The only force opposing the block's sliding as it slows down is friction with magnitude <em>f </em>. By Newton's second law, the net force in this direction is
∑ <em>F</em> = -<em>f</em> = <em>ma</em> = (4.00 kg) <em>a</em>
Assuming constant acceleration <em>a </em>, the acceleration applied by friction is such that
(1.5 m/s)² - (11 m/s)² = 2<em>a</em> (4.00 m)
Solve for the acceleration :
<em>a</em> = ((1.5 m/s)² - (11 m/s)²) / (8.00 m) ≈ -14.8 m/s²
Then the frictional force exerted a magnitude of
-<em>f</em> = (4.00 kg) (-14.8 m/s²)
<em>f</em> ≈ 59.4 N
and was directed opposite the block's motion.
Answer:
the pressure at the bottom of the tank is 12 kPa
Explanation:
The computation of the pressure at the bottom of the tank is as follows:
as we know that
Pressure = hdg
where
H = Height
d = density
Now the pressure is
= 1.5m × 800 × 10
= 12,000 Pa
= 12 kPa
Hence, the pressure at the bottom of the tank is 12 kPa
Answer: Option B: 1.3×10⁵ W
Explanation:
Work Done,
Where s is displacement in the direction of force and F is force.
where, v is the velocity.
It is given that, F = 5.75 × 10³N
v = 22 m/s
P = 5.75 × 10³N×22 m/s = 126.5 × 10³ W ≈1.3×10⁵W
Thus, the correct option is B
Answer:
It expands due to the elasticity of the rubber. but wen , the rubber is expanded to its maximum point and cannot be stretched further, it will burst because there will be no space for the rubber to accommodate the air.
Explanation: