Ask question

Ask question

All categories

1 year ago

14

A weightlifter lifts a 250-kg mass 0.5 meters above his head, how much PEg does the mass have (Note: g=9.8 m/s2)? Round your ans

wer to the nearest whole number. J

2 answers:

ValentinkaMS [17]1 year ago

8 0

Answer:

1225J

Explanation:

The Gravitational potential energy (PEG) gained by a mass lifted above the ground is given by

where

m is the mass

g = 9.8 m/s^2 is the acceleration due to gravity

h is the height at which the object has been lifted

In this problem, we have

m = 250 kg

h = 0.5 m

So, the PE of the object is

1225J

Maksim231197 [3]1 year ago

6 0

Answer:

1225 J

Explanation:

The Gravitational potential energy (PEG) gained by a mass lifted above the ground is given by

$PE=mgh$

where

m is the mass

g = 9.8 m/s^2 is the acceleration due to gravity

h is the height at which the object has been lifted

In this problem, we have

m = 250 kg

h = 0.5 m

So, the PE of the object is

$PE=(250 kg)(9.8 m/s^2)(0.5 m)=1225 J$

You might be interested in

A metal sphere with radius R1 has a charge Q1. Take the electric potential to be zero at an infinite distance from the sphere.

Airida [17]

Answer:

Part A : E = $\frac{1}{4\pi}$ ε₀ Q₁/R₁² Volt/meter

Part B : V = $\frac{1}{4\pi}$ ε₀ Q₁/R₁ Volt

Explanation:

Given that,

Charge distributed on the sphere is Q₁

The radius of sphere is R

₁

The electric potential at infinity is 0

<em>Part A</em>

The space around a charge in which its influence is felt is known in the electric field. The strength at any point inside the electric field is defined by the force experienced by a unit positive charge placed at that point.

If a unit positive charge is placed at the surface it experiences a force according to the Coulomb law is given by

F = $\frac{1}{4\pi}$ ε₀ Q₁/R₁²

Then the electric field at that point is

E = F/1

E = $\frac{1}{4\pi}$ ε₀ Q₁/R₁² Volt/meter

Part B

The electric potential at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against electric forces.

Thus, the electric potential at the surface of the sphere of radius R₁ and charge distribution Q₁ is given by the relation

V = $\frac{1}{4\pi}$ ε₀ Q₁/R₁ Volt

4 0

1 year ago

The length of a 60 W, 240 Ω light bulb filament is 60 cm Remembering that the current in the filament is proportional to the ele

faust18 [17]

Answer:

Finally current will be

i = 0.35 A

Explanation:

As we know that power of the bulb is given by the formula

$P = \frac{V^2}{R}$

now we have

$P = 60 W$

R = 240 ohm

so we have

$60 = \frac{V^2}{240}$

$V = 120 Volts$

now the current in the bulb is given as

$i = \frac{V}{R}$

$i = \frac{120}{240} = 0.5 A$

now when length of the filament is double

so the resistance of the wire also gets double

so we have

$P = \frac{V^2}{R}$

$60 = \frac{V^2}{480}$

$V = 169.7 volts$

now the current in the bulb is given as

$V = i R$

$169.7 = i(480)$

$i = 0.35 A$

8 0

1 year ago

What is the maximum negative displacement a dog could have if it started its motion at +3 m?

raketka [301]

Answer:

- 3 meter

Explanation:

A dog has started motion from +3 meter. ...(Given)

∴ maximum positive distance = + 3 meter

Magnitude of distance = 3

Maximum negative distance = (-) (magnitude of distance)

Maximum negative distance = (-) (3)

Maximum negative distance = -3 meter

Hence, maximum negative distance is -3 meter.

7 0

1 year ago

An air-track cart with mass m1=0.28kg and initial speed v0=0.75m/s collides with and sticks to a second cart that is at rest ini

arsen [322]

Kinetic energy is calculated through the equation,

KE = 0.5mv²

At initial conditions,

m₁: KE = 0.5(0.28 kg)(0.75 m/s)² = 0.07875 J

m₂ : KE = 0.5(0.45 kg)(0 m/s)² = 0 J

Due to the momentum balance,

m₁v₁ + m₂v₂ = (m₁ + m₂)(V)

Substituting the known values,

(0.29 kg)(0.75 m/s) + (0.43 kg)(0 m/s) = (0.28 kg + 0.43 kg)(V)

V = 0.2977 m/s

The kinetic energy is,
KE = (0.5)(0.28 kg + 0.43 kg)(0.2977 m/s)²
KE = 0.03146 J

The difference between the kinetic energies is 0.0473 J.

7 0

1 year ago

A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.

MakcuM [25]

Answer:

b. 9.5°C

Explanation:

$m_i$ = Mass of ice = 50 g

$T_i$ = Initial temperature of water and Aluminum = 30°C

$L_f$ = Latent heat of fusion = $3.33\times 10^5\ J/kg^{\circ}C$

$m_w$ = Mass of water = 200 g

$c_w$ = Specific heat of water = 4186 J/kg⋅°C

$m_{Al}$ = Mass of Aluminum = 80 g

$c_{Al}$ = Specific heat of Aluminum = 900 J/kg⋅°C

The equation of the system's heat exchange is given by

$m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C$

The final equilibrium temperature is 9.50022°C

4 0

1 year ago