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Jet001
3 months ago
9

If a woman weighs 125 lb, her mass expressed in kilograms is x kg, where x is

Physics
2 answers:
Sav [3.1K]3 months ago
8 0
The weight of the woman in pounds equals 125 lb
1 kg outweighs 1 pound, and 1 kilogram equals approximately 2.20462 pounds.  
Thus, converting 125 lb results in 125 / 2.20462 kilograms = 56.699 kg, which is less than 125.
Sav [3.1K]3 months ago
3 0
To tackle this issue, it's essential to understand the conversion of pounds to kilograms:
1 lb = 0.45 Kg
By applying a straightforward rule of three
1 lb ---> 0.45 Kg
125 lb ---> x
Solving for x yields:
x = ((125) / (1)) * (0.45) = 56.25 Kg.
 Response
 her mass in kilograms is 56.25 Kg.
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It's a snowy day and you're pulling a friend along a level road on a sled. You've both been taking physics, so she asks what you
Maru [3345]

Answer:

0.0984

Explanation:

The first diagram below illustrates a free body diagram that will aid in resolving this problem.

According to the diagram, the force's horizontal component can be expressed as:

F_X = F_{cos \ \theta}

Substituting 42° for θ and 87.0° for F

F_X =87.0 \ N \ *cos \ 42 ^\circ

F_X =64.65 \ N

Meanwhile, the vertical component is:

F_Y = Fsin \ \theta

Again substituting 42° for θ and 87.0° for F

F_Y =87.0 \ N \ *sin \ 42 ^\circ

F_Y =58.21 \ N

In resolving the vector, let A denote the components in mutually perpendicular directions.

The magnitudes of both components are illustrated in the second diagram provided and can be represented as A cos θ and A sin θ

The frictional force can be expressed as:

f = \mu \ N

Where;

\mu is the coefficient of friction

N = the normal force

Also, the normal reaction (N) is calculated as mg - F sin θ

Substituting F_Y \ for \ F_{sin \ \theta}. Normal reaction becomes:

N = mg \ - \ F_Y

By balancing the forces, the horizontal component of the force equals the frictional force.

The horizontal component is described as follows:

F_X = \mu \ ( mg - \ F_Y)

Rearranging the equation above to isolate \mu leads to:

\mu \ = \ \frac{F_X}{mg - F_Y}

Substituting in the following values:

F_X \ = \ 64.65 \ N

m = 73 kg

g = 9.8 m/s²

F_Y = \ 58.21 N

Thus:

\mu \ = \ \frac{64.65 N}{(73.0 kg)(9.8m/s^2) - (58.21 \ N)}

\mu = 0.0984

Therefore, the coefficient of friction is = 0.0984

5 0
3 months ago
A very small round ball is located near a large solid sphere of uniform density. The force that the large sphere exerts on the b
Softa [3030]

Respuesta:

Opción e

Explicación:

La Ley de Gravitación Universal indica que toda masa puntual atrae a otra masa puntual en el universo con una fuerza que se dirige en línea recta entre los centros de masa de ambos, siendo esta fuerza proporcional a las masas de los objetos y inversamente proporcional a su separación. Esta fuerza atractiva siempre es dirigida del uno hacia el otro. La ley es aplicable a objetos de cualquier masa, sin importar su tamaño. Dos objetos grandes pueden ser considerados masas puntuales si la distancia entre ellos es considerablemente mayor que sus dimensiones o si presentan simetría esférica. En tales casos, la masa de cada objeto puede ser modelada como una masa puntual en su centro de masa.

La misma fuerza actúa sobre ambas bolas.

5 0
4 months ago
Multiply the number 4.48E-8 by 5.2E-4 using Google. What is the correct answer in scientific notation?
Ostrovityanka [3204]

Answer:

2.32\times 10^{-11}

Explanation:

The first number is 4.48\times 10^{-8}.

The second number is 5.2\times 10^{-4}.

We must multiply these two numbers together.

4.48\times 10^{-8}\times 5.2\times 10^{-4}=(4.48\times 5.2)\times 10^{(-8-4)}\\\\=23.296\times 10^{-12}

In scientific notation: 2.32\times 10^{-11}

Therefore, this is the solution you are looking for.

8 0
4 months ago
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