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4 years ago

A worker pushes horizontally on a large crate with a force of 265 n and the crate is moved 4.5 M how much work was done

Physics

1 answer:

Vanyuwa [196]4 years ago

6 0

Answer:

the work done is 1192.5Nm

Explanation:

work done = force × distance

w.d = 265N × 4.5m

w.d = 1192.5Nm

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A + 8.42 nC point charge and a - 3.75 nC point charge are 2.73 cm apart. What is the electric field strength at the midpoint bet

aev [14]

Given:

The charge q1 = 8.42 nC

The charge q2 = -3.75 nC

The distance between the charges is d = 2.73 cm

To find the electric field strength at the midpoint between the two charges.

Explanation:

The distance between the charges and the midpoint is d' =d/2

The electric field strength can be calculated by the formula

$E\text{ = }\frac{k|q|}{d^{\prime2}}$

The electric field strength at the midpoint due to the charge q1 will be

$\begin{gathered} E_1=\frac{9\times10^9\times8.42\times10^{-9}\text{ C}}{(\frac{2.73}{2}\times10^{-2})^2} \\ =\text{ 4.07}\times10^5\text{ V/m} \end{gathered}$

The electric field strength at the midpoint due to the charge q2 will be

$\begin{gathered} E_2=\frac{9\times10^9\times3.75\times10^{-9}}{(\frac{2.73}{2}\times10^{-2})^2} \\ =1.8\times10^5\text{ V/m} \end{gathered}$

The electric field strength at the midpoint between the charges will be

$\begin{gathered} E=E_1+E_2 \\ =4.07\times10^5+1.8\times10^5 \\ =\text{ 5.87}\times10^5\text{ V/m} \end{gathered}$

The electric field strength at the midpoint between the two charges is 5.87 x 10^(5) m.

4 0

1 year ago

B) The student continues her investigation by loading the spring with different masses. The table shows her result.

liubo4ka [24]

Answer:

1) Id say the dependant is the distance

2) the force is calculated by dividing the mass by 100

Explanation:

the dependant variable is the variable you measure wich is in this case distance.

3 0

3 years ago

When two point charges are a distance dd part, the electric force that each one feels from the other has magnitude F.F . In orde

Harman [31]

Answer:

In order to make this force twice as strong, F' = 2 F, the distance would have to be changed to half i.e. r' = r/2.

Explanation :

The electric force between two point charges is directly proportional to the product of charges and inversely proportional to the square of the distance between charges. It is given by :

$F=\dfrac{kq_1q_2}{r^2}$

r is the separation between charges

$F\propto \dfrac{1}{r^2}$

$r=\sqrt{\dfrac{1}{F}}$

If F'= 2F

$r'=\dfrac{1}{\sqrt{2F} }$

In order to make this force twice as strong, F' = 2 F, the distance would have to be changed to half i.e. $r'=\dfrac{1}{\sqrt{2F} }$ . Hence, this is the required solution.

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3 years ago

The quantity of heat Q that changes the temperature L1Tof a mass mof a substance isgiven by Q = cmt:T, where c is the specific h

White raven [17]

Answer:

a) Q = 80,000 cal

b) Q = 100,000 cal

c) Q = 540,000 cal

d) Q = 720,000 cal

Explanation:

a)1 kg from 0⁰ Ice to 0⁰ water, the heat produced is latent heat of fusion

$Q_{l} = ML_{f}$ = 1 * 80

$Q_{l}$ = 80 kCal = 80,000 cal

b) 1 kg of O°C ice water to 1 kg of 100°C boiling water

Specific heat capacity, c = 1000cal/kg.C

$Q_{c} = mc \delta T\\Q_{c} = 1 * 1000 * (100 - 0)\\Q_{c} =100000 cal$

c) 1 kg of 100°C boiling water to 1 kg of 100°C steam

Latent heat of vaporization is needed for this conversion

$Q_{v} = ML_{v} \\L_{v} = 540 kCal/kg\\Q_{v} =1* 540 \\Q_{v} = 540 kCal = 540000 cal$

d) 1 kg of O°C ice to 1 kg of 100°C steam.

Q = $Q_{L} + Q_{c} + Q_{v}$

Q = 80,000 + 100,000 + 540,000

Q = 720,000 cal

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4 years ago

One indication that a planet may exist near a distant star is that ...

Bingel [31]

Shadows blocking part of the light from the star.

A quick warning though this only works on planets either close to the star or planets that are very large.

Also to ensure that the shadows are planets the shadows have to move or orbit around the star. IE The shadow moves

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3 years ago