Answer:
<h2>3.0 m/s²</h2>
Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
From the question we have
We have the final answer as
<h3>3.0 m/s²</h3>
Hope this helps you
Decreasing the trains velocity will DECREASE the kinetic energy
Answer:
<h2>
<u>Joule</u><u>:</u></h2>
1 Joule of work is said to be done when a force of 1 Newton is applied to move/displace a body by 1 metre.
1 Joule= 1 Newton × 1 metre
1 Newton is the amount of force required to accelerate body of mass 1 kg by 1m/s²
So units of N is kgm/s²
So,
1 Joule
=1kgm/s² × m
=1kgm²/s²
<h2><u>Erg</u><u>:</u></h2>
1 erg is the amount of work done by a force of 1 dyne exerted for a distance of one centimetre.
1 Erg =1 Dyne × 1 cm
1 dyne is the force required to cause a mass of 1 gram to accelerate at a rate of 1cm/s².
1 Erg=1 gmcm/s² × cm
1 Erg=1 gmcm/s² × cm=1gmcm²/s²
this is what you need to convert 1gmcm²/s² to 1kgm²/s²
<h3><u>
what you need to know for conversion</u></h3>
[1gm=0.001kg
1cm²
=1cm ×1cm
=0.01 m × 0.01 m
=0.0001m²
second remains constant
]
So,
1gmcm²/s²
=0.001kg×0.0001m²/s²
=0.001kg×0.0001m²/s² =0.0000001kgm²/s²
Hence,
<h3>
<u>1 Erg</u><u>=</u><u>0.0000001</u><u> </u><u>Joule</u></h3><h3>
<u>1</u><u> </u><u>Joule</u><u>=</u><u>1</u><u>0</u><u>,</u><u>0</u><u>0</u><u>0</u><u>,</u><u>0</u><u>0</u><u>0</u><u> </u><u>Erg</u></h3>
<h2>⇒15 J=15×10000000 Erg</h2><h2> =150000000 Erg</h2><h2>
=1.5×10⁶ Erg</h2>
Answer:
50 N
4.2 N
Explanation:
i) The force needed to balance the boom is 2400 N. If the weight of the counterbalance is 2350 N, then the downward force the park attendant must apply is 50 N.
ii) When the boom is resting on the end support, the normal force is:
∑τ = Iα
-W (0.50) + F (3.0) − N (6.0) = 0
-0.50 W + 3.0 F = 6.0 N
N = (-0.50 W + 3.0 F) / 6.0
N = (-0.50 × 2350 + 3.0 × 400) / 6.0
N ≈ 4.2
Answer:
A_f= 15,769 m²
Explanation:
Este es un ejercicio de dilatación térmica,
ΔA = (2α) A₀ ΔT
el arrea de recipiente
A₀ = L A
A₀ = 3,5 4,5
A₀= 15,75 m²
el coeficiente de dilatación térmica es alfa = 16,6 10⁻⁶ C⁻¹
calculemos
ΔA = 2 16,6 10⁻⁶ 15,75 ( 40 -4)
ΔA = 522,9 (36) 10⁻⁶
A= 1,88 10⁻² m2
el cambio de volumen es
ΔA = A_f – A₀
A_f = A₀ + ΔA
A_f= 15,75 +1,88 10⁻²
A_f= 15,769 m²
