2. Can you rearrange the equation F = kx to get k on one side of the equation

Joe drove at the speed of 45 miles per hour for a certain distance. He then drove at the speed of 55 miles per hour for the same

Snowcat [4.5K]

Answer:

$v_{avg} = 49.5 mph$

Explanation:

Let the distance moved by Joe is "d"

so the time taken by him to drove it by speed 45 mph is given as

$t_1 = \frac{d}{v_1}$

$t_1 = \frac{d}{45}$

now the same distance is traveled by him with speed 55 mph

so the time taken by him

$t_2 = \frac{d}{55}$

so total time taken by him for complete distance 2d

$t = t_1 + t_2$

$t = \frac{d}{45} + \frac{d}{55}$

$t = 0.0404 d$

now the average speed is given as

$v_{avg} = \frac{2d}{t}$

$v_{avg} = \frac{2d}{0.0404d}$

$v_{avg} = 49.5 mph$

5 0

3 years ago

Read 2 more answers

A block moving to the right on a level surface with friction is pulled by an increasing horizontal force also directed to the ri

KengaRu [80]

Answer:

e) True Yes both are constant

Explanation:

Let us propose the solution of the problem before reviewing the statements, we use Newton's second law

F - fr = m a

N- W = 0

N = mg

The equation for the force of friction is

fr = μ N

F - μ mg = m a

F = m (a- μ g)

Now let's review the claims

.a) False. Normal force and friction are constants.

.b) False. Both are constant.

.c) False. Both are constant.

d.) False

e) True Yes both are constant

6 0

3 years ago

A proton enters a region of constant magnetic field, perpendicular to the field and after being accelerated from rest by an elec

kicyunya [14]

Answer:

B = 1.1413 10⁻² T

Explanation:

We use energy concepts to calculate the proton velocity

starting point. When entering the electric field

Em₀ = U = q V

final point. Right out of the electric field

em_f = K = ½ m v²

energy is conserved

Em₀ = Em_f

q V = ½ m v²

v = $\sqrt{2qV/m}$

we calculate

v = $\sqrt{\frac{ 2 \ 1.6 \ 10^{-19} \ 300}{1.67 \ 0^{-27}} }$

v = $\sqrt{632.3353 \ 10^8}$

v = 25.15 10⁴ m / s

now enters the region with magnetic field, so it is subjected to a magnetic force

F = m a

the force is

F = q v x B

as the velocity is perpendicular to the magnetic field

F = q v B

acceleration is centripetal

a = v² / r

we substitute

qvB =1/2 m v² / r

B = v $\frac{m v}{2 q r}$

we calculate

B = $\frac{1.67 \ 10^{-27} 25.15 \ 10^4 }{1.6 \ 10^{-19} 0.23}$

B = 1.1413 10⁻² T

8 0

3 years ago

The atomic radius of metal X is 1.20 × 102 picometers (pm) and a crystal of metal X has a unit cell that is face-centered cubic.

dmitriy555 [2]

The density of the metal can be determined through the formula [n*MW]/ Na*[a^3] . substituting, we get,

<span>d = [n*MW]/ Na*[a^3]
</span><span>d = [4 atoms*42.3 g/mol]/ [6.022 x 1023atoms/mol* (sqrt 8 *1.20x10-10)^3]
</span>d = 0.719 g/cm3

4 0

3 years ago

In an electricity experiment, a 1.20 g plastic ball is suspended on a 59.0 cm long string and given an electric charge. A charge

stich3 [128]

Answer:

a.) 5.24 10⁻³ N . b) 0.013 N

Explanation:

a) In absence of other forces, the plastic ball is only subject to the force of gravity (downward) , and to the tension in the string, which are equal each other.

We are told that there exists an horizontal force , of an electric origin, that causes the ball to swing out to a 24º angle (respect the normal) and remain there, so there exists a new equilibrium condition.

In this situation, both the vertical and horizontal components of the external forces acting on the ball (gravity, tension and the electrical force) must be equal to 0.

The only force that has horizontal and vertical components, is the tension in the string.

We can apply Newton's 2nd Law to both directions, as follows:

T cos 24º - mg = 0

-T sin 24º + Fe = 0

where T= Tension in the string.

Fe = Electrical Force

mg = Fg = gravity force

⇒ T = mg/ cos 24º

Replacing in the horizontal forces equation:

-mg/cos 24º . sin 24º = -Fe

∴ Fe = mg. tg 24º = 0.0012 kg. 9.8 m/s². tg 24º = 5.24 10⁻³ N

b) In order to get the value of T, we can simply solve for T the vertical forces component equation , as follows:

T = mg/ cos 24º = 0.0012 kg. 9.8 m/s² / 0.914 = 0.013 N

6 0

3 years ago