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topjm [15]
2 years ago
13

A construction crane, like the one shown, has a power output of 1,500 watts. If it takes 200 seconds to lift the roof to the top

of the building, how much energy will the crane use
Physics
1 answer:
Likurg_2 [28]2 years ago
4 0

Answer:7.5j

Explanation:

P=w/s

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A brick is released with no initial speed from the roof of a building and strikes the ground in 1.80 s , encountering no appreci
Mars2501 [29]
<span>Kinematics is used in this problem. The mass does not matter here because the question is mass independent. vi = 0 vf = x d = ? d = vi + 1/2 a t^2 d = 0 + 1/2 (9.8) (1.8)^2 d = 15.9 m (counting sig figs)</span>
8 0
3 years ago
A particle moves in the xy plane with constant acceleration. At time zero, the particle is at x = 6.0 m, y = 10.0 m, and has vel
mihalych1998 [28]

Explanation:

Given that,

Position of the particle at t = 0,

y=(6i+10j)\ m

Velocity of the particle at t = 0

u=(1i+6j)\ m

Acceleration of the particle,

a=(5i+7j)\ m/s^2

Solution,

(a) Let v is the velocity at t = 10 s. Using the equation of kinematics as :

v=u+at

v=(1i+6j)+(5i+7j)10

v=(51i+76j)\ m/s

(b) Let y' is the position at t = 1 s. Again using second equation of kinematics as :

y'=y+ut+\dfrac{1}{2}at^2

y'=(6i+10j)+(1i+6j)1+\dfrac{1}{2}\times (5i+7j)1^2

y'=\dfrac{19}{2}i+\dfrac{39}{2}j

(c) Magnitude of y',

|y'|=\sqrt{(\dfrac{19}{2})^2+(\dfrac{39}{2})^2}

|y'| = 21.69 meters

Direction of the y',

tan\theta=\dfrac{y}{x}

tan\theta=\dfrac{39/2}{19/2}

\theta=64.02^{\circ}

Hence, this is the required solution.

4 0
2 years ago
Derek owns a farm that produces soybeans, avocados, and pecans. What is the word used to describe these items?
Novosadov [1.4K]

Answer:

crops

Explanation:

dairy are milk products and those aren’t

live stock is domesticated farm animals raised to produce resources

non-perishables are canned food and dried food. they can last a long time. they don’t need to be refrigerated or

7 0
3 years ago
Read 2 more answers
A planet of mass 7.00 1025 kg is in a circular orbit of radius 6.00 1011 m around a star. The star exerts a force on the planet
BARSIC [14]

Answer:

A) 1.88 * 10^17 m

B) 1.22 * 10^34 J

C) 1.95 * 10^34 J

Explanation:

Parameters given:

Mass of planet = 7.00 * 10^25 kg

Radius of orbit = 6.00 * 10^11 m

Force exerted on planet = 6.51 * 10^22 N

Velocity of planet = 2.36 * 10^4 m/s

A) The distance traveled by the planet is half of the circumference of the orbit (which is circular).

The circumference of the orbit is

C = 2 * pi * R

R = radius of orbit

C = 2 * 3.142 * 6.0 * 10¹¹

C = 3.77 * 10¹² m

Hence, distance traveled will be:

D = 0.5 * 3.77 * 10¹²

D = 1.88 * 10 ¹² m/s

B) Work done is given as:

W = F * D

W = 652 * 10²² * 1.88 * 10¹¹

W = 1.22 * 10³⁴ J

C) Change in Kinetic energy is given as:

K. E. = 0.5 * m * v²

K. E. = 0.5 * 7 * 10^25 * (2.36 * 10^4)²

K. E. = 1.95 * 10³⁴ J

7 0
3 years ago
What is torque in physics ?
irina [24]

Answer:

Torque is a measure of the force that can cause an object to rotate about an axis.

8 0
3 years ago
Read 2 more answers
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