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2 years ago

5

during a space shuttle launch about 830,000 kg of fuel is burned in 8 min. the fuel provides the shuttle with a constant thrust,

or forward force. how does newton's second law of motion explain why the shuttle's acceleration in creases as the fuel is burned?

1 answer:

kirill115 [55]2 years ago

8 0

The shuttles acceleration in the creases as the fuel is burned because the acceleration of the obect as produced by net force is directly proportional to the magnitude of the net force.

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1. A concave mirror has a focal length of 1.50 meters. What is the radius of curvature of the mirror? An object is placed 4.00 m

matrenka [14]

1) 3.0 m

2) 2.40 m

Explanation:

1)

A concave mirror is a reflecting surface that causes the reflection of the rays of light coming to the mirror, producing an image of the object facing the mirror.

There are two types of mirror:

- Concave mirror: this is curved inward - as a result, the rays of light coming from the object are reflected back into a single point, called focal point

- Convex mirror: this is curved outward - as a result, the rays of light coming from the object are reflected back into diverging direction, not into a single point

For a curved mirror, the radius of curvature is twice the focal length:

$R=2f$

Where

R is the radius of curvature

f is the focal length

In this problem,

f = 1.50 m

So, the radius of curvature is

$R=2(1.50)=3.0 m$

2)

The distance of the image from the mirror can be found by using the mirror equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length

p is the distance of the object from the mirror

q is the distance of the image from the mirror

IN this problem we have:

f = 1.50 m is the focal length

p = 4.00 m is the distance of the object from the mirror

Solving for q, we find:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{1.50}-\frac{1}{4.00}=0.416\\q=\frac{1}{0.416}=2.40 m$

8 0

2 years ago

A girl holds an arrow in her hands, ready to shoot the arrow at a dragon who is just outside her arrow’s range of 45 meters. The

gregori [183]

First, assume that the arrow leaves the bow with a velocity of 45 m/s above the horizontal with respect to the bow.

Since the bow is moving at 8.5 m/s 45º below the horizontal, find the initial velocity of the arrow with respect to the ground:

$\vec{v}_{AG}=\vec{v}_{AB}+\vec{v}_{BG}$

This equation reads:

<em>The velocity of the arrow with respect to the ground </em><em>is equal to</em><em> the velocity of the arrow with respect to the bow </em><em>plus</em><em> the velocity of the bow with respect to the gound.</em>

Notice that this is a vector equation. Then, the vertical and horizontal components of the velocities must be added separately:

$\begin{gathered} v_{AG-x}=v_{AB-x}+v_{BG-x} \\ v_{AG-y}=v_{AB-y}+v_{BG-y} \end{gathered}$

Find the vertical and horizontal components of the velocity of the arrow with respect to the bow and the velocity of the bow with respect to the ground:

$\begin{gathered} v_{AB-x}=v_{AB}\cos (\theta) \\ =45\frac{m}{s}\cdot\cos (30º) \\ =38.97\frac{m}{s} \end{gathered}$

$\begin{gathered} v_{AB-y}=v_{AB}\sin (\theta) \\ =45\frac{m}{s}\sin (30º) \\ =22.5\frac{m}{s} \end{gathered}$

Similarly, for the velocity of the bow with respect to the ground:

$\begin{gathered} v_{BG-x}=6.01\frac{m}{s} \\ v_{BG-y}=-6.01\frac{m}{s} \end{gathered}$

Then, the vertical and horizontal components of the initial velocity of the arrow with respect to the ground, are:

$\begin{gathered} v_{AG-x}=38.97\frac{m}{s}+6.01\frac{m}{s}=44.98\frac{m}{s} \\ \\ v_{AG-y}=22.5\frac{m}{s}-6.01\frac{m}{s}=16.49\frac{m}{s} \end{gathered}$

Use the horizontal component of the velocity to find how long it takes for the arrow to travel a horizontal distance x of 55 meters. Then, use that time to find the vertical position of the arrow.

Since the horizontal movement of the arrow is uniform, then:

$v_{AG-x}=\frac{x}{t}_{}$

Isolate t and substitute x=55m, v_{AG-x}=44.98 m/s:

$\begin{gathered} t=\frac{x}{v_{AG-x}} \\ =\frac{55m}{44.98\frac{m}{s}} \\ =1.2227s \end{gathered}$

The vertical motion of the arrow is a uniformly accelerated motion. Then, the vertical position is given by:

$y=v_{AG-y}t-\frac{1}{2}gt^2$

Replace v_{AG-y}=16.49 m/s, t=1.2227s and g=9.81 m/s^2 to find the vertical position of the arrow when the horizontal position is 55 meters. This matches the elevation of the dragon with respect to the girl when the girl shoots:

$\begin{gathered} y=(16.49\frac{m}{s})(1.2227s)-\frac{1}{2}(9.81\frac{m}{s^2})(1.2227s)^2 \\ =12.829\ldots m \\ \approx12.8m \end{gathered}$

Therefore, the dragon is 12.8 meters above the girl when the arrow is shoot.

6 0

5 months ago

Which question could be tested in a scientific manner?

andrey2020 [161]

The questions from this lot which could be tested in a scientific manner would be "what causes some people to be color-blind"
and
"what are the best shoes to wear when exercising"
Both of these questions can be tested in a scientific way through an experiment.

5 0

2 years ago

The brachialis attaches to the forearm .035 m from the elbow at an angle of 32 deg. If the brachialis produces 750 N of force, w

jek_recluse [69]

Answer:

$XY=636N$

Explanation:

From the question we are told that:

Distance $d=0.35m$

Angle $\theta=32\textdegree$

Force $F=750N$

Generally the equation for magnitude of the stabilizing component of the brachialis force is mathematically given by

$XY=Fcos\theta$

$XY=750cos 32\textdegree$

$XY=636N$

4 0

2 years ago

Background information about reflection and refraction of light

KIM [24]

Answer:

reflection :

angle of reflection and angle of incidence is equal
Incident ray and reflect d ray plus normal lies on the same plan on a same point
light reflect uniformly if it is incident on a plan surface

refraction :

light will refrect if light goes from rare medium to a dense medium or vice versa
after the light is refracted, it retunes to its original direction
angle of incidence is not equal to the angle of refraction

4 0

2 years ago