Describe how two isotopes of nitrogen differ from two nitrogen ions?

You are performing a knee extension exercise. You hold a 20kg weight at full knee extension. The weight is 0.4m from your knee j

dmitriy555 [2]

Answer:

The moment is -78.4 N-m (clockwise).

Explanation:

Given:

Mass of the object (m) = 20 kg

Distance of the object from the knee joint (d) = 0.4 m

Weight of leg is not considered.

Acceleration due to gravity (g) = 9.8 m/s²

Now, weight of the object is equal to the product of its mass and acceleration due to gravity. So,

Weight = Mass × Acceleration due to gravity

= $mg=20\times 9.8 =196\ N$

We know that, moment of a force about a point is defined as the product of force applied and the perpendicular distance between the point and the line of application of force.

Moment of the given weight about the knee joint is given as:

Moment about knee joint = Weight × Distance from knee joint to weight

Moment about knee joint = 196 × 0.4 = 78.4 Nm

Now, from the diagram below, we can observe that, the weight acts vertically down and thus the sense of rotation about the knee joint at point O is clockwise. So, moment is negative.

Therefore, the moment is -78.4 N-m (clockwise).

7 0

3 years ago

WILL GIVE BRAINLIEST!!!!!!!!!!!! 50 PTS!!!!!!!!!!!!!!! 10 SCIENCE QUESTIONS!!!!

Jet001 [13]

1.b 2.c 3.a 4.b 5.d 6.a 7.a 9.idk 10.c hope thes helps have a nice day and God bless

8 0

3 years ago

What did bohr contribute to modern atomic theory

ddd [48]

Bohr’s model showed that electron orbits had distinct radii.

5 0

3 years ago

Read 2 more answers

Two small metallic spheres, each of mass 1.0 g are suspended as pendulums by light strings from a common point. The spheres are

LUCKY_DIMON [66]

Answer:

Q = 1.52 X 10⁻⁷ C.

Explanation:

String is inclined at 10 degree with the vertical so the vertical component of T will balance the weight and the horizontal component will balance the force of repulsion F.

T Cos 10 = mg

T Sin 10 = F

Tan 10 = F / mg

F = mg Tan 10

= 10⁻³ x 9.8 x .176

= 1.728 x 10⁻³ .

Force of repulsion F = K Q² / R²

Distance between two charges = 2 x 1 x sin 10

= .347 m

1.728 X 10⁻³ = $[tex]\frac{9\times10^9\times Q^2}{(0.347)^2}$ [/tex]

Q² = 2.31 X 10⁻¹⁴

Q = 1.52 X 10⁻⁷ C.

5 0

3 years ago

A(n) 72.6 kg astronaut becomes separated from the shuttle, while on a space walk. She finds herself 48.1 m away from the shuttle

netineya [11]

Answer:

The time taken to get the shuttle is 6.91 minutes.

Explanation:

Given that.

Mass of an astronaut, $m_1=72.6\ kg$

Distance between astronaut and shuttle, d = 48.1 m

Mass of the camera, $m_2=0.704\ kg$

Speed of camera, v = 12 m/s

We need to find the time taken by the astronaut to reach her shuttle. Both shuttle and astronaut are at rest, their initial velocities will be equal to 0. Using the conservation of linear momentum as :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\m_1(0)+m_2(0)=m_1v_1+m_2v_2\\\\m_1v_1+m_2v_2=0\\\\v_1=\dfrac{-m_2v_2}{m_1}\\\\v_1=\dfrac{0.704\times 12}{72.6}\\\\v_1=0.116\ m/s$

The time taken to get the shuttle is given by :

$t=\dfrac{d}{v_1}\\\\t=\dfrac{48.1}{0.116}\\\\t=414.65\ s$

or

t = 6.91 minutes

So, the time taken to get the shuttle is 6.91 minutes.

3 0

3 years ago