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noname [10]
2 years ago
9

Betty weighs 420 n and she is sitting on a playground swing seat that hangs 0.40 m above the ground. tom pulls the swing back an

d releases it from rest when the seat is 1.0 m above the ground. how fast is betty moving when the swing passes through its lowest position
Physics
1 answer:
Svet_ta [14]2 years ago
8 0
Weight = mass * gravity
420 = mass * 9.8
mass of Betty = 42.857 kg

Difference in height = 1 - 0.45 = 0.55 meters

Total energy = Kinetic energy + potential energy

At the highest point, the kinetic energy is zero while the potential energy is maximum, therefore, we can get the total energy as follows:
Total energy = 0 + mgh
Total energy = 42.857*9.8*0.55 = 231 Joules

At the lowest point, the potential energy is zero while the kinetic energy is maximum. Therefore:
Total energy = 0.5 * m * (v)^2 + 0
231 = 0.5 * (42.857) * (velocity)^2
(velocity)^2 = 10.78
velocity = 3.28 meters/sec
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A system contains a perfectly elastic spring, with an unstretched length of 20 cm and a spring constant of 4 N/cm.
mote1985 [20]

Answer:

a) When its length is 23 cm, the elastic potential energy of the spring is

0.18 J

b) When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J

Explanation:

Hi there!

a) The elastic potential energy (EPE) is calculated using the following equation:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = stretched lenght.

Let´s calculate the elastic potential energy of the spring when it is stretched 3 cm (0.03 m).

First, let´s convert the spring constant units into N/m:

4 N/cm · 100 cm/m = 400 N/m

EPE = 1/2 · 400 N/m · (0.03 m)²

EPE = 0.18 J

When its length is 23 cm, the elastic potential energy of the spring is 0.18 J

b) Now let´s calculate the elastic potential energy when the spring is stretched 0.06 m:

EPE = 1/2 · 400 N/m · (0.06 m)²

EPE = 0.72 J

When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J

7 0
2 years ago
At a given point on a horizontal streamline in flowing air, the static pressure is â2.0 psi (i.e., a vacuum) and the velocity is
Nastasia [14]
At a point on the streamline, Bernoulli's equation is
p/ρ + v²/(2g) = constant
where
p = pressure
v = velocity
ρ = density of air, 0.075 lb/ft³ (standard conditions)
g = 32 ft/s²

Point 1:
p₁ = 2.0 lb/in² = 2*144 = 288 lb/ft²
v₁ = 150 ft/s

Point 2 (stagnation):
At the stagnation point, the velocity is zero.

The density remains constant.
Let p₂ = pressure at the stagnation point.
Then,
p₂ = ρ(p₁/ρ + v₁²/(2g))
p₂ = (288 lb/ft²) + [(0.075 lb/ft³)*(150 ft/s)²]/[2*(32 ft/s²)
     = 314.37 lb/ft²
     = 314.37/144 = 2.18 lb/in²

Answer: 2.2 psi

5 0
2 years ago
A charge of 4.9 x 10-11 C is to be stored on each plate of a parallel-plate capacitor having an area of 150 mm2 and a plate sepa
Triss [41]

Answer:2.53*10^-10F

Explanation:

C=£o£r*A/d

Where £ is the permitivity of a constant

£o= 8.85*10^-12f/m

£r=6.3

A=150mm^2=0.015m^2

d=3.3mm= 0.0033m

C=8.85*10^-12*6.3*0.015/0.0033

C=8.85*6.3*10^-12*0.015/0.0033

C=55.755*0.015^-12/0.003

C=8.36/3.3*10^-13+3

C=2.53*10^-10F

7 0
2 years ago
The U.S. Department of Energy had plans for a 1500-kg automobile to be powered completely by the rotational kinetic energy of a
navik [9.2K]

Answer:

230

Explanation:

\omega = Rotational speed = 3600 rad/s

I = Moment of inertia = 6 kgm²

m = Mass of flywheel = 1500 kg

v = Velocity = 15 m/s

The kinetic energy of flywheel is given by

K=\dfrac{1}{2}I\omega^2\\\Rightarrow K=\dfrac{1}{2}6\times 3600^2\\\Rightarrow K=38880000\ J

Energy used in one acceleration

K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}1500\times 15^2\\\Rightarrow K=168750\ J

Number of accelerations would be given by

n=\dfrac{38880000}{168750}\\\Rightarrow n=230.4

So the number of complete accelerations is 230

8 0
2 years ago
Each shot of the laser gun most favored by Rosa the Closer, the intrepid vigilante of the lawless 22nd century, is powered by th
mote1985 [20]

Answer:

U = 1794.005 × 10⁶ J

Explanation:

Data provided;

Capacitance of the original capacitor, C = 1.27 F

Potential difference applied to the original capacitor, V = 59.9 kV

= 59.9 × 10³ V

Now,

The Potential energy (U) for the capacitor is calculated as:

Potential energy of the original capacitor, U = \frac{\textup{1}}{\textup{2}}  × C × V²

on substituting the respective values, we get

U = \frac{\textup{1}}{\textup{2}}  × 1.27 × ( 59.9 × 10³ )²

or

U = 1794.005 × 10⁶ J

7 0
2 years ago
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