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3 years ago

In which situation is maximum work considered to be done by a force

Physics

2 answers:

madam [21]3 years ago

7 0

Answer:

The angle between the force and displacement is 0°.

Explanation:

Katena32 [7]3 years ago

3 0

Answer:

Work done is maximum when the movement of object is in line and direction of force.

Work done is maximum, when displacement takes place along the direction of force.

Work done is given by the equation

W = F.S

 W = F. S cos Θ

When cos Θ = 0° ; cos 0 = 1

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10. How much total work do you do when you lift a 50 kg microwave 1.0 m off the ground and then push it 1.0 m

Mariulka [41]

Work formula:

$W = Fd\cos(\theta)$

F = 50N, d = 1.0 m

When you lift something straight up, the angle of the force is 90º

cos(90º) is 0, so there's no work done when you lift the microwave off the ground

$W = (50N)(1.0)(0) = 0$

F = 50N, d = 1.0 m

When you push the microwave, the angle is 0º and cos(0º) is 1. So there is work done here:

$W = (50 N)(1.0m)(1)$

$W = 50$

total work = 50 joules

6 0

3 years ago

Which property of potential energy distinguishes it from kinetic energy

Bas_tet [7]

Answer:

Shape and position

Explanation:

Hope this helps! :)

7 0

2 years ago

A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s

Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g= $4.67\times 10^{-3} kg$

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1, $m_1=1177g=1.177kg$

Mass of block 2, $m_2=1626 g=1.626 kg$

Velocity of block 1, $v_1=0.681m/s$

(a)

Let velocity of the second block after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

$mv=m_1v_1+(m+m_2)v_2$

$4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2$

$1.66719=0.801537+1.63067v_2$

$1.66719-0.801537=1.63067v_2$

$0.865653=1.63067v_2$

$v_2=\frac{0.865653}{1.63067}$

$v_2=0.531m/s$

Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

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

$k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)$

$k_i=297.59 J$

Final kinetic energy after collision

$K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2$

$K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2$

$K_f=0.5028 J$

Now, he ratio of the total kinetic energy after the collision to that before the collision

= $\frac{k_f}{k_i}=\frac{0.5028}{297.59}$

=0.00169

5 0

3 years ago

List small/average stars

mario62 [17]

Answer:

Lol, you should do Nate, Bobby, Cindy, Joe, and Beth

Jk, if you want to be series and probably not fail go for these:

If it wants types of small/average stars, then go with

Small star names:

OGLE-TR-122B

Gliese 229 B

TRAPPIST-1

Teegarden's Star

Luyten 726-8 (A and B)

Proxima Centauri

Wolf 359 111400

Ross 248

Barnard's Star

CM Draconis B

Ross 154 167000

CM Draconis A

Kapteyn's Star

7 0

3 years ago

You make a(n) _____ by fastening together usually perpendicular parts with the ends cut at an angle.

nadezda [96]

Answer:

Miter joint

Explanation:

Made by fastening together usually perpendicular parts with the ends cut at an angle

8 0

3 years ago