All categories

3 years ago

Elements with positive valences usually ______ electrons

Physics

1 answer:

Leno4ka [110]3 years ago

7 0

The answer is donate, therefore elements with positive valences usually donate electrons

You might be interested in

During an auto accident, the vehicle’s air bags deploy and slow down the passengers more gently than if they had hit the windshi

Allushta [10]

Answer:

d = 0.38 m

Explanation:

As we know that the person due to the airbag action, comes to a complete stop, in 36 msec or less, and during this time, is decelerated at a constant rate of 60 g, we can find the initial velocity (when airbag starts to work), as follows:

vf = v₀ -a*t

If vf = 0, we can solve for v₀:

v₀ = a*t = 60*9.8 m/s²*36*10⁻³s = 21.2 m/s

With the values of v₀, a and t, we can find Δx, applying any kinematic equation that relates all of some of these parameters with the displacement.

Just for simplicity, we can use the following equation:

$vf^{2} -vo^{2} = 2*a*d$

where vf=0, v₀ =21.2 m/s and a= -588 m/s².

Solving for d:

$d = \frac{-vo^{2}}{2*a} = \frac{(21.2m/s)^{2} }{2*588 m/s2} =0.38 m$

⇒ d = 0.38 m

5 0

3 years ago

Read 2 more answers

A metal has a strength of 414 MPa at its elastic limit and the strain at that point is 0.002. Assume the test specimen is 12.8-m

ser-zykov [4K]

To solve this problem, we will start by defining each of the variables given and proceed to find the modulus of elasticity of the object. We will calculate the deformation per unit of elastic volume and finally we will calculate the net energy of the system. Let's start defining the variables

Yield Strength of the metal specimen

$S_{el} = 414Mpa$

Yield Strain of the Specimen

$\epsilon_{el} = 0.002$

Diameter of the test-specimen

$d_0 = 12.8mm$

Gage length of the Specimen

$L_0 = 50mm$

Modulus of elasticity

$E = \frac{S_{el}}{\epsilon_{el}}$

$E = \frac{414Mpa}{0.002}$

$E = 207Gpa$

Strain energy per unit volume at the elastic limit is

$U'_{el} = \frac{1}{2} S_{el} \cdot \epsilon_{el}$

$U'_{el} = \frac{1}{2} (414)(0.002)$

$U'_{el} = 414kN\cdot m/m^3$

Considering that the net strain energy of the sample is

$U_{el} = U_{el}' \cdot (\text{Volume of sample})$

$U_{el} = U_{el}'(\frac{\pi d_0^2}{4})(L_0)$

$U_{el} = (414)(\frac{\pi*0.0128^2}{4}) (50*10^{-3})$

$U_{el} = 2.663N\cdot m$

Therefore the net strain energy of the sample is $2.663N\codt m$

6 0

3 years ago

A book weighing 18.0 N is lifted from the floor and placed on a shelf 1.5 m high. What is the minimum amount of work required to

algol13

B 12j is the correct answer

6 0

3 years ago

A 1000-kg car traveling at 70 m/s takes 3 m to stop under full braking. the same car under similar road conditions, traveling at

azamat

We assume $a=const$ (acceleration is constant. We apply the equation
$v^2=v0^2+2as$ where s is the distance to stop $v=0(m/s)$ . We find the acceleration from this equation
$a=-v0^2/(2s)=-70^2/(2*3) =-816.7 (m/s^2)
$
We know the acceleration, thus we find the distance necesssary to stop when initial speed is $v=140 (m/s)$
$s=-v0^2/(2a) =140^2/(2*816.7)=12 (m)$

5 0

4 years ago

1. To get to school, a girl walks 1 km North in 15 minutes. She

yulyashka [42]

Answer:

Average velocity of the girl is 1.25 m/s.

Explanation:

3 0

3 years ago