Which ramp requires the least amount of force?

An object is dropped on Earth from a height of 15 m. How long did it take for the object to reach the ground?

Inga [223]

x=gt^2/2=>t^2=2x/g=30/9.81=3.05=>t=sqrt(3.05)=1.8 s

6 0

3 years ago

A particle is confined to the x-axis between x = 0 and x = 1 nm. The potential energy U = 0 inside this region and U is infinite

DiKsa [7]

Answer:

Explanation:

According to heisenberg uncertainty Principle

Δx Δp ≥ h / 4π , where Δx is uncertainty in position , Δp is uncertainty in momentum .

Given

Δx = 1 nm

Δp ≥ h /1nm x 4π

≥ 6.6 x 10⁻³⁴ / 10⁻⁹ x 4 π

≥ . 5254 x ⁻²⁵

h / λ ≥ . 5254 x ⁻²⁵

6.6 x 10⁻³⁴ /. 5254 x ⁻²⁵ ≥ λ

12.56 x 10⁻⁹ ≥ λ

longest wave length = 12.56 n m

6 0

3 years ago

Light is shining in the images. Which object reflects a lot of light, absorbs a little light, and transmits almost no light?

SVEN [57.7K]

I believe it is B, the object that reflects a lot of light, absorbs a little light, and transmit almost no light.

6 0

3 years ago

Read 2 more answers

10. Josie and Trey were working on their physics project and both built catapults. Trey's catapult shot a ball

SVEN [57.7K]

Answer:

Josie's ball faster than T

3 0

2 years ago

North Dakota Electric Company estimates its demand trend line (in millions of kilowatt hours) to be: D = 75.0 + 0.45Q, whe

Alborosie

Answer:

The demand forecast for winter is 96.36 millions KWH

The demand forecast for spring is 145.08 millions KWH

The demand forecast for summer is 169.89 millions KWH

The demand forecast for fall is 73.08 millions KWH

Explanation:

Given that,

The demand trend line is

$D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors$

We need to calculate the demand forecast for winter

Using given formula

$D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors$

Put the value into the formula

$D=(75.0+0.45\times101)\times0.80$

$D=96.36\ millions\ KWH$

We need to calculate the demand forecast for spring

Using given formula

$D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors$

Put the value into the formula

$D=(75.0+0.45\times102)\times1.20$

$D=145.08\ millions\ KWH$

We need to calculate the demand forecast for summer

Using given formula

$D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors$

Put the value into the formula

$D=(75.0+0.45\times103)\times1.40$

$D=169.89\ millions KWH$

We need to calculate the demand forecast for fall

Using given formula

$D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors$

Put the value into the formula

$D=(75.0+0.45\times104)\times0.60$

$D=73.08\ millions KWH$

Hence, The demand forecast for winter is 96.36 millions KWH

The demand forecast for spring is 145.08 millions KWH

The demand forecast for summer is 169.89 millions KWH

The demand forecast for fall is 73.08 millions KWH

3 0

3 years ago