A change in which of the following effects the weight of an object?

Which of the following produces a physical change ?

stellarik [79]

Answer:

B: leaving metal outside in the rain until rust forms an it's surface

4 0

1 year ago

Assuming a typical efficiency for energy use by the body, how many slices of pizza must you eat to walk for 2.5 h at a speed of

larisa86 [58]

Answer:

2.7 Pizzas.

Explanation:

The power required to walk through 5km in 1 hour is 380W.

A watt is basically Jules per second, then we need to standardized this measurement to second.

5km/hr is equal to,

$\frac{5km}{hr}*\frac{1hr}{3600s}*\frac{1000m}{1km}=1.389m/s$

Walking by 2.5 hours is equal to a distance of,

$d=v*t=1.389*(2.5*3600) = 12500m$

The total energy required then would be,

$E = \frac{380J}{1.389m/s}(12500)=3.4199*10^6J$

Then we know that one pizza slice gives $1260*10^3J$ of energy, the total pizza needed are,

$\eta = \frac{3.4199*10^6}{1260*10^3} = 2.7142$

<em>Then you need to buy 3 pizza.</em>

6 0

3 years ago

Which statements best characterize conductors? Check all that apply.

romanna [79]

It would be the first one and the third one

5 0

2 years ago

Read 2 more answers

At a time t = 2.80 s , a point on the rim of a wheel with a radius of 0.230 m has a tangential speed of 51.0 m/s as the wheel sl

bonufazy [111]

Answer:

Part A) the angular acceleration is α= 44.347 rad/s²

Part B) the angular velocity is 195.13 rad/s

Part C) the angular velocity is 345.913 rad/s

Part D ) the time is t= 7.652 s

Explanation:

Part A) since angular acceleration is related with angular acceleration through:

α = a/R = 10.2 m/s² / 0.23 m = 44.347 rad/s²

Part B) since angular acceleration is related

since

v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(3.4 s - 2.8 s) = 44.88 m/s

since

ω = v/R = 44.88 m/s/ 0.230 m = 195.13 rad/s

Part C) at t=0

v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(0 s - 2.8 s) = 79.56 m/s

ω = v/R = 79.56 m/s/ 0.230 m = 345.913 rad/s

Part D ) since the radial acceleration is related with the velocity through

ar = v² / R → v= √(R * ar) = √(0.23 m * 9.81 m/s²)= 1.5 m/s

therefore

v = v0 + a*(t-t0) → t =(v - v0) /a + t0 = ( 1.5 m/s - 51.0 m/s) / (-10.2 m/s²) + 2.8 s = 7.652 s

t= 7.652 s

4 0

2 years ago

A 2. 0 μf and a 4. 0 μf capacitor are connected in series across an 8. 0-v dc source. what is the charge on the 2. 0 μf capacito

Nezavi [6.7K]

voltage across 2.0μf capacitor is 5.32v

Given:

C1=2.0μf

C2=4.0μf

since two capacitors are in series there equivalent capacitance will be

[tex] \frac{1}{c} = \frac{1}{c1} + \frac{1}{c2} [/tex]

$c = \frac{c1 \times c2}{c1 + c2}$

$= \frac{2 \times 4}{2 + 4}$

=1.33μf

As the capacitance of a capacitor is equal to the ratio of the stored charge to the potential difference across its plates, giving: C = Q/V, thus V = Q/C as Q is constant across all series connected capacitors, therefore the individual voltage drops across each capacitor is determined by its its capacitance value.

Q=CV

given,V=8v

$= 1.33 \times 10 {}^{ - 6} \times 8$