A positive magnification means the image is inverted compared to the object

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Answer 1

Answer:

False

Explanation:

A positive magnification means the image is erect compared to the object. Magnifications with values greater than one represent images that are smaller than the object. A magnification of 1 (plus or minus) means that the image is the same size as the object. If m has a magnitude greater than 1 the image is larger than the object, and an m with a magnitude less than 1 means the image is smaller than the object. If the magnification is positive, the image is upright compared to the object; if m is negative, the image is inverted compared to the object.

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A 0.454-kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 21.0 N/m. The block rests on a frictionless surface. A 5.30×10?2-kg wad of putty is thrown horizontally at the block, hitting it with a speed of 8.97 m/s and sticking.Part AHow far does the putty-block system compress the spring?
A paper airplane moving 2.33 m/s has 0.161 J of KE. What is its mass? (Unit = kg)
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An airplane with a speed of 92.3 m/s is climbing upward at an angle of 51.1 ° with respect to the horizontal. When the plane's altitude is 532 m, the pilot releases a package. (a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth. (b) Relative to the ground, determine the angle of the velocity vector of the package just before impact. (a) Number Units (b) Number Units
Only one of three balls A, B, and C carries a net charge q. The balls are made from conducting material and are identical. One of the uncharged balls can become charged by touching it to the charged ball and then separating the two. This process of touching one ball to another and then separating the two balls can be repeated over and over again, with the result that the three balls can take on a variety of charges. Which one of the following distribution of charges could not possibly be achieved in this fashion, even if the process were repeated an infinite number of times?Why the answer is qA = 1/2q, qB=3/8q, qC=1/4q. Explain please.

Professional baseball pitchers deliver pitches that can reach the blazing speed of 100 mph (miles per hour). A local team has drafted an up-and-coming, left-handed pitcher who can consistently pitch at 42.91 m/s (96.00 mph) . Assuming a pitched ball has a mass of 0.1434 kg and has this speed just before a batter makes contact with it, how much kinetic energy does the ball have?

Answers

Answer: 132.02 J

Explanation:

By definition, the kinetic energy is written as follows:

KE = 1/2 m v²

In our question, we know from the question, the following information:

m = 0.1434 Kg

v= 42.91 m/s

Replacing in the equation for KE, we have:

KE = 1/2 . 0.1434 Kg. (42.91)² m²/s² ⇒ KE = 132.02 N. m = 132.02 J

Find the linear velocity of a point moving with uniform circular motion, if the point covers a distance s in the given amount of time t. s

Answers

Answer:

The linear velocity is represented by the following expression: $v = (s)/(t)$

Explanation:

From Rotation Physics we know that linear velocity of a point moving with uniform circular motion is:

$v = r\cdot \omega$ (Eq. 1)

Where:

$r$ - Radius of rotation of the particle, measured in meters.

$\omega$ - Angular velocity, measured in radians per second.

$v$ - Linear velocity of the point, measured in meters per second.

But we know that angular velocity is also equal to:

$\omega = (\theta)/(t)$ (Eq. 2)

Where:

$\theta$ - Angular displacement, measured in radians.

$t$ - Time, measured in seconds.

By applying (Eq. 2) in (Eq. 1) we get that:

$v = (r\cdot \theta)/(t)$ (Eq. 3)

From Geometry we must remember that circular arc ( $s$ ), measured in meters, is represented by:

$s = r\cdot \theta$

$v = (s)/(t)$

The linear velocity is represented by the following expression: $v = (s)/(t)$

Describe the objects that make up Saturn's rings. Your answer should include the range of sizes of objects in the rings, and the composition of the at least the outer layers of the objects.

Answers

Saturn's rings are made of billions of pieces of ice, dust and rocks. Some of these particles are as small as a grain of salt, while others are as big as houses.

Hercules X-1 is a pulsating X-ray source. The X-rays from this source sometimes completely disappear for 6 hours every 1.7 days because the neutron star has a 1.7-day orbital period around its companion star, and it is eclipsed for ____ hours once every orbital period.

Answers

Answer:

06 Hours

Explanation:

As per the details given in the question it self, the neutron star X-1 is revolving around its companion star. The orbital period is 1.7 years which means it will complete the revolution in 1.7 years. During the movement in the orbit we will be able to detect the x-rays except for the time when it goes behind the companion star and eclipsed by it as seen from Earth.

