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3 years ago

10

Please answer question B.

1 answer:

Flura [38]3 years ago

7 0

Answer: 3.92

Explanation:

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What airplanes did the ffa ground earlier this year after several crashes?

frez [133]

The FFA is the Future Farmers of America. It has no authority in aeronautical matters.

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4 years ago

Two ropes are attached to a 35 kg object. The first rope applies a force of 20 N and the second applies a force of 55 N. If the

Goshia [24]

Answer:

$a=1.672\ m.s^{-2}$

Explanation:

Given:

mass of the object, $m=35\ kg$

forces by two mutually perpendicular ropes of the attached to the object:

$F_x=20\ N$

$F_y=55\ N$

<u>Now we find the resultant force effect due to the two given forces:</u>

$F=\sqrt{F_x^2+F_y^2}$

$F=\sqrt{(20)^2+(55)^2}$

$F=58.52\ N$

<u>Now the acceleration will be due to this resultant force:</u>

$a=\frac{F}{m}$

$a=\frac{58.52}{35}$

$a=1.672\ m.s^{-2}$

3 0

3 years ago

Read 2 more answers

A 1250 kg car is stopped at a traffic light. A 3550 kg truck moving at 8.33 m/s hits the car from behind. If bumpers lock, how f

Radda [10]

Answer:

the two vehicles will be moving at a speed of 6.16 m/s

Explanation:

This is a case of completely inelastic collision, therefore, the conservation of momentum can be written as:

$m_1\,v_1+m_2\,v_2=(m_1+m_2)\,v_f$

which given the information provided results into:

$m_1\,v_1+m_2\,v_2=(m_1+m_2)\,v_f\\(1250)\,(0)+(3550)\,(8.33)=(1250+3550)\,v_f\\29571.5=4800\,v_f\\v_f=6.16\,\,m/s$

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3 years ago

if a string of light goes out when one of the bulbs is removed, are the lights probably connected in a series circuit or a paral

Minchanka [31]

Yes it is because other wise the light would stay on

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3 years ago

Read 2 more answers

Vector A with arrow, which is directed along an x axis, is to be added to vector B with arrow, which has a magnitude of 5.5 m. T

ohaa [14]

Answer:

Magnitude of vector A = 0.904

Explanation:

Vector A , which is directed along an x axis, that is

$\vec{A}=x_A\hat{i}$

Vector B , which has a magnitude of 5.5 m

$\vec{B}=x_B\hat{i}+y_B\hat{j}$

$\sqrt{x_{B}^{2}+y_{B}^{2}}=5.5\\\\x_{B}^{2}+y_{B}^{2}=30.25$

The sum is a third vector that is directed along the y axis, with a magnitude that is 6.0 times that of vector A $\vec{A}+\vec{B}=6x_A\hat{j}\\\\x_A\hat{i}+x_B\hat{i}+y_B\hat{j}=6x_A\hat{j}$

Comparing we will get

$x_A=-x_B\\\\y_B=6x_A$

Substituting in $x_{B}^{2}+y_{B}^{2}=30.25$

$\left (-x_{A} \right )^{2}+\left (6x_{A} \right )^{2}=30.25\\\\37x_{A}^2=30.25\\\\x_{A}=0.904$

So we have

$\vec{A}=0.904\hat{i}$

Magnitude of vector A = 0.904

8 0

4 years ago