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3 years ago

How does weight change as the gravitational acceleration changes and why?

Physics

1 answer:

seraphim [82]3 years ago

4 0

Answer

The correct formula for weight is F = m*g where g is the gravitational acceleration.

All over the earth's surface, g is slightly different. The mass does not change no matter where you are. We should be measuring out weights in Newtons, not in kg. So the unit of weight in the metric system is 9.8 about * mass in kg.

Stop reading. This is your answer.

============

Notes

It was hard enough to get people to change over to kg never mind newtons. Canada, which is on the metric system, still gives the price of food in pounds. Or if not, in grams if the container is small enough.

Cashews cost 19$ Canadian per 906 grams which is roughly 2 pounds.

Oranges are $1.27 a pound and that is the way they are listed in Walmart.

10 pounds of potatoes are 4.95 dollars.

I'm sure you get the point. We use kg for certain things and retain pounds for others.

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Earth surface are composed of 70% of water and 30% of lands or soil.

musickatia [10]

Answer:

idk

Explanation:

7 0

3 years ago

A 150kg motorcycle starts from rest and accelerates at a constant rate along with a distance of 350m. The applied force is 250N

mart [117]

Answer:

205N

Explanation:

The net force (F) is the difference between the applied force( $F_{A}$ ) and the kinetic frictional force( $F_{R}$ ). i.e

F = $F_{A}$ - $F_{R}$ -----------------(i)

Note that;

$F_{R}$ = μmg

Where;

μ = coefficient of kinetic friction

m = mass of the body

g = acceleration due to gravity = 10m/s²

Equation (i) then becomes;

F = $F_{A}$ - μmg -------------------(ii)

<em>Given from question;</em>

m = mass of motorcycle = 150kg

μ = 0.03

$F_{A}$ = 250N

Substitute these values into equation (ii) as follows;

F = 250 - (0.03 x 150 x 10)

F = 250 - (45)

F = 205N

Therefore, the net force applied to the motorcycle is 205N

3 0

3 years ago

QUESTION 3 ( MARKS]

horrorfan [7]

Answer:

392 N

Explanation:

Draw a free body diagram of the rod. There are four forces acting on the rod:

At the wall, you have horizontal and vertical reaction forces, Rx and Ry.

At the other end of the rod (point X), you have the weight of the sign pointing down, mg.

Also at point X, you have the tension in the wire, T, pulling at an angle θ from the -x axis.

Sum of the moments at the wall:

∑τ = Iα

(T sin θ) L − (mg) L = 0

T sin θ − mg = 0

T = mg / sin θ

Given m = 20 kg and θ = 30.0°:

T = (20 kg) (9.8 m/s²) / (sin 30.0°)

T = 392 N

7 0

3 years ago

A single-turn plane loop of wire with a cross-sectional area 200 cm2 is perpendicular to a magnetic field that increases uniform

BartSMP [9]

Answer:

Induced current, $I=2.87\times 10^{-3}\ A$

Explanation:

Given that,

Area of cross section of the wire, $A=200\ cm^2=0.02\ m^2$

Time, t = 2.2 s

Initial magnetic field, $B_i=0.2\ T$

Final magnetic field, $B_f=2.8\ T$

Resistance of the coil, R = 8 ohms

The expression for the induced emf is given by :

$\epsilon=-\dfrac{d\phi}{dt}$

$\phi$ = magnetic flux

$\epsilon=-\dfrac{d(BA)}{dt}$

$\epsilon=A\dfrac{d(B)}{dt}$

$\epsilon=A\dfrac{B_f-B_i}{t}$

$\epsilon=0.02\times \dfrac{2.8-0.2}{2.2}$

$\epsilon=-0.023\ volts$

So, the induced emf in the loop is 0.023 volts. The induced current can be calculated using Ohm's law as :

$\epsilon= IR$

$I=\dfrac{\epsilon}{R}$

$I=\dfrac{0.023}{8}$

$I=2.87\times 10^{-3}\ A$

So, the magnitude of the induced current in the loop of wire is $2.87\times 10^{-3}\ A$ . Hence, this is the required solution.

7 0

3 years ago

Which part of the electromagnetic spectrum has the longest wave lengths

Lana71 [14]

Radio Waves :)) i’m pretty confident in that

8 0

3 years ago