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4 years ago

If it takes 600 N to move a box 5 meters, how much work is done on the box?

Physics

1 answer:

Arada [10]4 years ago

6 0

Answer:3000joules

Explanation: work=force×distance

Work=600×5

Work=3000joules

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Meters for mass kilograms for volume cubic meters for density kilograms per cubic meter

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3 years ago

½H2(g) + ½I2(g) → HI(g), ΔH = + 26 kJ/mole 117 kJ/mole + C(s) + 2S(s) → CS2(g) The temperature of the surroundings will:

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From the above reaction the temperature of the surroundings will increase.

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3 years ago

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3 years ago

The first car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 9

kirill [66]

Answer:

Speed of the car 1 = $V_1=8.98m/s$

Speed of the car 2 = $V_2=17.96m/s$

Explanation:

Given:

Mass of the car 1 , M₁ = Twice the mass of car 2(M₂)

mathematically,

M₁ = 2M₂

Kinetic Energy of the car 1 = Half the kinetic energy of the car 2

KE₁ = 0.5 KE₂

Now, the kinetic energy for a body is given as

$KE =\frac{1}{2}mv^2$

where,

m = mass of the body

v = velocity of the body

thus,

$\frac{1}{2}M_1V_1^2=0.5\times \frac{1}{2}M_2V_2^2$

$\frac{1}{2}2M_2V_1^2=0.5\times \frac{1}{2}M_2V_2^2$

$2M_2V_1^2=0.5\times M_2V_2^2$

$2V_1^2=0.5\times V_2^2$

$4V_1^2= V_2^2$

$2V_1= V_2$ .................(1)

also,

$\frac{1}{2}M_1(V_1+9.0)^2=\frac{1}{2}M_2(V_2+9.0)^2$

$\frac{1}{2}2M_2(V_1+9.0)^2=\frac{1}{2}M_2(2V_1+9.0)^2$

$2(V_1+9.0)^2=(2V_1+9.0)^2$

$\sqrt{2}(V_1+9.0)=(2V_1+9.0)$

$(\sqrt{2}V_1+ \sqrt{2}\times 9.0)=(2V_1+9.0)$

$(\sqrt{2}V_1+ 12.72)=(2V_1+9.0)$

$(2V_1-\sqrt{2}V_1)=(12.72-9.0)$

$(0.404V_1)=(3.72)$

$V_1=8.98m/s$

and, from equation (1)

$V_2=2\times 8.98m/s = 17.96m/s$

Hence,

Speed of car 1 = $V_1=8.98m/s$

Speed of car 2 = $V_2=17.96m/s$

8 0

3 years ago

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Calculate the force of a particle with a net charge of 170 coulombs traveling at a speed of 135 meters/second perpendicular to t

amm1812

Answer:

F=1.14N j

Explanation:

The magnitude of the magnetic force over a charge in a constant magnetic field is given by the formula:

$|\vec{F}|=|q\vec{v} \ X\ \vec{B}|=qvsin\theta$ (|)

In this case v and B vectors are perpendicular between them. Furthermore the direction of the magnetic force is:

-i X k = +j

Finally, by replacing in (1) we obtain:

$\vec{F}=(170C)(135\frac{m}{s})(5.0*10^{-5}T)=1.14N\ \hat{j}$

hope this helps!

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3 years ago

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