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Alex777 [14]
2 years ago
11

d. The force is doubled and the object’s mass is halved? 18. ||| A man pulling an empty wagon causes it to accelerate at 1.4 m/s

2. What will the acceleration be if he pulls with the same force when the wagon contains a child whose mass is three times that of the wagon?
Physics
1 answer:
Harrizon [31]2 years ago
8 0

Answer:

a' = 0.35 m/s^2

Explanation:

Let say the empty wagon has mass "M"

now by newton's II Law we will have

F = Ma

now it is given that empty wagon is pulled with acceleration 1.4 m/s/s

now we will have

F = 1.4 M

now a child of mass three times the mass of wagon is sitting on the empty wagon

so here we have

F = (M + 3M) a

1.4 M = 4M a'

so we have

a' = 0.35 m/s^2

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A boy pulls his toy on a smooth horizontal surface with a rope inclined at 60 degrees to the horizontal. If the effective force
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I'm thinking that you're supposed to divide. So you would divide 5 into 60 and get 12
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1 year ago
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Two 7.0cm×7.0cm metal electrodes are spaced 1.0 mm apart and connected by wires to the terminals of a 9.0 v battery.
AleksandrR [38]

Answer:

q=390.285x10^{-12}C

Explanation:

Those kind of problems of electric physics is about capacitors, so the normal questions are:

What are the charge on each electrode?

Solve this you can get other information required in the problem or write down the other questions you need

7.0cm*\frac{1m}{100cm}=70x10^{-3}m

A=70x10^{-3}m*70x10^{-3}m=4.9x10^{-3}m62

Capacitance

C=E_o*\frac{A}{d}

E_o=8.85x10^{-12}\frac{C^2}{N*m^2}

C=8.85x10^{-12}\frac{C^2}{N*m^2}*\frac{4.9x10^{-3}m^2}{1x10^{-3}m}

C=43.365x10^{-12}F

C=43.365pF

The charge is find by the equation

q=C*V

q=43.365pF*9V

q=390.285x10^{-12}C

5 0
2 years ago
A dog is 60m away while moving at constant velocity of 10m/s towards you. Where is the dog after 4 seconds?
evablogger [386]
20m away

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4 0
2 years ago
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Consider a double-slit with a distance between the slits of 0.04 mm and slit width of 0.01 mm. Suppose the screen is a distance
scZoUnD [109]

Answer:

The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

Explanation:

Given that,

Distance between the slits = 0.04 mm

Width = 0.01 mm

Distance between the slits and screen = 1 m

Wavelength = 600 nm

We need to calculate the distance between the places where the intensity is zero due to the double slit effect

For constructive fringe

First minima from center

x_{1}=\dfrac{\lambda D}{2d}

Second minima from center

x_{2}=\dfrac{3\lambda D}{2d}

The distance between the places where the intensity is zero due to the double slit effect

\Delta x_{d}=x_{2}-x_{1}

\Delta x_{d}=\dfrac{3\lambda D}{2d}-\dfrac{\lambda D}{2d}

\Delta x_{d}=\dfrac{\lambda D}{d}

Put the value into the formula

\Delta x_{d}=\dfrac{600\times10^{-9}\times1}{0.04\times10^{-3}}

\Delta x_{d}=0.015 =15\times10^{-3}\ m

\Delta x_{d}=15\ mm

Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

8 0
2 years ago
What is the maximum negative displacement a dog could have if it started its motion at +3 m?
raketka [301]

Answer:

- 3 meter

Explanation:

A dog has started motion from +3 meter. ...(Given)

∴ maximum positive distance = + 3 meter

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Maximum negative distance = (-) (magnitude of distance)

Maximum negative distance = (-) (3)

Maximum negative distance = -3 meter

Hence, maximum negative distance is -3 meter.

7 0
2 years ago
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