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densk [106]
3 years ago
8

A monkey throws a banana peel from a height of 14 meters. It has a mass of 0.01 kilogram. What was the approximate potential ene

rgy of the banana peel before the monkey threw it?
Physics
1 answer:
liq [111]3 years ago
3 0
We need to use the equation for potencial energy

Ep= mgh

now we have to replace with the values we have

Ep=(14m)(9.81m/s²)(0.01kg)

Ep= 1.3734 Joules
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A particle moves along the x-axis so that its velocity at anytime is t greater than or equal to 0 is given by v(t)=1-sin(2pi t)
UkoKoshka [18]
A) We differentiate the expression for velocity to obtain an expression for acceleration:
v(t) = 1 - sin(2πt)
dv/dt = -2πcos(2πt)
a = -2πcos(2πt)

b) Any value of t can be plugged in as long as it is greater than or equal to 0. 

c) we integrate the expression of velocity to find an expression for displacement:
∫v(t) dt = ∫ 1 - sin(2πt) dt
x(t) = t + cos(2πt)/2π + c
x(0) = 0
0 = = + cos(0)/2π + c
c = -1/2π
x(t) = t + cos(2πt)/2π -1/2π
5 0
3 years ago
A sphere of volume 1.20×10−3m3 hangs from a cable. When the sphere is completely submerged in water, the tension in the cable is
KATRIN_1 [288]

Answer:

B = 62.9 N

Explanation:

This is an exercise on Archimedes' principle, where the thrust force equals the weight of the  liquid

         B = ρ g V

write the equilibrium equation

         T + B -W = 0

         B = W- T               (1)

use the density to write the weight

         ρ = m / V

        m = ρ V

         W = ρ g V

substitute in  1

         B = m g -T

         B = \rho_{body} g V - T

To finish the calculation, the density of the material must be known, suppose it is steel  \rho_{body} = 7850 kg / m³

calculate

         B = 7850 9.8 1.20 10⁻³ - 29.4

          B = 92.3 - 29.4

          B = 62.9 N

4 0
3 years ago
At a temperature of 320K, the gas in a cylinder has a volume of 40.0 liters. If the volume of the gas is decreased to 20.0 liter
Lerok [7]
Assuming the gas behaves ideally,
PV/T = constant. P will also be constant in this giving us:
V₁/T₁ = V₂/T₂
40/320 = 20/T₂
T₂ = 160 K
The answer is A.
6 0
3 years ago
Read 2 more answers
A capacitor with a capacitance of 50µf when connected to a battery of 400 V. The charge and energy stored on it is? a. 0.05 C an
nekit [7.7K]

Answer:

c. 0.02 C and 4 J

Explanation:

Applying,

Q = CV................ Equation 1

Where Q = Charge, C = Capacitance of the capacitor, V = Voltage.

From the question,

Given: C = 50 μF = 50×10⁻⁶ F, V = 400 V

Substitute these values into equation 1

Q = (50×10⁻⁶)(400)

Q = 0.02 C.

Also Applying

E = CV²/2............. Equation 2

Where E = Energy stored.

Therefore,

E = (50×10⁻⁶ )(400²)/2

E = 4 J

Hence the right option is c. 0.02 C and 4 J

3 0
2 years ago
GIZMO
CaHeK987 [17]
It is A or D but I believe A
5 0
3 years ago
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