All categories

3 years ago

Which object had more potential energy when it was lifted to a distance of 1000 centimeters? Show your calculation.

Physics

1 answer:

RSB [31]3 years ago

5 0

Explanation:

We Know That

POTENTIAL ENERGY= MASS*g*HEIGHT

When the objects are lifted to same height then the object with heavier mass would have the highest potential energy

You might be interested in

The thermal energy in a cup of tea is _____ the thermal energy in a pot of tea at the same temperature.

olga55 [171]

I think it would be 'less than'

8 0

3 years ago

Read 2 more answers

A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10

IRINA_888 [86]

Complete Question:

A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of 25 rad/s. The wheel is run at that angular velocity for 37 s and then power is shut off. The wheel decelerates uniformly at 1.5 rads/s2 until the wheel stops. In this situation, the time interval of angular deceleration (slowing down) is closest to

Answer:

t= 16.7 sec.

Explanation:

As we are told that the wheel is accelerating uniformly, we can apply the definition of angular acceleration to its value:

γ = (ωf -ω₀) / t

If the wheel was at rest at t-= 0.00 s, the angular acceleration is given by the following equation:

γ = ωf / t = 25 rad/sec / 10 sec = 2.5 rad/sec².

When the power is shut off, as the deceleration is uniform, we can apply the same equation as above, with ωf = 0, and ω₀ = 25 rad/sec, and γ = -1.5 rad/sec, as follows:

γ= (ωf-ω₀) /Δt⇒Δt = (0-25 rad/sec) / (-1.5 rad/sec²) = 16.7 sec

8 0

3 years ago

A 20.0 F capacitor initially charged to 30.0 uC is discharged through a 3.00 k resistor. How long does it take to reduce the cap

Maslowich

Answer:

t = 0.04159 s

Explanation:

when a capacitor and resister are connected in series. charge on its plates decreases exponentially

$q = qo * e^{\frac{-t}{RC}}$

where,

CHARGE q =15.0 C

INITIAL CHARGE q_0 = 30.0 uc

Resistor R = 20.0 F

capacitor C = 3.00

$15 = 30* e^{\frac{- t}{(20* 10^{-6} * 3* 1000)}$

t = 0.04159 s

3 0

4 years ago

m_Cu * sh_CuA system consists of a copper tank whose mass is 13 kilogram , 4 kilogram of liquid water, and an electrical resisto

notsponge [240]

Answer:

T₂ = 49.3°C

Explanation:

Applying law of conservation of energy to the system we get the following equation:

Energy Supplied by Resistor = Energy Absorbed by Tank + Energy Absorbed by Water

E = mC(T₂ - T₁) + m'C'(T'₂ - T'₁)

where,

E = Energy Supplied by Resistor = 100 KJ = 100000 J

m = mass of copper tank = 13 kg

C = Specific Heat of Copper = 385 J/kg.°C

T₂ = Final Temperature of Copper Tank

T₁ = Initial Temperature of Copper Tank = 27°C

T'₂ = Final Temperature of Water

T'₁ = Initial Temperature of Water = 50°C

m' = Mass of Water = 4 kg

C' = Specific Heat of Water = 4179.6 K/kg.°C

Since, the system will come to equilibrium finally. Therefor: T'₂ = T₂

Therefore,

(100000 J) = (13 kg)(385 J/kg.°C)(T₂ - 27°C) + (4 kg)(4179.6 J/kg.°C)(T₂ - 50°C)

100000 J = (5005 J/°C)T₂ - 135135 J + (16718.4 J/°C)T₂ - 835920 J

100000 J + 135135 J + 835920 J = (21723.4 J/°C)T₂

(1071055 J)/(21723.4 J/°C) = T₂

8 0

3 years ago

An elastic conducting material is stretched into a circular loop of 14.7 cm radius. It is placed with its plane perpendicular to

Hunter-Best [27]

Answer: 0.66 V

Explanation:

Given

Magnetic field, B = 0.963 T

Instantaneous rare = 74.5 cm/s = 0.745 m/s

radius, r = 14.7 cm = 0.147 m

We will use the formula

emf = dΦ/dt

emf = d(BA)/dt

emf = d(Bπr²)/dt

if B is constant, then we can say

emf = Bπ d(r²)/dt on differentiating, we have,

emf = Bπ (2r dr/dt)

emf = 2πrB dr/dt substituting each values, we have

emf = 2 * 3.142 * 0.147 * 0.963 * 0.745

emf = 0.66 V

Therefore, the induced emf in the loop at that instant is 0.66 V

3 0

3 years ago