Answer:
d = 4 d₀o
Explanation:
We can solve this exercise using the relationship between work and the variation of kinetic energy
W = ΔK
In that case as the car stops v_f = 0
the work is
W = -fr d
we substitute
- fr d₀ = 0 - ½ m v₀²
d₀ = ½ m v₀² / fr
now they indicate that the vehicle is coming at twice the speed
v = 2 v₀
using the same expressions we find
d = ½ m (2v₀)² / fr
d = 4 (½ m v₀² / fr)
d = 4 d₀o
Answer: 7022.2kg/m³, yes, I was cheated
Explanation:
Density of an object is defined as the ratio of the mass of the object to its volume. Mathematically;
Density = Mass/Volume
Note that the unit of both mass and volume must be standard unit.
Given mass = 0.0158kg
Dimension of the metal = 5mm×15mm×30mm
Note that 1mm = 0.001m
The volume of the metal will be
0.005×0.015×0.03
= 0.00000225m³
Density = 0.0158/0.00000225
Average density of the metal = 7022.2kg/m³
Since the standard density of Gold is 19,320kg/m³ and is higher than the density prescribed for me, it shows the I was cheated.
Take east to be the positive direction. Then the resultant force from adding <em>F</em>₁ and <em>F</em>₂ is
<em>F</em>₁ + <em>F</em>₂ = (-45 N) + 63 N = 18 N
which is positive, so it's directed east.
To this we add a third force <em>F</em>₃ such that the resultant is 12 N pointing west, making it negative, so that
18 N + <em>F</em>₃ = -12 N
<em>F</em>₃ = -30 N
So <em>F</em>₃ has a magnitude of 30 N and points west.
The general formula is: Momentum = (mass) x (speed)
I never like to just write a bunch of algebra without explaining it.
But in this particular case, there's really not much to say, and
I think the algebra will pretty well explain itself. I hope so:
Original momentum = (original mass) x (original speed)
New momentum = (2 x original mass) x (2 x original speed)
= (2) x (original mass) x (2) x (original speed)
= (2) x (2) x (original mass) x (original speed)
= (4) x (original mass) x (original speed)
= (4) x (original momentum).
Answer:
Time taken, 
Explanation:
It is given that, a small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread’s trajectory describes a cone as shown in attached figure.
From the figure,
The sum of forces in y direction is :
Sum of forces in x direction,
.............(1)
Also, 
Equation (1) becomes :
...............(2)
Let t is the time taken for the ball to rotate once around the axis. It is given by :
Put the value of T from equation (2) to the above expression:
On solving above equation :
Hence, this is the required solution.
