Answer:
Explanation:
When we apply a horizontal force of 76 N to a block, the block moves across the floor at a constant speed. So net force on the block is zero .
It implies that a force ( frictional ) acts on it which is equal to 76 N in opposite direction ( friction )
When we apply a greater force on it it starts moving with acceleration .
This time kinetic friction acts on it due to rough ground equal to 76 N .This is limiting friction ( maximum friction )
Net force on the body in later case
= 89 - 76
= 13 N
Force by ground on the block in horizontal direction = 76 N ( FRICTIONAL FORCE )
=
Answer:reflection by dust particles in air
The difference in the pressure between the inside and outside will be 369.36 N/m²
<h3>What is pressure?</h3>
The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure.
It is denoted by P. The pressure relative to the ambient pressure is known as gauge pressure.
The given data in the problem is;
dP is the change in the presure=?
Using Bernoulli's Theorem;
Hence, the difference in the pressure between the inside and outside will be 369.36 N/m²
To learn more about the pressure refer to the link;
brainly.com/question/356585
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Answer:
Explanation:
a. The equation of Lorentz transformations is given by:
x = γ(x' + ut')
x' and t' are the position and time in the moving system of reference, and u is the speed of the space ship. x is related to the observer reference.
x' = 0
t' = 5.00 s
u =0.800 c,
c is the speed of light = 3×10⁸ m/s
Then,
γ = 1 / √ (1 - (u/c)²)
γ = 1 / √ (1 - (0.8c/c)²)
γ = 1 / √ (1 - (0.8)²)
γ = 1 / √ (1 - 0.64)
γ = 1 / √0.36
γ = 1 / 0.6
γ = 1.67
Therefore, x = γ(x' + ut')
x = 1.67(0 + 0.8c×5)
x = 1.67 × (0+4c)
x = 1.67 × 4c
x = 1.67 × 4 × 3×10⁸
x = 2.004 × 10^9 m
x ≈ 2 × 10^9 m
Now, to find t we apply the same analysis:
but as x'=0 we just have:
t = γ(t' + ux'/c²)
t = γ•t'
t = 1.67 × 5
t = 8.35 seconds
b. Mavis reads 5 s on her watch which is the proper time.
Stanley measured the events at a time interval longer than ∆to by γ,
such that
∆t = γ ∆to = (5/3)(5) = 25/3 = 8.3 sec which is the same as part (b)
c. According to Stanley,
dist = u ∆t = 0.8c (8.3) = 2 x 10^9 m
which is the same as in part (a)
