From tables, the speed of sound at 0°C is approximately
V₁ = 331 m/s (in air)
V₃ = 5130 m/s (in iron)
Distance traveled is
d = 100 km = 10⁵ m
Time required to travel in air is
t₁ = d/V₁ = 10⁵/331 = 302.12 s
Time required to travel in iron is
t₂ = d/V₂ = 10⁵/5130 = 19.49 s
The difference in time is
302.12 - 19.49 = 282.63 s
Answer: 283 s (nearest second)
The force applied by the man is 60 N
Explanation:
We can solve this problem by applying Newton's second law, which states that:
(1)
where
is the net force acting on the child+cart
m is the mass of the child+cart system
a is their acceleration
In this problem, we have:
m = 30.0 kg is the mass
And there are two forces acting on the child+cart system:
- The forward force of pushing, F
- The force resisting the cart motion, R = 15.0 N
Therefore we can write the net force as
where R is negative since its direction is opposite to the motion
So eq.(1) can be rewritten as
And solving for F,
Learn more about Newton's second law:
brainly.com/question/3820012
#LearnwithBrainly
We solve this using special
relativity. Special relativity actually places the relativistic mass to be the
rest mass factored by a constant "gamma". The gamma is equal to 1/sqrt
(1 - (v/c)^2). <span>
We want a ratio of 3000000 to 1, or 3 million to 1.
</span>
<span>Therefore:
3E6 = 1/sqrt (1 - (v/c)^2)
1 - (v/c)^2 = (0.000000333)^2
0.99999999999999 = (v/c)^2
0.99999999999999 = v/c
<span>v= 99.999999999999% of the speed of light ~ speed of light
<span>v = 3 x 10^8 m/s</span></span></span>
Answer:
a) a= 8.33 m/s², T = 12.495 N
, b) a = 2.45 m / s²
Explanation:
a) this is an exercise of Newton's second law. As the upper load is secured by a cable, it cannot be moved, so the lower load is determined by the maximum acceleration.
We apply Newton's second law to the lower charge
fr₁ + fr₂ = ma
The equation for the force of friction is
fr = μ N
Y Axis
N - W₁ –W₂ = 0
N = W₁ + W₂
N = (m₁ + m₂) g
Since the beams are the same, it has the same mass
N = 2 m g
We replace
μ₁ 2mg + μ₂ mg = m a
a = (2μ₁ + μ₂) g
a = (2 0.30 + 0.25) 9.8
a= 8.33 m/s²
Let's look for cable tension with beam 2
T = m₂ a
T = 1500 8.33
T = 12.495 N
b) For maximum deceleration the cable loses tension (T = 0 N), so as this beam has less friction is the one that will move first, we are assuming that the rope is horizontal
fr = m₂ a₂
N- w₂ = 0
N = W₂ = mg
μ₂ mg = m a₂
a = μ₂ g
a = 0.25 9.8
a = 2.45 m / s²
The mass of the object doesn't matter. The change in its momentum is equal to the impulse that changed it ... 15 N-sec.
