Answer:
k = 26.25 N/m
Explanation:
given,
mass of the block= 0.450
distance of the block = + 0.240
acceleration = a_x = -14.0 m/s²
velocity = v_x = + 4 m/s
spring force constant (k) = ?
we know,
x = A cos (ωt - ∅).....(1)
v = - ω A cos (ωt - ∅)....(2)
a = ω²A cos (ωt - ∅).........(3)

now from equation (3)



k = 26.25 N/m
hence, spring force constant is equal to k = 26.25 N/m
Okay, here is my stab at this, I hope it helps!
You know the bullet's initial velocity, V₀ = 450 m/s
You know the final velocity, V = 220 m/s
You also know how long the bullet accelerates (actually decelerates), 14cm, or .14 m
With this information, you learn that you need this equation.
V² = V₀² + 2a (x - x₀), because we have all the information except a, which is the acceleration. So putting it into the equation, it looks like this.
(220m/s)² = (450m/s)² +2a(.14m - 0m)
I'll let you solve the rest, but here are some hints. Your answer will be really big because the bullet slows down really quickly in a really small distance, and you answer will be negative, because this acceleration is causing the bullet to go slower, which is also called deceleration. Hope that helps!
Answer:
6.78 X 10³ N/C
Explanation:
Electric field near a charged infinite plate
= surface charge density / 2ε₀
Field will be perpendicular to the surface of the plate for both the charge density and direction of field will be same so they will add up.
Field due to charge density of +95.0 nC/m2
E₁ = 95 x 10⁻⁹ / 2 ε₀
Field due to charge density of -25.0 nC/m2
E₂ = 25 x 10⁻⁹ / 2ε₀
Total field
E = E₁ + E₂
= 95 x 10⁻⁹ / 2 ε₀ + 25 x 10⁻⁹ / 2ε₀
= 6.78 X 10³ N/C
Answer:
Explanation:
a )
one kg of coal gives energy of 27 x 10⁶ J
75 kg of coal gives energy of 27 x 10⁶ x 75 J
So rate which energy is coming out of coal per second
= 27 x 10⁶ x 75 J
= 2025 x 10⁶ J /s
2025 million watts .
b ) energy output = 800 million watts
efficiency = (800 / 2025) x 100
= 39.5 % .
Answer:
The centripetal force acting on the child is 39400.56 N.
Explanation:
Given:
Mass of the child is, 
Radius of the barrel is, 
Number of revolutions are, 
Time taken for 10 revolutions is, 
Therefore, the time period of the child is given as:

Now, angular velocity is related to time period as:

Now, centripetal force acting on the child is given as:

Therefore, the centripetal force acting on the child is 39400.56 N.