Answer:183.94 N
Explanation:
Given
Time Period 
mass of child 
Radius 
Velocity 
Now Centripetal Force will be Balanced by Frictional Force
Centripetal Force 
therefore Friction Force is 183.94 N
Answer:
Vd = 2.42 ×10⁻⁴ m/s
Explanation:
Given: A = 3.00×10⁻⁶ m², I = 7.00 A, ρ = 2.70 g/cm³
To find Drift Velocity Vd=?
Sol
the formula is Vd = I/nqA (n is the number of charge per unit volume)
n = No. of electron in a mole ( Avogadro's No.) / Volume
Volume = Molar mass / density ( molar mass of Al =27 g)
V = 27 g / 2.70 g/cm³ = 10 cm³ = 1 × 10 ⁻⁵ m³
n= (6.02 × 10 ²³) / (1 × 10 ⁻⁵ m³)
n= 6.02 × 10 ²⁸
Now
Vd = (7A) / ( 6.02 × 10 ²⁸ × 1.6 × 10⁻¹⁹ C × 3.00×10⁻⁶ m²)
Vd = 2.42 ×10⁻⁴ m/s
Distance=20m
time=18s
Initial velocity=0m/s
Final velocity=10m/s
The equation we use to find average velocity is v=d/t (average velocity=distance/time)
V=20m/18s
V (average velocity)=1.11m/s
Answer: Truck 2, because it has a larger mass than truck 1 (and the same velocity)
Explanation:
We can write the kinetic energy of an object with mass M and velocity V as:
K = (M/2)*V^2
Let's calculate the kinetic energy for each truck:
Now, we can see that both trucks move at the same velocity, but truck 1 has a mass of 5000 pounds, while truck 2 has also a mass of 5000 pounds and an extra of 900 pounds.
Then the total mass of truck 2 is 5000 pounds + 900 pounds = 5900 pounds.
Then both trucks have the same velocity, and truck 2 has a larger mass than truck 1, this implies that truck 2 will have a larger kinetic energy than truck 1.
Answer:
piece of submerged iron
Explanation:
The buoyant force is given by:
where
is the density of the liquid (in this case, water)
is the submerged part of the object
g is the acceleration due to gravity
In this problem, the liquid used in the two examples is the same, water, so is the same; and g is the same as well. Therefore, the only difference between the two examples is .
For the piece of wood, we are told that the object is floating, so only a fraction of its total volume of 25.0 cm3 will be submerged: therefore the submerged volume will be less than 25.0 cm3. On the contrary, the problem tells us that the piece of iron is submerged, so the submerged volume in this case is 25.0 cm3. Therefore, the submerged volume for the iron is greater than for the piece of wood, so the buoyant force of the piece of iron is greater.
