<span>Atoms with the same atomic number but different atomic mass are called:
<span>Isotopes</span>
</span>
Answer:
25 N; 250 W.
Explanation: Work Done
=
Force
×
Displacement
×
cos
(
The angle between Force and Displacement
)
So, Let's Assume the Force to be
x
Newtons.
So, According to The Sum,
×
x
x
⋅
100
=
2500
⇒
x
=
25
00
1
00
=
25
So, The force was
25
N
.
And, We also know,
Power
=
Work Done
Time
So,
The Power of the Machine =
2500
10
Watts
=
250
Watts
According to newton's 3rd law of motion,
For every action, there is equal and opposite reaction. So if we move a body against a rough surface, there were be reaction against the force applied.
So using conservation of energy, we know:
Work done to move a body = Work done against Friction
So, Force applied * distance moved = coefficient of Friction * Normal Reaction * distance moved
For a body moving against a normal surface, Normal Reaction (R) = mg
or, mass * acceleration * distance (s) = ∪ * R * distance(s)
or, mass * (v^2/2s) = ∪ * mass * gravity
Now, s = stopping distance = v²/ 2∪g
so, using given value,∪=0.05,
s = v2/2*0.05*g
We know, g = 10, so s = v²/(2*0.05*10) = v²
where v = initial velocity
Answer:
1 kg
Explanation:
The container has negligible mass and no heat is loss to the surrounding.
Mass of ice = 0.4kg, initial temperature of ice = -29oC, final temperature of the mixture = 26oC, mass of water (m2) = ?kg, initial temperature of water = 80oC, c ( specific heat capacity of water ) = 4200J/kg.K, Lf = heat of fusion of water = 3.36 × 10^5 J/kg
Using the formula:
Quantity of heat gain by ice = Quantity of heat loss by water
Quantity of heat gain by ice = mass of ice × heat of fusion of ice + mass of water × specific heat capacity of water = (0.4 × 3.36 × 10^ 5) + (0.4 × 4200 × (26- (-29) = 13.44 × 10^4 + 9.24 × 10^ 4 = 22.68 × 10^4 J
Quantity of heat loss by water = m2cΔT
Quantity of heat loss by water = m2 ×4200× (80 - 26) = m(226800)
since heat gain = heat loss
22.68 × 10^4 = 226800 m2
divide both side by 226800
226800 / 226800 = m2
m2 = 1 kg
Answer: 
Explanation:
Given
Mass of hockey puck 
Initial velocity of hockey puck is 
First a horizontal force of 
is applied to the right for 
acceleration associated with it is
Using equation of motion i.e.
(b) When a force of 
is applied for 0.05s
Using equation of motion i.e.
