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Rudiy27
3 years ago
13

a farmer moves along the boundary of a square field of side 10 m in 40 s.what will be the magnitude of displacement of the farme

r at the end of 2 minutes 20 seconds from his initial position ?
Physics
1 answer:
melamori03 [73]3 years ago
7 0

Answer:

i think 80 m.

Explanation:

You might be interested in
What was the name of the voyages taken by Zheng he during the Ming dynasty on behalf of China
telo118 [61]

Answer:

Ming treasure voyages

Explanation:

6 0
3 years ago
a net force of 219 N is exterted on a rock. the rock has an acceleration of 3m/s^2 due to this force. what is the mass of the ro
Sonbull [250]

Answer:

<h2>73 kg</h2>

Explanation:

The mass of the object can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

From the question we have

m =  \frac{219}{3}  \\

We have the final answer as

<h3>73 kg</h3>

Hope this helps you

6 0
3 years ago
You want to estimate the diameter of a very small circular pinhole that you've made in a piece of aluminum foil. to do so, you s
Leya [2.2K]
Central maximum = d* wavelength/ D
thus
12*10-^3 = 3.4*6.32*10-^7/D
D = 3.4*6.32*10-^7/12*10-^3
D = 1.79*10-^4 m
3 0
3 years ago
Two engineering students, John with a weight of 96 kg and Mary with a weight of 48 kg, are 30 m apart. Suppose each has a 0.04%
djyliett [7]

Answer:

6.8370869499\times 10^{20}\ N

Explanation:

N_A = Avogadro's number = 6.022\times 10^{23}

e = Charge of electron = 1.6\times 10^{-19}\ C

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

Z = Atomic number of water = 18

M = Molar mass of water = 0.018 kg/mol

m = Mass of person

The charge is given by

q=imbalance\times n\times e

Total number of protons and electrons in each sphere

n=\dfrac{mN_AZe}{M}

q=imbalance\times \dfrac{mN_AZe}{M}

q_1=0.0004\times \dfrac{96\times 6.022\times 10^{23}\times 18\times 1.6\times 10^{-19}}{0.018}\\\Rightarrow q_1=3699916.8\ C

q_2=0.0004\times \dfrac{48\times 6.022\times 10^{23}\times 18\times 1.6\times 10^{-19}}{0.018}\\\Rightarrow q_1=1849958.4\ C

Electrical force is given by

F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{8.99\times 10^{9}\times 3699916.8\times 1849958.4}{30^2}\\\Rightarrow F=6.8370869499\times 10^{20}\ N

The electrostatic force of attraction between them is 6.8370869499\times 10^{20}\ N

4 0
3 years ago
When a parachutist jumps from an airplane, he eventually reaches a constant speed, called the terminal speed. Once he has reache
Masteriza [31]

Answer:

b) True.    the force of air drag on him is equal to his weight.

Explanation:

Let us propose the solution of the problem in order to analyze the given statements.

The problem must be solved with Newton's second law.

When he jumps off the plane

     fr - w = ma

Where the friction force has some form of type.

     fr = G v + H v²

Let's replace

     (G v + H v²) - mg = m dv / dt

We can see that the friction force increases as the speed increases

At the equilibrium point

      fr - w = 0

      fr = mg

      (G v + H v2) = mg

For low speeds the quadratic depended is not important, so we can reduce the equation to

     G v = mg

     v = mg / G

This is the terminal speed.

Now let's analyze the claims

a) False is g between the friction force constant

b) True.

c) False. It is equal to the weight

d) False. In the terminal speed the acceleration is zero

e) False. The friction force is equal to the weight

3 0
3 years ago
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