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labwork [276]
3 years ago
14

Mr. Hershman has purchased a farm that is in the shape of a rectangle. The dimensions of the piece of land are 4.3 km by 4.85 km

. How many square kilometers did he​ buy, and how many hectares did he​ purchase?
Physics
1 answer:
koban [17]3 years ago
8 0

Explanation:

Given that,

Mr. Hershman has purchased a farm that is in the shape of a rectangle. The dimensions of the piece of land are 4.3 km by 4.85 km. The area of the rectangle is given by :

A=l\times b

A=4.3\times 4.85

A=20.855\ km^2

So, she must buy 20.855\ km^2 of area.

Since, 1\ km^2=100\ hectares

So, the area of rectangle is 2085.5 hectares. Hence, this is the required solution.

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Question 25 Answer: Destructive interference occurs.

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4 0
2 years ago
What is the difference between a good conductor and a good insulator?
brilliants [131]

Answer:

Explanation:

In a conductor, electric current can flow freely, in an insulator it cannot.

Metals such as copper typify conductors, while most non-metallic solids are said to be good insulators, having extremely high resistance to the flow of charge through them.

Most atoms hold on to their electrons tightly and are insulators.

4 0
2 years ago
A 0.03 kg golf ball is hit off the tee at a speed of 34 m/s. The golf club was in contact with the ball for 0.003 s. What is the
Liula [17]

Answer:

The average force on ball by the golf club is 340 N.

Explanation:

Given that,

Mass of the golf ball, m = 0.03 kg

Initial speed of the ball, u = 0

Final speed of the ball, v = 34 m/s

Time of contact, \Delta t=0.003\ s

We need to find the average force on ball by the golf club. We know that the rate of change of momentum is equal to the net external force applied such that :

F=\dfrac{\Delta p}{\Delta t}\\\\F=\dfrac{mv-mu}{\Delta t}\\\\F=\dfrac{mv}{\Delta t}\\\\F=\dfrac{0.03\ kg\times 34\ m/s}{0.003\ s}\\\\F=340\ N

So, the average force on ball by the golf club is 340 N.

4 0
3 years ago
What is the magnitude of the electric force between a proton and an electron when they are at a distance of 4.09 angstrom from e
Zinaida [17]

Answer:

F=1.38*10^{-9}N

Explanation:

According to Coulomb's law, the magnitude of the electric force between two point charges is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them:

F=\frac{kq_1q_2}{d^2}

Here k is the Coulomb constant. In this case, we have q_1=-e, q_2=e and d=4.09*10^-10m. Replacing the values:

F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(-1.6*10^{-19}C)(1.6*10^{-19}C)}{(4.09*10^{-10})^2}\\F=-1.38*10^{-9}N

The negative sign indicates that it is an attractive force. So, the magnitude of the electric force is:

F=1.38*10^{-9}N

5 0
3 years ago
a spring gun initially compressed 2cm fires a 0.01kg dart straight up into the air. if the dart reaches a height it 5.5m determi
Vikki [24]

Answer:

2697.75N/m

Explanation:

Step one

This problem bothers on energy stored in a spring.

Step two

Given data

Compression x= 2cm

To meter = 2/100= 0.02m

Mass m= 0.01kg

Height h= 5.5m

K=?

Let us assume g= 9.81m/s²

Step three

According to the principle of conservation of energy

We know that the the energy stored in a spring is

E= 1/2kx²

1/2kx²= mgh

Making k subject of formula we have

kx²= 2mgh

k= 2mgh/x²

k= (2*0.01*9.81*5.5)/0.02²

k= 1.0791/0.0004

k= 2697.75N/m

Hence the spring constant k is 2697.75N/m

7 0
3 years ago
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