By heat or change in matter,
there can be different reactions that create heat (like exothermic or endothermic reactions)
or movement can give off energy too,
energy is basically just heat
Answer:

Explanation:
<u>2-D Projectile Motion</u>
In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are






The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:


Using the formulas for 

Simplifying

Dividing by 

Rearranging



Answer:
The angular frequency of the block is ω = 5.64 rad/s
Explanation:
The speed of the block v = rω where r = amplitude of the oscillation and ω = angular frequency of the oscillation.
Now ω = v/r since v = speed of the block = 62 cm/s and r = the amplitude of the oscillation = 11 cm.
The angular frequency of the oscillation ω is
ω = v/r
ω = 62 cm/s ÷ 11 cm
ω = 5.64 rad/s
So, the angular frequency of the block is ω = 5.64 rad/s
Answer:
451.13 J/kg.°C
Explanation:
Applying,
Q = cm(t₂-t₁)............... Equation 1
Where Q = Heat, c = specific heat capacity of iron, m = mass of iron, t₂= Final temperature, t₁ = initial temperature.
Make c the subject of the equation
c = Q/m(t₂-t₁).............. Equation 2
From the question,
Given: Q = 1500 J, m = 133 g = 0.113 kg, t₁ = 20 °C, t₂ = 45 °C
Substitute these values into equation 2
c = 1500/[0.133(45-20)]
c = 1500/(0.133×25)
c = 1500/3.325
c = 451.13 J/kg.°C
Answer:
yeah this statement is tru
Explanation:
it is because the speed of the bullet is more than the speed of rolling ball .so from this reason we cannot catch a bullet.