Answer:
5 m/s
Explanation:
Horizontal distance traveled, x = 2 m
vertical distance traveled, y = 4/5 m
Let the speed of cup as it leaves the counter is v and it takes time t to hit the ground.
Use second equation of motion in vertical direction
Here acceleration in vertical direction is 9.8 m/s^2.
So,
t = 0.4 second
Now in horizontal direction the acceleration in zero.
Horizontal distance = horizontal velocity x time
x = v t
2 = v (0.4)
v = 5 m/s
Thus, the horizontal velocity of cup as it leaves the counter is 5 m/s.
Answer:
The property to form polar covalent bond and hydrogen bonds
Explanation:
Liquid water exhibits the property of surface tension, functions as solvent for ionic and polar covalent bonded molecules and have high specific heat.
All these properties can be credited to the property of water to form hydrogen bonds with liquid water molecules where the hydrogen bonding results when hydrogen atom of a water molecule is attracted to a more electro negative oxygen atom of other water molecule in the neighborhood while polar covalent bond formation results in the the water molecule as it is a polar covalent molecule as it contains hydrogen atom and an electronegative oxygen atom which results in the charge distribution in the molecule.
A force is a push or pull acting upon an object as a result of its interaction with another object. There are a variety of types of forces. a variety of force types were placed into two broad category headings on the basis of whether the force resulted from the contact or non-contact of the two interacting objects.
Contact Forces
Action-at-a-Distance Forces
Frictional Force
Gravitational Force
Tensional Force
Electrical Force
Normal Force
Magnetic Force
Air Resistance Force
Applied Force
Spring Force
These are types of individual forces
Applied Force
Gravitational Force
Normal Force
Frictional Force
Air Resistance Force
Tensional Force
Spring Force
Complete Question:
A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of 25 rad/s. The wheel is run at that angular velocity for 37 s and then power is shut off. The wheel decelerates uniformly at 1.5 rads/s2 until the wheel stops. In this situation, the time interval of angular deceleration (slowing down) is closest to
Answer:
t= 16.7 sec.
Explanation:
As we are told that the wheel is accelerating uniformly, we can apply the definition of angular acceleration to its value:
γ = (ωf -ω₀) / t
If the wheel was at rest at t-= 0.00 s, the angular acceleration is given by the following equation:
γ = ωf / t = 25 rad/sec / 10 sec = 2.5 rad/sec².
When the power is shut off, as the deceleration is uniform, we can apply the same equation as above, with ωf = 0, and ω₀ = 25 rad/sec, and γ = -1.5 rad/sec, as follows:
γ= (ωf-ω₀) /Δt⇒Δt = (0-25 rad/sec) / (-1.5 rad/sec²) = 16.7 sec
