Answer:
Contact force is a force that requires a contact between two bodies and is ubiquitous in nature
Explanation:
The force is distributed in two categories i.e
Contact Forces are the forces which is requires a contact between two bodies to occur.
The Types of contact forces are given below:
- Tension Force
- Normal Force
- Air Resistance Force
- Applied Force
- Spring Force
Some common daily examples of the contact force are
- Frictional force between the tyres of a moving vehicle and the road.
- Air flowing in opposite direction of a moving object.
- Pushing a table with hand and the friction between its base and floor.
- Stretching a rubber band with hands etc.
Also non-contact forces are Gravitational force, Electrical force and Magnetic force.
Answer:
Lost pigment of marker when dipped in alcohol
Explanation:
dependent viable = output
so it's the output of what happens after the input.
- she put the marker in the water which is the independent variable, that's the input
- the output or the result of that decision is having lost pigment in the marker
Answer:
Explanation:
2Al(s) + 3 2 O2(g) → Al2O3(s) And given the stoichiometry ...and EXCESS dioxygen gas...we would get 6.25⋅ mol of alumina. the which represents a mass... ...6.25 ⋅ mol ×101.96 ⋅ g ⋅ mol−1 molar mass of alumina ≡ 637.25 ⋅ g.
Answer:
A) secondary amide
Explanation:
When carboxylic acid reacts with a primary amine, a condensation reaction takes place with the elimination of a water molecule .
For example, ethanol reacts with methylamine which is a primary amine gives N-Methylacetamide and a water molecule as:
The bond formed which is
O
||
-- C ---NH ---
is known as secondary amide group as only one hydrogen is attached to nitrogen atom in the amide bond.
Note: The question is incomplete. The complete question is given below :
Suppose a substance has a heat of fusion equal to 45 cal/g and a specific heat of 0.75 cal/g°C in the liquid state. If 5.0 kcal of heat are applied to a 50 g sample of the substance at a temperature of 24°C, what will its new temperate be? What state will the sample be in? (melting point of the substance = 27°C; specific heat of the solid =0.48 cal/g°C; boiling point of the substance = 700°C)
Explanation:
1.a) Heat energy required to raise the temperature of the substance to its melting point, H = mcΔT
Mass of solid sample = 50 g; specific heat of solid = 0.75 cal/g; ΔT = 27 - 24 = 3 °C
H = 50 × 0.75 × 3 = 112.5 calories
b) Heat energy required to convert the solid to liquid at its melting point at 27°C, H = m×l, where l = 45 cal/g
H = 50 × 45 = 2250 cal
c) Total energy used so far = 112.5 cal + 2250 cal = 2362.5 calories.
Amount of energy left = 5000 - 2362.5 = 2637.5 cal
The remaining energy is used to heat the liquid
H = mcΔT
Where specific heat of the liquid, c = 0.75 cal/g/°C, H = 2637.5 cal, ΔT = temperature change
2637.5 = 50 × 0.75 x ΔT
ΔT = 2637.5 / ( 50*0.75)
ΔT = 70.3 °C
Final temperature of sample = (70.3 + 27) °C = 97.3 °C
The substance will be in liquid state at a temperature of 97.3 °C
i hope that this eg gonna help u
