The resultant acceleration of the box is 0.055 m/s² in the direction of the greater force.
The given parameters:
- Mass of the box, m = 31 kg
- Two applied forces, = 8.3 N and 6.6 N
<h3>Resultant force on the box</h3>
The resultant force on the box is calculated as follows;
<h3>Resultant acceleration of the box</h3>
The resultant acceleration of the box is calculated as follows;
Thus, the resultant acceleration of the box is 0.055 m/s² in the direction of the greater force.
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Answer:
3.71 m/s in the negative direction
Explanation:
From collisions in momentum, we can establish the formula required here which is;
m1•u1 + m2•v2 = m1•v1 + m2•v2
Now, we are given;
m1 = 1.5 kg
m2 = 14 kg
u1 = 11 m/s
v1 = -1 m/s (negative due to the negative direction it is approaching)
u2 = -5 m/s (negative due to the negative direction it is moving)
Thus;
(1.5 × 11) + (14 × -5) = (1.5 × -1) + (14 × v2)
This gives;
16.5 - 70 = -1.5 + 14v2
Rearranging, we have;
16.5 + 1.5 - 70 = 14v2
-52 = 14v2
v2 = - 52/14
v2 = 3.71 m/s in the negative direction
Answer:
a) True. The number of photoelectrons is proportional to the amount (intensity) of the incident beam. From the expression above we see that threshold frequency cannot emit electrons.
b) λ = c / f
Therefore, as the wavelength increases, the frequency decreases and therefore the energy of the photoelectrons emitted,
c) threshold energy
h f =Ф
Explanation:
It's photoelectric effect was fully explained by Einstein by the expression
Knox = h f - fi
Where K is the kinetic energy of the photoelectrons, f the frequency of the incident radiation and fi the work function of the metal
a) True. The number of photoelectrons is proportional to the amount (intensity) of the incident beam. From the expression above we see that threshold frequency cannot emit electrons.
b) wavelength is related to frequency
λ = c / f
Therefore, as the wavelength increases, the frequency decreases and therefore the energy of the photoelectrons emitted, so there is a wavelength from which electrons cannot be removed from the metal.
c) As the work increases, more frequency radiation is needed to remove the electrons, because there is a threshold energy
h f =Ф
Answer:
Acceleration a ≤ 3 m/s^2
the greatest acceleration that the truck can have without losing its load is 3 m/s^2
Explanation:
For the truck to accelerate without losing its load.
Acceleration force of truck must be less than or equal to the maximum friction force between the truck bed and the load.
Fa ≤ F(friction)
But;
Fa = mass × acceleration
Fa = ma
ma ≤ F(friction)
a ≤ (F(friction))/m ......1
Given;
Fa = mass × acceleration
Fa = ma
mass m = 800 kg
F(friction) = 2400 N
Substituting the given values into equation 1;
a ≤ F(friction)/m
a ≤ 2400N/800kg
a ≤ 3 m/s^2
the greatest acceleration that the truck can have without losing its load is 3 m/s^2
