Answer:
flagpole
Explanation:
if it is about electricity then its flagpole
Answer:
t = 4.17 [s]
Explanation:
We know that work is defined as the product of force by distance.
W = F*d
where:
F = force [N] (units of Newtons)
d = distance = 6.34 x 10⁴ [mm] = 63.4 [m]
In order to find the force, we must determine the weight of the box, the weight can be determined by means of the product of mass by gravitational acceleration.
w = m*g
where:
m = mass = 1.47 x 10⁴ [g] = 14.7 [kg]
g = gravity acceleration = 9.81 [m/s²]
w = 14.7*9.81
w = 144.2 [N]
Therefore the work can be calculated.
W = w*d
W = 144.2*63.4
W = 9142.72 [J] (units of Joules)
Power is now defined in physics as the relationship of work at a given time
P = W/t
where:
P = power = 2190 [W]
t = time [s]
Now clearing t, we have.
t = W/P
t = 9142.72/2190
t = 4.17 [s]
Answer:
The graphs are attached
Explanation:
We are told that he starts with a constant speed of 25 m/s for a distance of 100 m.
At constant velocity, v = distance/time
time(t) = distance(d)/velocity(v)
t1 = 100/25
t1 = 4 s
Now, we are told that he applies his brakes and accelerates uniformly to a stop just as he reaches a wall 50m away.
It means, he decelerate and final velocity is zero.
Thus;
v² = u² + 2as
0² = 25² + 2a(50)
25² = - 100a
625 = - 100a
a = - 625/100
a = - 6.25 m/s²
v = u + at
0 = 25 + (-6.25t)
25 = 6.25t
t = 25/6.25
t = 4 s
With the values gotten, kindly find attached the distance-time and velocity-time graphs.
Answer
given,


mass of book = 0.305 Kg
so, from the diagram attached below




computing horizontal component




θ = 62.35°
Answer:
The amount of work that must be done to compress the gas 11 times less than its initial pressure is 909.091 J
Explanation:
The given variables are
Work done = 550 J
Volume change = V₂ - V₁ = -0.5V₁
Thus the product of pressure and volume change = work done by gas, thus
P × -0.5V₁ = 500 J
Hence -PV₁ = 1000 J
also P₁/V₁ = P₂/V₂ but V₂ = 0.5V₁ Therefore P₁/V₁ = P₂/0.5V₁ or P₁ = 2P₂
Also to compress the gas by a factor of 11 we have
P (V₂ - V₁) = P×(V₁/11 -V₁) = P(11V₁ - V₁)/11 = P×-10V₁/11 = -PV₁×10/11 = 1000 J ×10/11 = 909.091 J of work