surface, which takes advantage of the symmetry of the problem. We select a cylinder with the faces perpendicular to the coaxial.
The flow on the faces is zero, since the field goes in the radial direction of the cylinders.
The area of the cylinder is the length of the circle along the length of the cable
dA = 2π dr L
A = 2π r L
They indicate that the distance at which we must calculate the field is
r = 5 R₁
r = 5 1.3
r = 6.5 mm
The radius of the outer shell is
r₂ = 10 R₁
r₂ = 10 1.3
r₂ = 13 mm
r₂ > r
When comparing these two values we see that the field must be calculated between the two housings.
Gauss's law states that the charge is on the outside of the Gaussian surface does not contribute to the field, the charged on the inside of the surface is
λ = q / L
Qint = λ L
Let's replace
E 2π r L = λ L /ε₀
E = 1 / 2piε₀ λ / r
Let's calculate
E = 1 / 2pi 8.85 10⁻¹² 3.4 10-12 / 6.5 10-3
E = 9.4 10⁶ N / C
The field goes from the inner cylinder to the outside