The force of friction is 19.1 N
Explanation:
According to Newton's second law, the net force acting on the bag is equal to the product between its mass and its acceleration:

where
is the net force
m is the mass
a is the acceleration
The bag is moving at constant speed, so its acceleration is zero:

Therefore the net force is zero as well:

Here we are interested only in the forces acting along the horizontal direction, therefore the net force is given by:

where
is the horizontal component of the applied force, with
F = 22.5 N

is the force of friction
And solving for
, we find

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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:
The water will flow at a speed of 3,884 m/s
Explanation:
Torricelli's equation
v = 
*v = liquid velocity at the exit of the hole
g = gravity acceleration
h = distance from the surface of the liquid to the center of the hole.
v =
= 3,884 m/s
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
Answer:
it is essential that the charge on the plates are of the same magnitude, but in the opposite direction
Explanation:
The configuration of parallel plates is called a capacitor and is widely used to create constant electric fields inside.
To obtain this field it is essential that the charge on the plates are of the same magnitude, but in the opposite direction
This is so that the fields created by each plate can be added inside and subtracted from the outside of the plates