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Lyrx [107]
1 year ago
11

Four students were loading boxes of food collected during a food drive. The force that each student exerted while lifting and th

e distance each box was lifted are listed in the table.
A 3-column table with 4 rows. The first column labeled student has entries Chet, Mika, Sara, Bill. The second column labeled Force (Newtons) has entries 50, 40, 30, 60. The third column labeled Distance (meters) has entries 1.0, 2.0, 1.5, 0.5.

Which lists the students in order from the greatest amount of work done to the least? (Work: W = Fd)
Physics
2 answers:
Aleonysh [2.5K]1 year ago
8 0

Answer:

The correct answer is B. Mika, Chet, Sara, Bill

Explanation:

Look at the picture

hope this helps :)

Fudgin [204]1 year ago
7 0

Answer:

it is B

Explanation:

Edge2020

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An apple falls from an apple tree growing on a 20° slope. The apple hits the ground with an impact velocity of 16.2 m/s straight
EleoNora [17]

Apple hits the surface with speed 16.2 m/s

The angle made by the apple velocity with normal to the incline surface is given as 20 degree

now the component of velocity which is parallel to the surface and perpendicular to the surface is given as

v_{perpendicular} = v cos20

v_{parallel} = v sin20

so here we have

v_{parallel} = 16.2 sin20

v_{parallel} = 5.5 m/s

<em>so its velocity along the incline plane will be 5.5 m/s</em>

7 0
2 years ago
Identical cannon balls are fired with the same force, one each from four cannons having respective bore lengths of 1.0 meter, 2.
Gala2k [10]

To answer this question, we must bear in mind the following considerations that are mentioned in the statement:

The cannon balls are identical and shoot with the same force

The force acting on the cannonball increases with the length of the hole.

You want to know which cannon will have the least momentum on the ball.

Then, the force on the ball increases as the barrel length increases and the impulse depends on the magnitude of the force, then, the cannon that will have the minimum impulse will be the 1 meter one.

The answer is option B.


4 0
2 years ago
If the blocks are released from rest, which way does the 10 kg block slide, and what is its acceleration? enter a positive value
anyanavicka [17]
<span>Let m1=10kg and m2=5kg and for our calculations assume right is positive and up is positive (note: for block hanging, the x axis is vertical so tilt your head to help) For m1 Sigma Fx = ma T - m1gsin35 = m1a where T = tension For m2 m2g - T = m2a Add equation together m1a + m2a = T-m1gsin35 + m2g - T a(m1 + m2) = m2g - m1gsin35 a= (5*9.8 - 10*9.8*sin35)/(10 + 5) a= -0.48m/s/s So the system is moving in the opposite direction of our set coordinate system where we said right positive, its negative so its moving left therefore down the ramp</span>
6 0
2 years ago
Read 2 more answers
Two tiny particles having charges 20.0 μC and 8.00 μC are separated by a distance of 20.0 cm What are the magnitude and directio
Alecsey [184]

Answer:

The magnitude and direction of electric field midway between these two charges is 10.8\times10^{5}\ N/C along AB.

Explanation:

Given that,

First charge q_{1}= 20\mu C

second charge q_{2}= 8\mu C

Distance = 20 cm

We need to calculate the electric field

For first charge,

Using formula of electric field

E_{1}= \dfrac{kq_{1}}{r^2}

Put the valueinto the formula

E_{1}=\dfrac{9\times10^{9}\times20\times10^{-6}}{10\times10^{-2}}

E_{1}=18\times10^{5}\ N/C

Direction of electric field along AB

We need to calculate the electric field

For second charge,

Using formula of electric field

E_{2}= \dfrac{kq_{2}}{r^2}

Put the valueinto the formula

E_{2}=\dfrac{9\times10^{9}\times8\times10^{-6}}{10\times10^{-2}}

E_{2}=7.2\times10^{5}\ N/C

Direction of electric field along AO

We need to calculate the net electric field at midpoint

E_{net}=E_{1}-E_{2}

E_{net}=(18-7.2)\times10^{5}\ N/C

E_{net}=10.8\times10^{5}\ N/C

Direction of net electric field along AB

Hence, The magnitude and direction of electric field midway between these two charges is 10.8\times10^{5}\ N/C along AB.

8 0
2 years ago
A helicopter, starting from rest, accelerates straight up from the roof of a hospital. The lifting force does work in raising th
kobusy [5.1K]

Answer:

24,267.6 watts

Explanation:

from the question we are given the following:

mass (m) = 810 kg

final velocity (v) = 7 m/s

initial velocity (u) = 0 m/s

time (t) = 3.5 s

final height (h₁) = 8.2 m

initial height (h₀) = 0 m

acceleration due to gravity (g) = 9.8 m/s^{2}

find the power

power = \frac{work done}[time}

and

work done = change in kinetic energy (K.E) + change in potential energy (P.E)

work done = (0.5 mv^{2} - 0.5 mu^{2} ) + ( mgh₁ - mgh₀)

since u and h₀ are zero the work done now becomes

work done = (0.5 mv^{2}) + ( mgh₁ )                    

work done = (0.5 x 810 x 7^{2}) + ( 810 x 9.8 x 8.2)

work done = 84, 936.6 joules

recall that power = \frac{work done}[time}

power = \frac{84,936.6}[3.5}

power = 24,267.6 watts

7 0
1 year ago
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