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Paraphin [41]
3 years ago
11

A net force, the magnitude of which is 3800 N, accelerates a 1260-kg vehicle for 10.0 s. The vehicle travels 50.0 m during this

time. What power output does this represent? (g=10 m/s^2) *
Physics
1 answer:
Novay_Z [31]3 years ago
7 0

Answer:

SEE EXPLANATION

Explanation:

p =  \frac{fd}{t}  \\ where \: \\p  = power \\  f = force \\ d = distance \\ and \: t = time \\  \\ p =  \frac{3800 \times 50}{10}  \\ p =  \frac{190000}{10}  \\ p = 19000w

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KATRIN_1 [288]

Explanation:

It is given that, the position of a particle as as function of time t is given by :

r(t)=(8t+9)i+(2t^2-8)j+6tk

Let v is the velocity of the particle. Velocity of an object is given by :

v=\dfrac{dr(t)}{dt}

v=\dfrac{d[(8t+9)i+(2t^2-8)j+6tk]}{dt}

v=(8i+4tj+6k)\ m/s

So, the above equation is the velocity vector.

Let a is the acceleration of the particle. Acceleration of an object is given by :

a=\dfrac{dv(t)}{dt}

a=\dfrac{d[8i+4tj+6k]}{dt}

a=(4j)\ m/s^2

At t = 0, v=(8i+0+6k)\ m/s

v(t)=\sqrt{8^2+6^2} =10\ m/s

Hence, this is the required solution.

7 0
3 years ago
A 12.0 g sample of gas occupies 19.2 L at STP. what is the of moles and molecular weight of this gas?​
lubasha [3.4K]

At STP, 1 mole of an ideal gas occupies a volume of about 22.4 L. So if <em>n</em> is the number of moles of this gas, then

<em>n</em> / (19.2 L) = (1 mole) / (22.4 L)   ==>   <em>n</em> = (19.2 L•mole) / (22.4 L) ≈ 0.857 mol

If the sample has a mass of 12.0 g, then its molecular weight is

(12.0 g) / <em>n</em> ≈ 14.0 g/mol

4 0
2 years ago
If it takes 600 N to move a box 5 meters, how much work is done on the box?
Anika [276]

Answer:

<h2>3000 J</h2>

Option C is the correct option.

Explanation:

Given,

Force = 600 N

Distance = 5 meters

Work = ?

Now,

Work = Force \times distance

= 600  \times 5

Calculate the product

= 3000 \: Joule

Hope this helps...

Good luck on your assignment..

3 0
3 years ago
Read 2 more answers
Please help
iVinArrow [24]
Answer is 4,400,000 kg • m/s
4 0
3 years ago
Two 1.0 cm * 2.0 cm rectangular electrodes are 1.0 mm apart. What charge must be placed on each electrode to create a uniform el
kvv77 [185]

Answer:

The number of electrons that must be moved from one electrode to the other to accomplish this is 1.4 X 10⁹ electrons.

Explanation:

<u>Step 1:</u> calculate the charge on each electrode

Given;

Electric field strength = 2.0 X 10⁶ N/C

The distance between the electrode = 1mm = 1 X 10⁻³ m

Electric field strength (E) = Force (F)/Charge (q)

E =\frac{Kq}{r^2}

where;

E is the electric field strength = 2.0 X 10⁶ N/C

K is coulomb's constant = 8.99 X 10⁹ Nm²/C²

r is the distance between the electrodes = 1 X 10⁻³ m

q is the charge in each electrode = ?

q = \frac{Er^2}{K} = \frac{(2X10^6)(1X10^{-3})^2}{8.99 X10 ^9} = 0.2225 X 10⁻⁹ C

The charge on each electrode is 0.2225 X 10⁻⁹ C

<u>Step 2:</u> calculate the number of electrons to be moved from one electrode to the other.

1 electron contains 1.602 X 10⁻¹⁹ C

So, 0.2225 X 10⁻⁹ C will contain how many electrons ?

= (0.2225 X 10⁻⁹)/(1.602 X 10⁻¹⁹)

= 1.4 X 10⁹ electrons

Therefore, the number of electrons that must be moved from one electrode to the other to accomplish this is 1.4 X 10⁹ electrons.

8 0
3 years ago
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