. The gravitational force experienced by an object is . mass . weight .push .pull

PLEASE SOLVE QUICKLY!!! Solve for A, B, and C from graph

hram777 [196]

A = 59.35cm

B = 196.56g

C = 74.65g

Explanation:

We know,

$x = \frac{L}{\frac{W}{F} +1}$

and L = x+y

1.

Total length, L = 100cm

Weight of Beam, W = 71.8g

Center of mass, x = 49.2cm

Added weight, F = 240g

Position weight placed from fulcrum, y = ?

$L-y = \frac{L}{\frac{W}{F}+1 } \\100 - y = \frac{100}{\frac{71.8}{49.2}+1 } \\100 - y = \frac{100}{1.46+1}\\\\100 - y = \frac{100}{2.46} \\100-y = 40.65\\\\y = 59.35cm$

Therefore, position weight placed from fulcrum is 59.35cm

2.

Total length, L = 100cm

Center of mass, x = 47.8 cm

Added weight, F = 180g

Position weight placed from fulcrum, y = 12.4cm

Weight of Beam, W = ?

$47.8 = \frac{100}{\frac{W}{180}+1 }\\\47.8 = \frac{100}{\frac{W+180}{180} } \\\\47.8 = \frac{100 X 180}{W+180}\\ \\47.8W + 47.8 X 180 = 18000\\47.8W = 18000 - 8604\\W = \frac{9396}{47.8}\\ W = 196.56g$

Therefore, weight of the beam is 196.56g

3.

Total length, L = 100cm

Center of mass, x = 50.8 cm

Position weight placed from fulcrum, y = 9.8cm

Weight of Beam, W = 72.3g

Added weight, F = ?

$50.8 = \frac{100}{\frac{72.3}{F}+1 }\\\ 50.8 = \frac{100}{\frac{72.3+F}{F} } \\\\50.8 = \frac{100 X F}{72.3+F}\\ \\50.8 X 72.3 + 50.8 X F = 100F\\\\3672.84 = 100F-50.8F\\3672.84 = 49.2F\\F = 74.65g$

Therefore, Added weight F is 74.65g

A = 59.35cm

B = 196.56g

C = 74.65g

4 0

3 years ago

002 (part 1 of 2) 10.0 points

Charra [1.4K]

Answer:

120.03 points on it and its 1234.4KG on the earth

5 0

2 years ago

Galileo Galilei was the first scientist to perform experiments in order to test his ideas. He was also the first astronomer to s

arsen [322]

Explanation:

Geocentric model said that the Earth is at the center of the universe and everything revolves around it. It was considered to be stationary. Galileo proved this model incorrect with the help of his astronomical observations. Some of the key observation that he used to support the heliocentric model were:

1. He proposed the theory that the tides on the Earth occur because of its motion.

2. He observed the phases of the Venus which meant that the Venus revolved around the Sun and not the Earth.

3. He observed other planets and thus noted that they also move around the Sun and not Earth.

4. He discovered the Moons of other planets.

6 0

3 years ago

Which of the following actions would make a pulse travel faster along a stretched string? More than one answer may be correct. I

MaRussiya [10]

Answer:

Option d and e are correct.

Explanation:

The expression for velocity of pulse in a stretched string can be given as follows

v = $\sqrt{\frac{T}{m} }$

where T is tension in the string , m is mass of string per unit length.

Use of lighter string of the same length, under the same tension amounts to higher m so velocity will decrease. Hence option d is correct.

Similarly, v is directly proportional to square root of tension. So if we increase tension , velocity also increases. So option e ) is correct.

5 0

4 years ago

Una atracción de feria consiste en lanzar un trineo de 2 kg por una rampa ascendente que forma un ángulo de 30º con la horizonta

Arada [10]

Answer:

La velocidad con la que se debe lanzar el trineo es aproximadamente 9.96 m/s

Explanation:

La masa del trineo, m = 2 kg

El ángulo de inclinación del trineo, θ = 30 °

El coeficiente de fricción, μ = 0,15

La altura a la que debe ascender el trineo, h = 4 m

La reacción normal del trineo en la rampa, N = m · g · cos (θ)

La fuerza de fricción $F_f$ = N × μ

Dónde;

g = Th aceleración debida a la gravedad ≈ 9.81 m/s²

∴ $F_f$ = 0.15 × 2 kg × 9.81 m/s² × cos (30°) ≈ 2.55 N

La longitud de la rampa que se mueve el trineo, l = h/(sin(θ))

∴ l = 4/(sin(30°)) = 8

La longitud de la rampa que se mueve el trineo, l = 8 m

El trabajo realizado sobre la fricción, $W_f$ = $F_f$ × l

$W_f$ = 2.55 × 8 ≈ 20.4

El trabajo realizado sobre la fricción, $W_f$ ≈ 20.4 Julios

El trabajo realizado para levantar el trineo a 4 metros de altura, P.E. = m·g·h

∴ P.E. = 2 kg × 9.81 m/s² × 4 m ≈ 78.48 Joules

El trabajo realizado para levantar el trineo a 4 metros de altura, P.E. ≈ 78.48 Joules

El trabajo total requerido para levantar el trineo 4 metros por la rampa, $W_t$ = $W_f$ + P.E.

$W_t$ = 20.4 J + 78.84 J ≈ 99.24 J

Energía cinética, K.E. = 1/2 × m × v²

La energía cinética necesaria para mover el trineo por la rampa, K.E. se da de la siguiente manera;

K.E. = 1/2 × 2 kg × v² ≈ 99.24 J

∴ v² ≈ 99.24 J/(1/2 × 2 kg)

La velocidad con la que se debe lanzar el trineo para que ascienda 4 metros por la rampa, v ≈ 9.96 m/s

4 0

3 years ago

. The gravitational force experienced by an object is. mass. weight.push .pull​

. The gravitational force experienced by an object is. mass. weight.push .pull