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3 years ago

9

Devonte pushes a wheelbarrow with 830 W of power. How much work is required to get the wheelbarrow across the yard in 11 s? Roun

d your answer to two significant figures.

2 answers:

zaharov [31]3 years ago

8 0

9,100 is the answer if you do the work so i failed for yall you welcome

Maslowich3 years ago

6 0

Answer: To get the wheelbarrow across the yard in 11 s Is required 9,100 J of work.

Explanation:

Hi, to solve this problem we have to apply the formula:

Power (w) = work (J) / time (sec)

So, replacing the variables with the values given:

830 w = work / 11sec

Isolating work we have:

830 w / 11 sec = work

Work = 9,130 J = 9,100 j (rounded to two significant figures)

To get the wheelbarrow across the yard in 11 s Is required 9,100 J of work.

Feel free to ask for more if it´s necessary or if you did not understand something.

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Cuál es la diferencia entre fricción beneficiosa y perjudicial? Pon 5 ejemplos de cada uno.

Rus_ich [418]

Answer:

Difference is given below.

Explanation:

The main difference between beneficial and harmful friction is that beneficial friction is necessary for performing different activities while on the other hand, harmful effect of friction is destroy different parts of products and machines. examples of beneficial friction are walking, holding things, rubbing hands to produce heat, running etc whereas examples of harmful friction are destruction of sole, slipping, tearing of machine's part, Wet roads and Mudslides etc.

7 0

3 years ago

A car starts from rest and experiences a constant acceleration of 5 m/s2 until it reaches a speed of 650 meters

Zigmanuir [339]

Answer: Time.

Explanation:

Hey there!!!

The reason is,

According to the question we have,

Initial velocity = 0m/s { as it starts from rest}.

Final velocity = 650 m/s

acceleration = 5m/s^2.

Looking on this formula,

$a = \frac{v - u}{t}$

Where "v" represents final velocity, "u" represents initial velocity and "a" represents acceleration. We dont have is time "t".

So, it's obvious that tine is an unknown variable.

<em><u>Hope</u></em><em><u> </u></em><em><u>it helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

8 0

3 years ago

What force pulls all objects with mass toward<br> one another?

Vilka [71]

Gravity

Hope this helps:)

5 0

3 years ago

A pirate ship shoots a cannon towards an oncoming ship. The cannon ball has a mass of 25 kg and the ship has a mass of 2500 kg.

mars1129 [50]

Answer:

25 m/s in the opposite direction with the ship recoil velocity.

Explanation:

Assume the ship recoil velocity and velocity of the cannon ball aligns. By the law of momentum conservation, the momentum is conserved before and after the shooting. Before the shooting, the total momentum is 0 due to system is at rest. Therefore, the total momentum after the shooting must also be 0:

$m_sv_s + m_bv_b = 0$

where $m_s = 2500 kg, m_b = 25 kg$ are masses of the ship and ball respectively. $v_s = 0.25 m/s, v_b$ are the velocities of the ship and ball respectively, after the shooting.

$2500*0.25 + 25*v_b = 0$

$25v_b = -2500*0.25$

$v_b = -2500*0.25/25 = -25 m/s$

So the cannon ball has a velocity of 25 m/s in the opposite direction with the ship recoil velocity.

3 0

3 years ago

10. If the mass of the Earth is... increased by a factor of 2, then the Fgrav is ______________ by a factor of _______. ... incr

12345 [234]

Answer:

If the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3, then Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4

Explanation:

In order to solve this question, we must take into account that the force of gravity is given by the following formula:

$F_{g0}=G \frac{mM_{E0}}{r^{2}}$

So if the mass of the earth is increased by a factor of 2, this means that:

$M_{Ef}=2M_{E0}$

so:

$F_{gf}=G \frac{2mM_{E0}}{r^{2}}$

Therefore:

$\frac{F_{gf}}{F_{g0}}=\frac{G \frac{2mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}$

When simplifying we end up with:

$\frac{F_{gf}}{F_{g0}}=2$

so if the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3

So if the mass of the earth is increased by a factor of 2, this means that:

$M_{Ef}=3M_{E0}$

so:

$F_{gf}=G \frac{3mM_{E0}}{r^{2}}$

Therefore:

$\frac{F_{gf}}{F_{g0}}=\frac{G \frac{3mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}$

When simplifying we end up with:

$\frac{F_{gf}}{F_{g0}}=3$

so if the mass of the Earth is increased by a factor of 3, then the Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4

So if the mass of the earth is decreased by a factor of 4, this means that:

$M_{Ef}=\frac{M_{E0}}{4}$

so:

$F_{gf}=G \frac{mM_{E0}}{4r^{2}}$

Therefore:

$\frac{F_{gf}}{F_{g0}}=\frac{G \frac{mM_{E0}}{4r^{2}}}{G \frac{mM_{E0}}{r^{2}}}$

When simplifying we end up with:

$\frac{F_{gf}}{F_{g0}}=\frac{1}{4}$

so if the mass of the Earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4.

4 0

3 years ago