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Mandarinka [93]
3 years ago
14

As you found out while solving Part C, the only force acting on Tarzan that does work on him is gravity. An important property o

f gravity is that the work done by this force on a body moving along a curved path is the same as if the body were displaced vertically. Therefore, you can evaluate your previous work using the kinematics equations for free fall. What vertical distance Δy does a free-falling particle travel from the moment it starts to the moment it reaches a speed of 7.9 m/s if it starts from rest? Work out your solution using one of the equations for vertical motion with constant acceleration, specifically,v2f=v2i+2aΔywhere vi and vf are, respectively, the particle’s initial and final speed, and a is the particle’s acceleration.
Physics
1 answer:
pshichka [43]3 years ago
4 0

Answer:

The vertical distance traveled is 3.18 m.

Explanation:

In order to find the vertical distance traveled by a body in free fall motion, starting from rest to a velocity of 7.9 m/s. For this purpose, we use Newton's third equation of motion,

2as = Vf² - Vi²

where,

a = g = 9.8 m/s²

s = vertical distance = Δy = ?

Vf = final velocity = 7.9 m/s

Vi = initial velocity = 0 m/s

Therefore,

2(9.8 m/s²)Δy = (7.9 m/s)² - (0 m/s)²

Δy =  (7.9 m/s)²/(2)(9.8 m/s²)

<u>Δy =  3.18 m</u>

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Answer:

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Explanation:

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3 years ago
3. Light travels from the Sun to Earth in 8.3 min. Given that the speed of light is 3.00108 m/s, what is the distance in meters
nasty-shy [4]

Answer:

13

Explanation:

13.0120481928 it is the distance

4 0
3 years ago
Two balloons (m = 0.021 kg) are separated by a distance of d = 16 m. They are released from rest and observed to have an instant
evablogger [386]

(a) 2.56\cdot 10^{-5} C

According to Newton's second law, the force experienced by each balloon is given by:

F = ma

where

m = 0.021 kg is the mass

a = 1.1 m/s^2 is the acceleration

Substituting, we found:

F=(0.021)(1.1)=0.0231 N

The electrostatic force between the two balloons can be also written as

F=k\frac{Q^2}{r^2}

where

k is the Coulomb's constant

Q is the charge on each balloon

r = 16 m is their separation

Since we know the value of F, we can find Q, the magnitude of the charge on each balloon:

Q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(0.0231)(16)^2}{9\cdot 10^9}}=2.56\cdot 10^{-5} C

(b) 1.6\cdot 10^{14} electrons

The magnitude of the charge of one electron is

e=1.6\cdot 10^{-19}C

While the magnitude of the charge on one balloon is

Q=2.56\cdot 10^{-5} C

This charge can be written as

Q=Ne

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N=\frac{Q}{e}=\frac{2.56\cdot 10^{-5}}{1.6\cdot 10^{-19}}=1.6\cdot 10^{14}

5 0
3 years ago
A light platform is suspended from the ceiling by a spring. A student with a mass of 90 kg climbs onto the platform. When it sto
Ilya [14]
Refer to the diagram shown.

When the student climbs onto the platform, the spring stretches by 0.82 m to reach the equilibrium position.
The mass of the student is m = 90 kg, so his weight is
mg = (90 kg)*(9.8 m/s²) = 882 N

By definition, the spring constant is
k = (882 N)/(0.82 m) = 1075.6 N/m

When the spring is stretched by x from the equilibrium position, the restoring force is
F = - k*x.

If damping is ignored, the equation of motion is
F = m * acceleration
or
m \frac{d^{2}x}{dt^{2}} = -kx \\ \frac{d^{2}x}{dt^{2}} + \frac{k}{m} x = 0

Define ω² = k/m = 11.751 => ω = 3.457.
Then the solution of the ODE is
x(t) = c₁ cos(ωt) + c₂ sin(ωt)

x'(t) = -c₁ω sin(ωwt) + c₂ω cos(ωt)
When t=0, x' =0, therefore c₂ = 0

The solution is of the form
x(t) = c₁ cos(ωt)
When t = 0, x = 0.32 m. Therefore c₁ = 0.32

The motion is
x(t) = 0.32 cos(3.457t)
The single amplitude is 0.32 m, and the double amplitude is 0.64 m.

Answer: 
0.32 m (single amplitude), or
0.64 m (double amplitude)

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