HELP PLEASE 20 ;OINTS

1. ________________electricity is the type of electricity commonly used in homes and businesses throughout the world.

amid [387]

Number 1 is letter A

3 0

3 years ago

Assume that the stopping distance of a van varies directly with the square of the speed. A van traveling 40 miles per hour can s

Daniel [21]

Answer:

d = 100.8 ft

Explanation:

As we know that initial speed of the van is 40 miles then the stopping distance is given as 70 feet

here we know that

$v_f^2 - v_i^2 = 2 ad$

so here we have

$0^2 - 40^2 = 2 a (70 feet)$

now again if the speed is increased to 48 mph then let say the stopping distance is "d"

so we will have

$0^2 - 48^2 = 2 a (d)$

now divide the above two equations

$\frac{40^2}{48^2} = \frac{70 feet}{d}$

$d = 100.8 ft$

4 0

3 years ago

HELP MY GRADES-

belka [17]

If you count the number of seconds between the flash of lightning and the sound of thunder, and then divide by 5, you'll get the distance in miles to the lightning: 5 seconds = 1 mile, 15 seconds = 3 miles, 0 seconds = very close. Keep in mind that you should be in a safe place while counting.

5 0

2 years ago

A sprinter must average 24.0 mi/h to win a 100-m dash in 9.30 s. What is his wavelength at this speed if his mass is 84.5 kg?

crimeas [40]

Answer:

Wavelength λ = 7.31 × 10^-37 m

Explanation:

From De Broglie's equation;

λ = h/mv

Where;

λ = wavelength in meters

h = plank's constant = 6.626×10^-34 m^2 kg/s

m = mass in kg

v = velocity in m/s

Given;

v = 24 mi/h

Converting to m/s

v = 24mi/h × 0.447 m/s ×1/(mi/h)

v = 10.73m/s

m = 84.5kg

Substituting the values into the equation;

λ = (6.626×10^-34 m^2 kg/s)/(84.5kg × 10.73m/s)

λ = 7.31 × 10^-37 m

7 0

3 years ago

Read 2 more answers

How do i find stretch? The problem in questioning has already given me the elastic energy and k-value, but I have no idea how to

finlep [7]

Answer:

Stretch can be obtained using the Elastic potential energy formula.

The expression to find the stretch (x) is $x=\sqrt{\frac{2\times EPE}{k}}$

Explanation:

Given:

Elastic potential energy (EPE) of the spring mass system and the spring constant (k) are given.

To find: Elongation in the spring (x).

We can find the elongation or stretch of the spring using the formula for Elastic Potential Energy (EPE).

The formula to find EPE is given as:

$EPE=\frac{1}{2}kx^2$

Rewriting the above expression in terms of 'x', we get:

$x=\sqrt{\frac{2\times EPE}{k}}$

Example:

If EPE = 100 J and spring constant, k = 2 N/m.

Elongation or stretch is given as:

$x=\sqrt{\frac{2\times EPE}{k}}\\\\x=\sqrt{\frac{2\times 100}{2}}\\\\x=\sqrt{100}=10\ m$

Therefore, the stretch in the spring is 10 m.

So, stretch in the spring can be calculated using the formula for Elastic Potential Energy.

6 0

2 years ago