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You are standing on a large sheet of frictionless ice and holding a large rock. In order to get off the ice, you throw the rock

kondor19780726 [428]

Answer:

0.4778 m/s

Explanation:

To solve this question, we will make use of law of conservation of momentum.

We are given that the rock's velocity is 12 m/s at 35°. Thus, the horizontal component of this velocity is;

V_x = (12 m/s)(cos(35°)) = 9.83 m/s.

Thus, the horizontal component of the rock's momentum is;

(3.5 kg)(9.83 m/s) = 34.405 kg·m/s.

Since the person is not pushed up off the ice or down into it, his momentum will have no vertical component and so his momentum will have the same magnitude as the horizontal component of the rock's momentum.

Thus, to get the person's speed, we know that; momentum = mass x velocity

Mass of person = 72 kg and we have momentum as 34.405 kg·m/s

Thus;

34.405 = 72 x velocity

Velocity = 34.405/72

Velocity = 0.4778 m/s

6 0

3 years ago

The private papers of what scientist had to be decontaminated for two years in the 1990's before being put on file at the Nation

SashulF [63]

The answer is Marie Skłodowska Curie (AKA Marie Curie). She lived her life awash in ionizing radiation. She would be carrying bottles of the radium and polonium in the pocket of her coat and put them in her desk drawer.
So even after a century, her papers are still radioactive. Since the most general isotope of radium, which is radium-226, has a half life of 1,601 years.

3 0

2 years ago

A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arr

Paha777 [63]

Answer:

The question is incomplete, below is the complete question "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 − (5.00 N)k and the corresponding torque about the origin is vector tau = (4 N · m)i hat + (10 N · m)j + (11N · m)k.

Determine Fx."

$F_{x}=-1N.m$

Explanation:

We asked to determine the "x" component of the applied force. To do this, we need to write out the expression for the torque in the in vector representation.

torque=cross product of force and position . mathematically this can be express as

$T=r*F$

Where

$F=F_{x}i+(7N)j-(5N)k$ and the position vector

$r=(2m)i-(3m)j+(2m)k$

using the determinant method to expand the cross product in order to determine the torque we have

$\left[\begin{array}{ccc}i&j&k\\2&-3&2\\ F_{x} &7&-5\end{array}\right]\\\\$

by expanding we arrive at

$T=(18-14)i-(-12-2F_{x})j+(12+3F_{x})k\\T=4i-(-12-2F_{x})j+(12+3F_{x})k\\\\$

since we have determine the vector value of the toque, we now compare with the torque value given in the question

$(4Nm)i+(10Nm)j+(11Nm)k=4i-(-12-2F_{x})j+(12+3F_{x})k\\$

if we directly compare the j coordinate we have

$10=-(-12-2F_{x})\\10=12+2F_{x}\\ 10-12=2F_{x}\\ F_{x}=-1N.m$

8 0

3 years ago

A woman with a mass of 52.0 kg is standing on the rim of a large disk that is rotating at an angular velocity of 0.470 rev/s abo

Charra [1.4K]

Explanation:

It is given that,

Mass of the woman, m₁ = 52 kg

Angular velocity, $\omega=0.47\ rev/s=2.95\ rad/s$

Mass of disk, m₂ = 118 kg

Radius of the disk, r = 3.9 m

The moment of inertia of woman which is standing at the rim of a large disk is :

$I={m_1r^2}$

$I={52\times 3.9^2}$

I₁ = 790.92 kg-m²

The moment of inertia of of the disk about an axis through its center is given by :

$I_2=\dfrac{m_2r^2}{2}$

$I_2=\dfrac{118\times (3.9)^2}{2}$

I₂ =897.39 kg-m²

Total moment of inertia of the system is given by :

$I=I_1+I_2$

$I=790.92+897.39$

I = 1688.31 kg-m²

The angular momentum of the system is :

$L=I\times \omega$

$L=1688.31 \times 2.95$

$L=4980.5\ kg-m^2/s$

So, the total angular momentum of the system is 4980.5 kg-m²/s. Hence, this is the required solution.

8 0

3 years ago

An open cart is rolling to the left on a horizontal surface. A package slides down a chute and lands in the cart. Which quantity

Marat540 [252]

Answer: the horizontal component of total momentum

Explanation:

Since the open cart is rolling to the left on the horizontal surface, the quantity that has the same value just before and just after the package lands in the cart is the horizontal component of total momentum.

Momentum, is the product of the mass of a particle and the velocity of the particle. The change of momentum depends on the force which acts on it. The addition of the the individual momenta is the total momentum.

8 0

2 years ago