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3 years ago

5

If force 1 has a magnitude of 5 N (forward) and force 2 has a magnitude of 30 N (forward), what is the net force acting on the o

bject?
A.) 25 N forward
B.) 35 N forward
C.) 25 N backwards
D.) 35 N backwards

1 answer:

Phoenix [80]3 years ago

5 0

D I am pretty sure thanks

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Hallar la distancia que recorre una movil al cabo

Kryger [21]

Answer:

Distancia, S = 136 metros

Explanation:

Dados los siguientes datos;

Aceleración, a = 3 m/s²

Velocidad inicial, u = 5 m/s

Tiempo, t = 8 segundos

Para encontrar la distancia recorrida, usaríamos la segunda ecuación de movimiento;

S = ut + ½at² Sustituyendo en la fórmula, tenemos;

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3 years ago

Ethan is using a machine to mix cement. The work input is 10 and the work output is 8. What is the efficiency of the

Arturiano [62]

Efficiency or $\eta$ is calculated by formula: $\eta=\frac{W_{output}}{W_{input}}$ where $\eta\leq1$ also efficiency has no unit.

Now put in the data and calculate the efficiency.

$\eta=\frac{8}{10}=\boxed{0.8}$

The efficiency of the machine is 0.8

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3 years ago

a stone is thrown horizonttaly from a cliff of a hill with an initial velocity of 30m/s it hits the ground at a horizontal dista

ELEN [110]

Answer:

a) Time = 2.67 s

b) Height = 35.0 m

Explanation:

a) The time of flight can be found using the following equation:

$x_{f} = x_{0} + v_{0_{x}}t + \frac{1}{2}at^{2}$ (1)

Where:

$x_{f}$ : is the final position in the horizontal direction = 80 m

$x_{0}$ : is the initial position in the horizontal direction = 0

$v_{0_{x}}$ : is the initial velocity in the horizontal direction = 30 m/s

a: is the acceleration in the horizontal direction = 0 (the stone is only accelerated by gravity)

t: is the time =?

By entering the above values into equation (1) and solving for "t", we can find the time of flight of the stone:

$t = \frac{x_{f}}{v_{0}} = \frac{80 m}{30 m/s} = 2.67 s$

b) The height of the hill is given by:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}$

Where:

$y_{f}$ : is the final position in the vertical direction = 0

$y_{0}$ : is the initial position in the vertical direction =?

$v_{0_{y}}$ : is the initial velocity in the vertical direction =0 (the stone is thrown horizontally)

g: is the acceleration due to gravity = 9.81 m/s²

Hence, the height of the hill is:

$y_{0} = \frac{1}{2}gt^{2} = \frac{1}{2}9.81 m/s^{2}*(2.67 s)^{2} = 35.0 m$

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3 years ago

In the Olympic shot-put event, an athlete throws the shot with an initial speed of 12.0 m/s at a 40.0 ∘ angle from the horizonta

Gre4nikov [31]

Answer:

Explanation:

given,

initial speed of the shot = 12.0 m/s

angle = 40°

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v_y = 12 sin 40° = 7.71 m/s

time to reach maximum height =

= $\dfrac{v_y}{g}$

= $\dfrac{7.71}{9.8}$

= 0.787 s

$h = v_y t + \dfrac{1}{2}gt^2$

h = 7.71 × 0.787 - 0.5 × 9.81 × 0.787²

h = 3.03 m

the maximum height attain = 3.03 + 1.8 = 4.83 m

now free fall from the maximum height

$h =\dfrac{1}{2}gt^2$

$4.83 = \dfrac{1}{2}\times 9.8 \times t^2$

t = 0.9928 s

total time = 0.9928 + 0.787 = 1.7798 s

range =

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4 years ago