The answer:
the relationship between elementary charge, potential difference and electrical potential energy is given by
E= qV
E: lectrical potential energy
q: elementary charge
V: potential difference
but we have e=abs val(q)=3
so we have E= qV=3ex4.5V=<span>13.5 eV
</span>
the answer is <span>(4)13.5 eV</span>
By using the formula density(p)= mass(kg)/volume(m^3) and gravitational acceleration (g)
Mass=2.5N/9.8m/s^2=0.26kg
Volume=0.1x 0.03^2 x3.142=0.009m^3
Density=28.89kg/m^-3
Answer:
Explanation:
Let the bigger crate be in touch with the ground which is friction less. In the first case both m₁ and m₂ will move with common acceleration because m₁ is not sliding over m₂.
1 ) Common acceleration a = force / total mass
= 234 / ( 25 +91 )
= 2.017 m s⁻².
2 ) Force on m₁ accelerating it , which is nothing but friction force on it by m₂
= mass x acceleration
= 25 x 2.017
= 50.425 N
The same force will be applied by m₁ on m₂ as friction force which will act in opposite direction.
3 ) Maximum friction force that is possible between m₁ and m₂
= μ_s m₁g
= .79 x 25 x 9.8
= 193.55 N
Acceleration of m₁
= 193 .55 / 25
= 7.742 m s⁻²
This is the common acceleration in case of maximum tension required
So tension in rope
= ( 25 +91 ) x 7.742
= 898 N
4 ) In case of upper crate sliding on m₂ , maximum friction force on m₁
= μ_k m₁g
= .62 x 25 x 9.8
= 151.9 N
Acceleration of m₁
= 151.9 / 25
= 6.076 m s⁻².
Answer:
(A) Original speed= 9.22 m/s
(B) Acceleration= -1.0099 m\s^2
Explanation:
A truck covers 40m in 7.10 secs
The truck slows down at a uniform velocity of 2.05 m/s
(A) The original speed can be calculated as follows
Vo= 2(40)/7.10 - 2.05
= 80/7.10 - 2.05
= 11.2676 - 2.05
= 9.22m/s
(B) The acceleration can be calculated as follows
a= Vf-Vo/t
= 2.05-9.22/7.10
= -7.17/7.10
= -1.0099m/s^2
