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3 years ago

Problem 3.2 Part A Draw the vector C⃗ =A⃗ +B⃗ .(Figure 1)

Physics

1 answer:

Elodia [21]3 years ago

5 0

I can't answer this question without a figure. I've found a similar problem as shown in the first picture attached. When adding vectors, you don't have to add the magnitudes only, because vectors also have to factor in the directions. To find the resultant vector C, connect the end tails of the individual vectors. <em>The red line (second picture) represents the vector C.</em>

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Determine how many times per second it would move back and forth across a 6.0-m-long room on the average, assuming it made very

timofeeve [1]

Answer:

The right solution is "24.39 per sec".

Explanation:

According to the question,

⇒ $v=\frac{502.1}{\sqrt{3} }$

$=289.9 \ m/s$

The time will be:

⇒ $t=\frac{d}{v}$

$=\frac{2\times 6}{289.9}$

$=\frac{12}{289.9}$

$=0.041 \ sec$

hence,

⇒ $N=\frac{1}{t}$

$=\frac{1}{0.041}$

$=24.39 \ per \ sec$

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2 years ago

________ is the average kinetic energy of each atom.<br><br> radiation<br> temperature<br> potential

solong [7]

I think it's temperature

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3 years ago

A 950-kg car strikes a huge spring at a speed of 22m/s (fig. 11-54), compressing the spring 5.0m. (a) what is the spring stiffne

alukav5142 [94]

(a) The spring stiffness constant of the spring is 18,392 N/m.

(b) The time the car was in contact with the spring before it bounces off in the opposite direction is 0.23 s.

<h3>Kinetic energy of the car</h3>

The kinetic energy of the car is calculated as follows;

K.E = ¹/₂mv²

K.E = ¹/₂ x 950 x 22²

K.E = 229,900 J

<h3>Stiffness constant of the spring</h3>

The stiffness constant of the spring is calculated as follows;

K.E = U = ¹/₂kx²

k = 2U/x²

k = (2 x 229,900)/(5)²

k = 18,392 N/m

<h3>Force exerted on the spring</h3>

F = kx

F = 18,392 x 5

F = 91,960 N

<h3>Time of impact</h3>

F = mv/t

t = mv/F

t = (950 x 22)/(91960)

t = 0.23 s

Learn more about spring constant here: brainly.com/question/1968517

#SPJ4

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1 year ago

Two astronauts, each with a mass of 50 kg, are connected by a 7 m massless rope. Initially they are rotating around their center

kiruha [24]

Answer:

The angular velocity is $w_f = 1.531 \ rad/ s$

Explanation:

From the question we are told that

The mass of each astronauts is $m = 50 \ kg$

The initial distance between the two astronauts $d_i = 7 \ m$

Generally the radius is mathematically represented as $r_i = \frac{d_i}{2} = \frac{7}{2} = 3.5 \ m$

The initial angular velocity is $w_1 = 0.5 \ rad /s$

The distance between the two astronauts after the rope is pulled is $d_f = 4 \ m$

Generally the radius is mathematically represented as $r_f = \frac{d_f}{2} = \frac{4}{2} = 2\ m$

Generally from the law of angular momentum conservation we have that

$I_{k_1} w_{k_1}+ I_{p_1} w_{p_1} = I_{k_2} w_{k_2}+ I_{p_2} w_{p_2}$

Here $I_{k_1 }$ is the initial moment of inertia of the first astronauts which is equal to $I_{p_1}$ the initial moment of inertia of the second astronauts So

$I_{k_1} = I_{p_1 } = m * r_i^2$

Also $w_{k_1 }$ is the initial angular velocity of the first astronauts which is equal to $w_{p_1}$ the initial angular velocity of the second astronauts So

$w_{k_1} =w_{p_1 } = w_1$

Here $I_{k_2 }$ is the final moment of inertia of the first astronauts which is equal to $I_{p_2}$ the final moment of inertia of the second astronauts So

$I_{k_2} = I_{p_2} = m * r_f^2$

Also $w_{k_2 }$ is the final angular velocity of the first astronauts which is equal to $w_{p_2}$ the final angular velocity of the second astronauts So