When a pair of balanced forces acts on an object, the net force that results is.

the wattage ratting of a light bulb is the power it sonsumers when it is connected accross a 120 V potential difference what is

igor_vitrenko [27]

Answer:

240 Ω

Explanation:

Resistance: This can be defined as the opposition to the flow of current in an electric field. The S.I unit of resistance is ohms (Ω).

The expression for resistance power and voltage is give as,

P = V²/R.......................... Equation 1

Where P = Power, V = Voltage, R = Resistance

Making R the subject of the equation,

R = V²/P.................... Equation 2

Given: V = 120 V, P = 60 W.

Substitute into equation 2

R = 120²/60

R = 240 Ω

Hence the resistance of the bulb = 240 Ω

8 0

3 years ago

A 50 cm^3 block of iron is removed from an 800 degrees Celsius furnance and immediately dropped into 200 mL of 20 degrees Celsiu

NNADVOKAT [17]

Answer:

% of water boils away= 12.64 %

Explanation:

given,

volume of block = 50 cm³ removed from temperature of furnace = 800°C

mass of water = 200 mL = 200 g

temperature of water = 20° C

the density of iron = 7.874 g/cm³ ,

so the mass of iron(m₁) = density × volume = 7.874 × 50 g = 393.7 g

the specific heat of iron C₁ = 0.450 J/g⁰C

the specific heat of water Cw= 4.18 J/g⁰C

latent heat of vaporization of water is L_v = 2260 k J/kg = 2260 J/g

loss of heat from iron is equal to the gain of heat for the water

$m_1\times C_1\times \Delta T = M\times C_w\times \Delta T + m_2\times L_v$

$393.7\times 0.45\times (800-100) = 200\times 4.18\times(100-20) + m_2\times 2260$

m₂ = 25.28 g

25.28 water will be vaporized

% of water boils away = $\dfrac{25.28}{200}\times 100$

% of water boils away= 12.64 %

5 0

3 years ago

An electric motor operates a pump that irri-

PolarNik [594]

Answer:

a. 5A

b. 39.60%

Explanation:

The computation is shown below:

a. The current does it draw is

= v ÷ R

= 110v ÷ 22

= 5A

b. Now the efficiency of the motor is

n = mgh ÷ vlt

= (10,000 × 9.8 × 8) ÷ (5 × 3600 × 110)

= 784000 J ÷ 1,980,000

= 39.60%

hence, the above formulas are applied & the same is relevant

7 0

3 years ago

Assume that the weights of quarters are normally distributed with a mean of 5.67 g and a standard deviation 0.070 g. A vending m

Igoryamba

Answer:

The percentage of legal quarters that will be rejected is 1.954 %.

Explanation:

Here we are required to find

P(X<5.48 or X>5.82)

We have μ = 5.67 and σ = 0.07 where the range is between 5.48 and 5.82 we each z-score given by

$z =\frac{x-\mu}{\sigma}$

Therefore for 5.48 we have $z_{5.48} = \frac{5.48 -5.67}{0.07}$ = -2.714

and for 5.82 we have $z_{5.82} = \frac{5.82 -5.67}{0.07}$ = 2.143

Therefore we have the area of interest on the normal distribution chart given by

P(Z<-2.714) + P(Z > 2.143) = P(Z<-2.714) + (1 - P(Z < 2.143))

+ 0.00336 + 1 - 0.98382 = 0.01954

= 0.01954 × 100 = 1.954 %

Therefore 1.954 percentage of legal quarters will be rejected.

3 0

3 years ago

When can you apply the idea that momentum is constant to solve a problem?

wlad13 [49]

It seems that you have missed the necessary options for us to answer this question, so I had to look for it. Anyway, here is the answer. You can apply the idea that <span>momentum is constant to solve a problem when the system is not isolated but the time interval when the external forces are exerted is very small. Hope this helps.</span>

3 0

3 years ago