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3 years ago

How much power is used if a force of 54 newtons is used to pull a wagon a distance of 10 meters in 6 seconds?

Physics

2 answers:

Debora [2.8K]3 years ago

8 0

Answer : The amount of power used is, 90 W

Solution :

Formula used for power :

$P=\frac{W}{t}\\\\As,W=F\times d\\\\So,P=\frac{F\times d}{t}$

where,

P = power

W = work done

F = applied force = 54 N

d = displacement = 10 m

t = time taken = 6 s

Now put all the given values in the above formula, we get the amount of power used.

$P=\frac{(54N)\times (10m)}{6s}=90watts=90W$

Therefore, the amount of power used is, 90 W

AnnZ [28]3 years ago

7 0

First, calculate the work done: 54 x 10 = 540J.
Now calculate power: 540 / 6 = 90.
The answer is 90W.

Hope this helps.

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Suppose that you have a 680 Ω, a 720 Ω and a 1.20 kΩ resistor. (a) What is the maximum resistance you can obtain by combining th

Delvig [45]

Explanation:

As the given data is as follows.

$R_{1} = 680 \ohm$ ohm $\ohm$ , $R_{2} = 720 \ohm$ ohm,

$R_{3} = 1.2 k\ohm$ = 1200 $\ohm$ (as 1 k ohm = 1000 m)

(a) We will calculate the maximum resistance by combining the given resistances as follows.

Max. Resistance = $R_{1} + R_{2} + R_{3}$

= $(680 + 720 + 1200) \ohm$ ohm

= 2600 ohm

or, = 2.6 $k\ohm$ ohm

Therefore, the maximum resistance you can obtain by combining these is 2.6 $k\ohm$ ohm.

(b) Now, the minimum resistance is calculated as follows.

Min. Resistance = $\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}$

= $\frac{1}{680} + \frac{1}{720} + \frac{1}{1200}$

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3 0

2 years ago

The results of a recent television survey of American TV households revealed that 86 out of every 100 TV households have at leas

Lelechka [254]

Answer:

$q = \dfrac{14}{100}$

Explanation:

given,

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now, Calculating probability that at least one remote control

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7 0

3 years ago

A sample of copper has a volume of 23.4 cm3 if the density of copper is 8.9 gcm3 what is the coppers mass?

murzikaleks [220]

The answer is: " 208 g " .
_____________________________________________
Explanation:
__________________________________________
The formula/ equation for density is:
__________________________________________
D = m / V ; That is, "mass divided by volume" ;

Density is expressed as:
__________________________________________
   "mass per unit volume"; in which the "mass" is expressed in units of "g" ("grams") ; and the "unit volume" is expressed in units of:
"cm³ " or "mL";
_____________________________________________
{Note the exact equivalent: 1 cm³ = 1 mL }.
____________________________________________
→ The formula is: " D = m / V " ;
___________________________________________
in which:

"D" refers to the "density" (see above), which is: "8.9 g/cm³ " (given);

"m" refers to the "mass" , in units of "g" (grams), which is unknown; and we want to find this value;

"V" refers to the "volume", in units of "cm³ " ;
which is: "23.4 cm³ " (given);
_________________________________________________
We want to find the mass, "m" ; so we take the original equation/formula for the density:
_________________________________________________
D = m / V ;
_________________________________________________________
And we rearrange; to isolate "m" (mass) on ONE side of the equation; and then we plug in our known/given values;
to solve for "m" (mass); in units of "g" (grams) ;
___________________________________________________
Multiply each side of the equation by "V" ;
____________________________________________________
V * { D = m / V } ; to get:
____________________________________________________
V * D = m ;   ↔ m = V * D ;
___________________________________________________
Now, we plug in the given values for "V" (volume) and "D" (density) ; to solve for the mass, "m" ;
______________________________________________________
m = V * D ;

m = (23.4 cm³) * (8.9 g / 1 cm³) = (23.4 * 8.9) g = 208.26 g ;

→ Round to "208 g" (3 significant figures);
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The answer is: " 208 g " .
_____________________________________________________

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3 years ago