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aliina [53]
3 years ago
5

How do you give an example for finding net force?

Physics
1 answer:
Crazy boy [7]3 years ago
7 0
--  6 people all trying to push a car out of snow

--  a Tug-o-War with 30 people of different sizes pulling on each end of the rope

--  you and your sister both pulling on the same doll (or Transformer)

--  lifting a book up from the table to a high shelf
    taking a book down from a high shelf to the table
    (one force is you; another force is gravity)

--  grabbing your big dog by his collar and trying to bring him inside

--  three people at the table all grab the ketchup bottle at the same time
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PLEASE HELP ASAP!! Will give brainliest!!<br><br> Which item is necessary to make an electromagnet?
vichka [17]
To make something electrical you need electricity and the wires will turn a magnet into an electromagnet.
or for a short answer: <span>1) soft iron core 2) coil of insulated wire 3)source of electricity</span>
8 0
3 years ago
A tree loses water to the air by the process of transpiration at the rate of 110 g/h. This water is replaced by the upward flow
Ksivusya [100]

Answer:

The answer is 1.87nm/s.

Explanation:

The 110g/hr  water loss must be replaced by 110g/hr of sap. 110g of sap corresponds to a volume of  

110g \div \dfrac{1040*10^3g}{1*10^6cm^3}  = 106cm^3

thus rate of sap replacement is

106cm^3/hr = 106*10^{-6}m^3/3600s  = 2.94*10^{-8}m^3/s

The volume of sap in the vessel of length x is

V = Ax,

where A is the cross sectional area of the vessel.

For 2000 such vessels, the volume is

V = 2000Ax

taking the derivative of both sides we get:

\dfrac{dV}{dt} = 2000A \dfrac{dx}{dt}

on the left-hand-side \dfrac{dx}{dt} is the velocity v of the sap, and on right-hand-side \dfrac{dV}{dt}  = 2.94*10^{-8}m^3/s; therefore,

2.94*10^{-8}m^3/s=2000Av

and since the cross-sectional area is

A = \pi (\dfrac{100*10^{-3}m}{2} )^2 = 7.85*10^{-3}m^2;

therefore,

2.94*10^{-8}m^3/s =2000(7.85*10^{-3}m^2)v

solving for v we get:

v = \dfrac{2.94*10^{-8}m^3/s}{2000(7.85*10^{-3}m^2)}

\boxed{v =1.875*10^{-9}m/s = 1.875nm/s}

which is the upward speed of the sap in each vessel.

8 0
4 years ago
Ele is playing with a ball in a bus that moves in a straight line with constant velocity What can you say about the motion of th
cupoosta [38]

Answer:

The ball will fall back and land to Elle's hands.

Explanation:

The bus move in a straight line with constant velocity means that there is no change of direction and no acceleration. Inertia can change the direction of the ball and acceleration can change its velocity. Since these two factors is not present in this scenario, the ball only has vertical movement. Thus the ball will land where it was thrown, in Elle's hands.

6 0
3 years ago
In terms of saturation (unsaturated, saturated, super-saturated). How would you classify the following?
miskamm [114]

<u>Answer:</u>

<em>1. A NaCl solution with a concentration of 50g/100mL of water at 40°C:</em> The NaCl solution with a given concentration is saturated at this temperature .As the temperature increases the solution will more dissolves.

<em>2. A sugar solution with a concentration of 200g/100mL of water at 40°C: </em>The sugar solution with a given concentration is saturated at this temperature. As the temperature increases the solution will more dissolves.

<em>3. A sugar solution with a concentration of 240g/100mL of water at 40°C:</em> The sugar solution with a given concentration is saturated at given temperature.

7 0
3 years ago
The magnetic field at the center of a 1.40-cm-diameter loop is 2.50 mT . PART A) What is the current in the loop?
NISA [10]

Explanation:

It is given that,

Diameter of loop, d = 1.4 cm

Radius of loop, r = 0.7 cm = 0.007 m

Magnetic field, B=2.5\ mT=2.5\times 10^{-3}\ T

(A) Magnetic field of a current loop is given by :

B=\dfrac{\mu_oI}{2r}

I is the current in the loop

I=\dfrac{2Br}{\mu_o}

I=\dfrac{2\times 2.5\times 10^{-3}\times 0.007}{4\pi \times 10^{-7}}

I = 27.85 A

(B) Magnetic field at a distance r from a wire is given by :

B=\dfrac{\mu_o I}{2\pi r}

r=\dfrac{\mu_o I}{2\pi B}

r=\dfrac{4\pi \times 10^{-7}\times 27.85}{2\pi \times 2.5\times 10^{-3}}

r = 0.00222 m

r=2.2\times 10^{-3}\ m

Hence, this is the required solution.

3 0
3 years ago
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