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mel-nik [20]
3 years ago
9

A pulley is able to lift a mass of 25 kg 0.30 m with an applied force of 50 N over a distance of 1.5 m. What is the ideal mechan

ical advantage of the pulley system?
Physics
1 answer:
spin [16.1K]3 years ago
7 0

The ideal mechanical advantage of the pulley system is 3

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what is the magnitude of g at a height above earth surface where free fall acceleration equal 6.5m/s2
ZanzabumX [31]
The magnitude of gravity is expressed in terms of its acceleration. So the magnitude of ' g ' at that altitude is exactly 6.5 m/s^2.
8 0
3 years ago
Water enters a baseboard radiator at 180 °F and at a flow rate of 2.0 gpm. Assuming the radiator releases heat into the room at
beks73 [17]

Answer:

Temperature of water leaving the radiator = 160°F

Explanation:

Heat released = (ṁcΔT)

Heat released = 20000 btu/hr = 5861.42 W

ṁ = mass flowrate = density × volumetric flow rate

Volumetric flowrate = 2 gallons/min = 0.000126 m³/s; density of water = 1000 kg/m³

ṁ = 1000 × 0.000126 = 0.126 kg/s

c = specific heat capacity for water = 4200 J/kg.K

H = ṁcΔT = 5861.42

ΔT = 5861.42/(0.126 × 4200) = 11.08 K = 11.08°C

And in change in temperature terms,

10°C= 18°F

11.08°C = 11.08 × 18/10 = 20°F

ΔT = T₁ - T₂

20 = 180 - T₂

T₂ = 160°F

8 0
3 years ago
A series circuit has a capacitor of 0.25 × 10−6 F, a resistor of 5 × 103 Ω, and an inductor of 1 H. The initial charge on the ca
svetoff [14.1K]

Answer:

q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

Explanation:

Given that L=1H, R=5000\Omega, \ C=0.25\times10^{-6}F, \ \ E(t)=12V, we use Kirchhoff's 2nd Law to determine the sum of voltage drop as:

E(t)=\sum{Voltage \ Drop}\\\\L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{1}{C}q=E(t)\\\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+\frac{1}{0.25\times10^{-6}}q=12\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+4000000q=12\\\\m^2+5000m+4000000=0\\\\(m+4000)(m+1000)=0\\\\m=-4000  \ or \ m=-1000\\\\q_c=c_1e^{-4000t}+c_2e^{-1000t}

#To find the particular solution:

Q(t)=A,\ Q\prime(t)=0,Q\prime \prime(t)=0\\\\0+0+4000000A=12\\\\A=3\times10^{-6}\\\\Q(t)=3\times10^{-6},\\\\q=q_c+Q(t)\\\\q=c_1e^{-4000t}+c_2e^{-1000t}+3\times10^{-6}\\\\q\prime=-4000c_1e^{-4000t}-1000c_2e^{-1000t}\\q\prime(0)=0\\\\-4000c_1-1000c_2=0\\c_1+c_2+3\times10^{-6}=0\\\\#solving \ simultaneously\\\\c_1=10^{-6},c_2=-4\times10^{-6}\\\\q=10^{-6}e^{-4000t}-4\times10^{-6}e^{-1000t}+3\times10^{-6}\\\\q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

Hence the charge at any time, t is q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

6 0
3 years ago
At the end of the Shock Wave roller coaster ride, the 6000-kg train of cars (includes passengers) is allowed from a speed of 20
Angelina_Jolie [31]

Answer:

56250 N

Explanation:

mass, m = 6000 kg

initial speed, u = 20 m/s

final speed, v = 5 m/s

distance, s = 20 m

Use third equation of motion

v^{2}=u^{2}+2as

5 x 5 = 20 x 20 + 2 a x 20

25 = 400 + 40 a

a = - 9.375 m/s^2

Braking force, F = mass x acceleration

F = 6000 x 9.375

F = 56250 N

6 0
3 years ago
tyhe mass of an object is 117 g adding 1200j of heat will raise the temperture of the object by 12 celsius what is the specifc h
Sati [7]

Answer:

C 0.85 j/g*k

Explanation:

The specific heat capacity of a material is given by:

C_s = \frac{Q}{m \Delta T}

where

Q is the amount of heat supplied to the object

m is the mass of the object

\Delta T is the increase in temperature of the object

For the object in this problem, we have

m = 117 g is the mass

Q = 1200 J is the heat supplied

\Delta T=12^{\circ} is the increase in temperature

Substituting into the formula, we find the specific heat:

C_s = \frac{1200 J}{(117 g)(12^{\circ})}=0.85 J/gC

8 0
3 years ago
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