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2 years ago

If 3.0 newtons are required to lift a carton 12 meters, how much work is done? 0.0 J 4.0 J 15 J 36 J

Physics

2 answers:

svlad2 [7]2 years ago

7 0

The answer is actually 36 J.

Hope this helped :)

Brainliest?

statuscvo [17]2 years ago

3 0

Answer:

36 J is the answer

Explanation:

i just had this problem on a test

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makkiz [27]

Answer:

It's C. Length

Explanation:

Just think of a ruler. Hope I helped! :)

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2 years ago

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A car with an initial velocity of 5.0 is acelerated at 3.0for 4.0seconds what os the velocity of tje car ofter the 4.0seconds

sasho [114]

Vf =Vi+at
Vf= 5 + 3*4
Vf=5+12
Vf=17m/sec

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3 years ago

What is the temperature outside of a tree?

ozzi

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2 years ago

What is the net torque on the square plate, with sides 0.2 m, from each of the three forces? F1=18 N, F2=26 N, and F3=14 N. Use

marshall27 [118]

Answer:

The net torque on the square plate is 2.72 N-m.

Explanation:

Given that,

Side = 0.2 m

Force $F_{1}=18\ N$

Force $F_{2}=26\ N$

Force $F_{3}=14\ N$

We need to calculate the torque due to force F₁

Using formula of torque

$\tau_{1}=-F_{1}d_{1}$

$\tau_{1}=-F_{1}\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{1}=-18\times\dfrac{0.2}{2}$

$\tau_{1}=-1.8\ N-m$

We need to calculate the torque due to force F₂

Using formula of torque

$\tau_{2}=F_{2}d_{2}$

$\tau_{2}=F_{2}\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{2}=26\times\dfrac{0.2}{2}$

$\tau_{2}=2.6\ N-m$

We need to calculate the torque due to force F₃

Using formula of torque

$\tau_{3}=F_{3}d_{3}$

$\tau_{3}=(F_{3}\sin45+F_{3}\cos45)\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{3}=0.1(14\sin45+14\cos45)$

$\tau_{3}=1.92\ N-m$

We need to calculate the net torque on the square plate

$\tau=\tau_{1}+\tau_{2}+\tau_{3}$

$\tau=-1.8+2.6+1.92$

$\tau=2.72\ N-m$

Hence, The net torque on the square plate is 2.72 N-m.

3 0

3 years ago

PLEASE help me quickly

Anna007 [38]

Explanation:

its either a or d however i say the best choice is d

6 0

2 years ago