When a pendulum is swinging, the velocity is highest at which point?

All matter is in constant motion<br><br><br> True or False?

svlad2 [7]

True the answer is true

4 0

3 years ago

Read 2 more answers

Covalent Bonds usually bond atoms of which type of elements together?

anzhelika [568]

<h2>Answer: Nonmetals to Nonmetals </h2>

Covalent bonds hold non-metallic atoms together. These atoms have many electrons in their outermost level and have a tendency to gain electrons rather than to yield them.

In this case the bond is formed by sharing a pair of electrons between the two atoms, one from each atom. Then, this pair of shared electrons is common to the two atoms and holds them together, so that both atoms acquire more stability.

Therefore the correct answer is C.

5 0

4 years ago

The drawing shows a model for the motion of the human forearm in throwing a dart. Because of the force M applied by the triceps

Serga [27]

Answer:

464.3 N

Explanation:

Given parameters are:

I = 0.065 $kg*m^2$

L = 0.025 m

R = 0.28 m

$v_0$ = 0 m/s

$v_f$ = 5 m/s

t = 0.1 s

$v_f=v_0+at=at$

Hence, $a=v_f/t$

We must connect two torque equations to find the answer.

$\tau=LM=I\alpha$

Where $\alpha =\frac{a}{R} =\frac{v_f}{Rt}$

Hence, $LM=I\frac{v_f}{Rt}$

Thus, $M = \frac{Iv_f}{LRt} = \frac{0.065*5}{0.025*0.28*0.1} =464.3$ N

5 0

4 years ago

A copper wire has a radius of 3.5 mm. When forces of a certain equal magnitude but opposite directions are applied to the ends o

denpristay [2]

Answer:

The tensile stress on the wire is 550 MPa.

Explanation:

Given;

Radius of copper wire, R = 3.5 mm

extension of the copper wire, e = 5.0×10⁻³ L

L is the original length of the copper wire,

Young's modulus for copper, Y = 11×10¹⁰Pa.

Young's modulus, Y is given as the ratio of tensile stress to tensile strain, measured in the same unit as Young's modulus.

$Y =\frac{Tensile \ stress}{Tensile \ strain} \\\\Tensile \ stress = Y*Tensile \ strain\\\\But, Tensile \ strain = \frac{extension}{original \ Length} = \frac{5.0*10^{-3} L}{L} = 5.0*10^{-3}\\\\Tensile \ stress = 11*10^{10} *5.0*10^{-3} \ = 550*10^6 \ Pa$