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2 years ago

Describe how an inclined plane increases the force without changing the amount of work done

1 answer:

6 0

An inclined plane decreases the amount of force needed to move an object but increases the distance the onject needs to be moved. Since work = distance x force, whe amount of work stays the same.

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A Micro –Hydro turbine generator is accelerating uniformly from an angular velocity of 610 rpm to its operating angular velocity

Salsk061 [2.6K]

Answer:

Angular displacement of the turbine is 234.62 radian

Explanation:

initial angular speed of the turbine is

$\omega_i = 2\pi f_1$

$\omega_1 = 2\pi(\frac{610}{60})$

$\omega_1 = 63.88 rad/s$

similarly final angular speed is given as

$\omega_f = 2\pi f_2$

$\omega_2 = 2\pi(\frac{837}{60})$

$\omega_2 = 87.65 rad/s$

angular acceleration of the turbine is given as

$\alpha = 5.9 rad/s^2$

now we have to find the angular displacement is given as

$\theta = \omega t + \frac{1}{2}\alpha t^2$

$\theta = (63.88)(3.2) + (\frac{1}{2})(5.9)(3.2^2)$

$\theta = 234.62 radian$

3 0

2 years ago

At what speed do a bicycle and its rider, with a combined mass of 100 kg, have the same momentum as a 1400 kg car traveling at 6

Ilia_Sergeevich [38]

Momentum of car

Given: Mass m= 1,400 Kg; V = 6.0 m/s

Formula: P = mv

P = (1,400 Kg)(6.0 m/s)

P = 8,400 Kg.m/s

Velocity of the rider to have the same momentum as a car.

Mass of rider and bicycle m = 100 Kg

P = mv

V = P/m

V = 8,400 Kg.m/s/100 Kg

V = 84 m/s

7 0

2 years ago

Two neutral metal spheres on wood stands are touching. A negatively charged rod is held directly above the top of the left spher

Paha777 [63]

Answer:

The right sphere is negatively charged, the left sphere is charged positively.

Explanation:

When a negatively charged rod is held above the top of left sphere, the rod will attract positive charges and repel negative charges. As the sphere are initially touching each other so positive charges from the both spheres will moves toward the rod. When we separate the spheres positive charges from right sphere have already moved toward the rod i.e. left sphere, creating a deficiency of positive charges in the right sphere and excessiveness of positive charges in left sphere , hence the right sphere will remain negatively charged and left sphere will remain positively charged.

4 0

2 years ago

Which of the graphs describes the motion of a person who first rode her bicycle at constant speed and then rested?

sergeinik [125]

Answer:

Its graph 1

Explanation:

She started at the origin and kept riding her bike until she stopped which causes the line to go staright because she's not moving.

5 0

1 year ago

Long Jump: inital center of mass height of 1.08 m, final center of mass height of 0.42 m, projection velocity of 8.7 m/s, projec

sammy [17]

Answer:

1) The maximum jump height is reached at A. $0.337s$

2) The maximum center of mass height off of the ground is B. $1.64m$

3) The time of flight is C. $0.834s$

4) The distance of jump is B. $7.49m$

Explanation:

First of all we need to decompose velocity in its rectangular components, so

$v_{xi}=8.7m/s(cos 22.3\°)=8.05m/s= constant\\v_{yi}=8.7m/s(sin 22.3\°)=3.3m/s$

1) We use, $v_{fy}=v_{iy}-gt$ , as we clear it for $t$ and using the fact that $v_{fy}=0$ at max height, we obtain $t=\frac{v_{iy}}{g} =\frac{3.3m/s}{9,8m/s^{2}} =0.337s$

2) We can use the formula $y_{max}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ for $t=0.337s$ , so

$y_{max}=1.08m+(3.3m/s)(0.337s)-\frac{(9.8m/s^{2})(0.337)^{2}}{2}=1.64m$

3) We can use the formula $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find total time of fligth, so $0.42=1.08+3.3t-\frac{(9.8)t^{2}}{2}\\0=-4.9t^{2}+3.3t+0.66$ , as it is a second-grade polynomial, we find that its positive root is $t=0.834s$

4) Finally, we use $x=v_{x}t=8.05m/s(0.834s)=6.71m$ , as it has an additional displacement of $0.77m$ due the leg extension we obtain,

$x=6.71m+0.77m=7.48m$ , aprox $x=7.49m$

5 0

3 years ago