All categories

2 years ago

1. If the angle of the ramp were increased from 30˙ to 45˙, how would this change the weight of the box? Explain.

Physics

1 answer:

yawa3891 [41]2 years ago

4 0

When the angle of the ramp increases, the weight of the box acting perpendicular to the ramp decreases.

<h3>Normal reaction of the box</h3>

The normal reaction of the box is due to weight of the box acting perpendicular to the ramp.

Fn = Wcosθ

<h3>when the angle of the ramp = 30⁰</h3>

Fn = Wcos(30)

Fn = 0.866W

<h3>when the angle of the ramp = 45⁰</h3>

Fn = W x cos(45)

Fn = 0.7071W

Thus, when the angle of the ramp increases, the weight of the box acting perpendicular to the ramp decreases.

Learn more about normal reaction here: brainly.com/question/18292235

#SPJ1

You might be interested in

Muscular Strength exercises focus on high______and low___

Advocard [28]

Answer:

d d d d dd d d d d d dd d d d d dd d

4 0

3 years ago

Which statement about thermal energy is true?

NNADVOKAT [17]

Answer:Thermal energy is a measure of a substance 's specific heat

Explanation:

7 0

3 years ago

Read 2 more answers

150 ml of water is heated with a burning marshmallow From 20 c to 93.5 C

kap26 [50]

93.5 if it’s wrong sorry sis I need my homework done too

7 0

3 years ago

Bonus: (It's not that hard, you just have to pay attention to units.) The Saturn V rocket first stage

agasfer [191]

$v = 2.45×10^3\:\text{m/s}$

Explanation:

Newton's 2nd Law can be expressed in terms of the object's momentum, in this case the expelled exhaust gases, as

$F = \dfrac{d{p}}{d{t}}$ (1)

Assuming that the velocity remains constant then

$F = \dfrac{d}{dt}(mv) = v\dfrac{dm}{dt}$

Solving for $v,$ we get

$v = \dfrac{F}{\left(\frac{dm}{dt}\right)}\;\;\;\;\;\;\;(2)$

Before we plug in the given values, we need to convert them first to their appropriate units:

The thrust <em>F</em><em> </em> is

$F = 7.5×10^6\:\text{lbs}×\dfrac{4.45\:\text{N}}{1\:\text{lb}} = 3.34×10^7\:\text{N}$

The exhaust rate dm/dt is

$\dfrac{dm}{dt} = 15\dfrac{T}{s}×\dfrac{2000\:\text{lbs}}{1\:\text{T}}×\dfrac{1\:\text{kg}}{2.2\:\text{lbs}}$

$\;\;\;\;\;= 1.36×10^4\:\text{kg/s}$

Therefore, the velocity at which the exhaust gases exit the engines is

$v = \dfrac{F}{\left(\frac{dm}{dt}\right)} = \dfrac{3.34×10^7\:\text{N}}{1.36×10^4\:\text{kg/s}}$

$\;\;\;= 2.45×10^3\:\text{m/s}$

6 0

2 years ago

What can the magnetic field be used for

deff fn [24]

Answer:

the magnetic field can be used to make electricity

Explanation:

Moving a magnet around a coil of wire, or moving a coil of wire around a magnet, pushes the electrons in the wire and creates an electrical current. Electricity generators essentially convert kinetic energy (the energy of motion) into electrical energy

8 0

3 years ago