Answer:
20 cm
Explanation:
Te electric potential enery U = kq₁q₂/r were q₁ = 5 nC = 5 × 10⁻⁹ C and q₂ = -2 nC = -2 × 10⁻⁹ C and r = √(x - 2)² + (0 - 0)² +(0 - 0)² = x - 2. U = -0.5 µJ = -0.5 × 10⁻⁶ J, k = 9 × 10⁹ Nm²/C².
So r = kq₁q₂/U
x - 2 = kq₁q₂/U
x = 0.02 + kq₁q₂/U m
x = 0.02 + 9 × 10⁹ Nm²/C² × 5 × 10⁻⁹ C × -2 × 10⁻⁹ C/-0.5 × 10⁻⁶ J
x = 0.02 - 90 × 10⁻⁹ Nm²/-0.5 × 10⁻⁶ J
x = 0.02 + 0.18 = 0.2 m = 20 cm
Answer:
The load has a mass of 2636.8 kg
Explanation:
Step 1 : Data given
Mass of the truck = 7100 kg
Angle = 15°
velocity = 15m/s
Acceleration = 1.5 m/s²
Mass of truck = m1 kg
Mass of load = m2 kg
Thrust from engine = T
Step 2:
⇒ Before the load falls off, thrust (T) balances the component of total weight downhill:
T = (m1+m2)*g*sinθ
⇒ After the load falls off, thrust (T) remains the same but downhill component of weight becomes m1*gsinθ .
Resultant force on truck is F = T – m1*gsinθ
F causes the acceleration of the truck: F= m*a
This gives the equation:
T – m1*gsinθ = m1*a
T = m1(a + gsinθ)
Combining both equations gives:
(m1+m2)*g*sinθ = m1*(a + gsinθ)
m1*g*sinθ + m2*g*sinθ =m1*a + m1*g*sinθ
m2*g*sinθ = m1*a
Since m1+m2 = 7100kg, m1= 7100 – m2. This we can plug into the previous equation:
m2*g*sinθ = (7100 – m2)*a
m2*g*sinθ = 7100a – m2a
m2*gsinθ + m2*a = 7100a
m2* (gsinθ + a) = 7100a
m2 = 7100a/(gsinθ + a)
m2 = (7100 * 1.5) / (9.8sin(15°) + 1.5)
m2 = 2636.8 kg
The load has a mass of 2636.8 kg
A simple electromagnet consisting of a coil of wire wrapped around an iron core. <u><em>A core of ferromagnetic material like iron serves to increase the magnetic field created.</em></u> The strength of magnetic field generated is proportional to the amount of current through the winding.
your answer is b :)
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The period of the second pendulum is 0.9 s
Explanation:
The period of a simple pendulum is given by the equation
where
L is the length of the pendulum
g is the acceleration of gravity at the location of the pendulum
For the first pendulum, we have
L = 0.64 m
T = 1.2 s
Therefore we can find the value of g at that location:
Now we can find the period of the second pendulum at the same location, which is given by
where we have
L = 0.36 m (length of the second pendulum)
Substituting,
#LearnwithBrainly
When the system is experiencing a uniformly accelerated motion, there are a set of equations to work from. In this case, work is energy which consist solely of kinetic energy. That is, 1/2*m*v2. First, let's find the final velocity.
a = (vf - v0)/t
2.6 = (vf - 0)/4
vf = 10.4 m/s
Then W = 1/2*(2100 kg)*(10.4 m/s)2
W = 113568 J = 113.57 kJ
