PHYSICS COLLEGE

A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arrow = (2.00 m)i hat − (3.00 m)j + (2.00 m)k, the force is F with arrow = Fxi hat + (7.00 N)j − (5.80 N)k and the corresponding torque about the origin is vector tau = (3.40 N · m)i hat + (2.80 N · m)j + (0.800 N · m)k. Determine Fx.

Answers

Answer 1

Answer:

Answer:

-4.40

Explanation:

explanation is in attachment

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A metal block of mass 3 kg is falling downward and has velocity of 0.44 m/s when it is 0.8 m above the floor. It strikes the top of a relaxed vertical spring of length 0.4 m. The spring constant is 2000 N/m. After striking the spring, the block rebounds. What is the maximum height above the floor that the block reaches after the impact
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A 2,000 kg car travels with a tangentialvelocity of 12 m/s around a circular
track with a radius of 30 meters. What
is the car's rate of centripetal
acceleration?

Answers

The car's rate of centripetal acceleration in the circular path is 4.8 m/s².

The given parameters;

mass of the car, m = 2,000 kg
velocity of the car, v = 12 m/s
radius of the circular path, r = 30 m

The centripetal acceleration of the car is calculated as follows;

$a_c = (v^2)/(r)$

where;

v is the tangential speed of the car
r is the radius of the circular path

Substitute the given parameters and solve for the centripetal acceleration;

$a_c = (12^2)/(30) \n\na_c = 4.8 \ m/s^2$

Thus, the car's rate of centripetal acceleration is 4.8 m/s².

Learn more here:brainly.com/question/11700262

a= v²/R
a = 12²/30 =4.8 m/s²

A cat leaps to catch a bird. If the cat's jump was at 60.0° off the ground and its initial velocity was point of its trajectory? 0.30 m 3.44 m/s, what is the highest O 13.76 m 0.45 m 0.90 m

Answers

Answer:

The highest of its trajectory = 0.45 m

Option C is the correct answer.

Explanation:

Considering vertical motion of cat:-

Initial velocity, u = 3.44 sin60 = 2.98 m/s

Acceleration , a = -9.81 m/s²

Final velocity, v = 0 m/s

We have equation of motion v² = u² + 2as

Substituting

v² = u² + 2as

0² = 2.98² + 2 x -9.81 x s

s = 0.45 m

The highest of its trajectory = 0.45 m

Option C is the correct answer.

A block of mass 0.221 kg is placed on top of a light, vertical spring of force constant 5365 N/m and pushed downward so that the spring is compressed by 0.097 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise?

Answers

Answer:

The maximum height above the point of release is 11.653 m.

Explanation:

Given that,

Mass of block = 0.221 kg

Spring constant k = 5365 N/m

Distance x = 0.097 m

We need to calculate the height

Using stored energy in spring

$U=(1)/(2)kx^2$ ...(I)

Using gravitational potential energy

$U' =mgh$ ....(II)

Using energy of conservation

$E_(i)=E_(f)$

$U_(i)+U'_(i)=U_(f)+U'_(f)$

$(1)/(2)kx^2+0=0+mgh$

$h=(kx^2)/(2mg)$

Where, k = spring constant

m = mass of the block

x = distance

g = acceleration due to gravity

Put the value in the equation

$h=(5365*(0.097)^2)/(2*0.221*9.8)$

$h=11.653\ m$

Hence, The maximum height above the point of release is 11.653 m.

A block of mass m slides with a speed vo on a frictionless surface and collides with another mass M which is initially at rest. The two blocks stick together and move with a speed of vo /3. In terms of m, mass M is most nearly_____.

Answers

To solve this problem we will apply the concepts related to the conservation of momentum. Momentum can be defined as the product between mass and velocity. We will depart to facilitate the understanding of the demonstration, considering the initial and final momentum separately, but for conservation, they will be later matched. Thus we will obtain the value of the mass. Our values will be defined as

$m_1 = m$

$m_2 = M$

$v_(1i) =v_0$

$v_(2i) = 0$

Initial momentum will be

$P_i = m_iv_(1i)+m_2v_(2i)$

$P_i = mv_0$

After collision

$v_(1f) = v_(2f) = (v_0)/(3)$

Final momentum

$P_f = (m_1+m_2)((v_0)/(3))$

$P_f = (m+M)((v_0)/(3))$

From conservation of momentum

$P_f = P_i$

Replacing,

$(m+M)((v_0)/(3))=mv_0$

$(m+M)(1)/(3) = m$

$m+M=3m$

$M=3m-m$

$M=2m$

Look at the Kunoichi's of Naruto but as a Gang. Who do you think looks the baddest out of the group(but in a good way)?

My opinion is Hinata...just saying

Answers

Answer:

hinata for sure

Explanation:

seems reasonable

Find the speed of light in carbon tetrachlorideethyl alcohol. The refraction index is 1.461 using 3 x 10^8 m/s as the speed of light in vacuum. Answer in units of m/s.

Answers

Answer:

2.05 x 10^8 m /s

Explanation:

c = 3 x 10^8 m/s

μ = c / v

where, μ is the refractive index, c be the velocity of light in air and v be the velocity of light in the medium.

μ = 1.461

1.461 = 3 x 10^8 / v

v = 3 x 10^8 / 1.461

v = 2.05 x 10^8 m /s

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