Using the Pythagorean’s theorem to find the resultant, we find 5’2 +3’2=x’2
X=square root of 34=5.8unitd
Answer:
a. the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N
b. this force compare to the weight of the muon; the force is 1.38 × 10⁸ greater than muon
Explanation:
F= ma
v²=u² -2aS
(1.56 ✕ 10⁶)²=(2.40 ✕ 10⁶)²-2a(1220)
a=1.36×10⁹m/s²
recall
F=ma
F = 1.88 ✕ 10⁻²⁸ kg × 1.36×10⁹m/s²
F= 2.55 × 10⁻¹⁹N
the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N
b. this force compare to the weight of the muon
F/mg= 2.55 × 10⁻¹⁹/ (1.88 ✕ 10⁻²⁸ × 9.8)
= 1.38 × 10⁸
Explanation:
This is a good example of Doppler effect
Given
Source velocity Vs=20m/s
Apparent frequency Fa= 5.2 Hz
Recall speed of sound V=340m/s
The frequency of the horn Fh=?
But mathematically Fh=(V-Vs/V)*Fa
(340-20)/340*5.2=4.89Hz
Answer:
Explanation:
Net external force that exerted on the block is given as
here we know that
now we have
so we have
now the force exerted by bigger block on smaller block is given as
now we know that two blocks will exert equal and opposite force on each other
so here the force exerted by 2 kg block on 4 kg block will be
Density = 3/5 = 0.6g/cm^3. Since the density is less than the density of water, which is 1, the object will float.