Since the x-rays disappear completely for around 6 hours. This clearly means that eclipse period is 06 hours.

A spinning wheel on a fireworks display is initially rotating in a counterclockwise direction. The wheel has an angular acceleration of -4.46 rad/s2. Because of this acceleration, the angular velocity of the wheel changes from its initial value to a final value of -31.4 rad/s. While this change occurs, the angular displacement of the wheel is zero. (Note the similarity to that of a ball being thrown vertically upward, coming to a momentary halt, and then falling downward to its initial position.) Find the time required for the change in the angular velocity to occur.

Answers

Answer:

The time for the change in the angular velocity to occur is 14.08 secs

Explanation:

From the question,

the angular acceleration is - 4.46 rad/s²

Angular acceleration is given by the formula below

$\alpha =(\omega -\omega _(o) )/(t - t_(o) )$

Where $\alpha$ is the angular acceleration

$\omega$ is the final angular velocity

$\omega _(o)$ is the initial angular velocity

$t$ is the final time

$t_(o)$ is the initial time

From the question

$\alpha$ = - 4.46 rad/s²

$\omega _(o)$ = 0 rad/s (starting from rest)

$\omega$ = -31.4 rad/s

$t_(o)$ = 0 s

Now, we will determine t

From $\alpha =(\omega -\omega _(o) )/(t - t_(o) )$ , then

$-4.46 = (-31.4 - 0)/(t - 0)$

$-4.46 = (-31.4)/(t)$

$t = (-31.4)/(-4.46)$

t = 7.04 secs

This is the time spent in one direction,

Since the angular displacement of the wheel is zero ( it returned to its initial position), then the time required for the change in the angular velocity will be twice this time, that is 2t

Hence,

The time is 2×7.04 secs = 14.08 secs

This is the time for the change in the angular velocity to occur.

A physics student standing on the edge of a cliff throws a stone vertically downward with an initial speed of 10.0 m/s. The instant before the stone hits the ground below, it is traveling at a speed of 30.0 m/s. If the physics student were to throw the rock horizontally outward from the cliff instead, with the same initial speed of 10.0 m/s, what is the magnitude of the velocity of the stone just before it hits the ground? Ignore any effects of air resistance.

Answers

Answer:

vf = 30 m/s : (the magnitude of the velocity of the stone just before it hits the ground)

Explanation:

Because the stone moves with uniformly accelerated movement we apply the following formulas:

vf²=v₀²+2*g*h Formula (1)

Where:

h: displacement in meters (m)

v₀: initial speed in m/s

vf: final speed in m/s

g: acceleration due to gravity in m/s²

Free fall of the stone

Data

v₀ = 10 m/s

vf = 30.0 m/s

g = 9,8 m/s²

We replace data in the formula (1) to calculate h:

vf²=v₀²+2*g*h

(30)² = (10)² + (2)(9.8)*h

(30)²- (10)²= (2)(9.8)*h

h =( (30)²- (10)²) /( 2)(9.8)

h = 40.816 m

Semiparabolic movement of the stone

Data

v₀x = 10 m/s

v₀y = 0 m/s

g = 9.8 m/s²

h= 40.816 m

We replace data in the formula (1) to calculate vfy :

vfy² = v₀y² + 2*g*h

vfy² = 0 + (2)(9.8)( 40.816)

$v_(fy)=√(2*9.8*40.816) = 28.284 (m)/(s)$

$v_(f)=\sqrt{v_(ox)^2+v_(fy)^2}=√((10)^2+(28.284)^2) = 30(m)/(s)$

The magnitude of the velocity of the stone just before it hits the ground is 30 m/s.

The given parameters;

initial vertical velocity of the stone, $v_y_0$ = 10 m/s

final vertical velocity of the stone, $v_y_f$ = 30 m/s

The height traveled by the stone before it hits the ground is calculated as;

$v_y_f^2 = v_y_0^2 + 2gh\n\nh = (v_y_f^2- v_y_0^2)/(2g) \n\nh = ((30)^2 - (10)^2)/(2* 9.8) \n\nh = 40.82 \ m$

If the the stone is projected horizontally with initial velocity of 10 m/s;

the initial vertical velocity = 0

Final vertical velocity of the stone is calculated as follow;

$v_y_f^2 = v_y_0^2 + 2gh\n\nv_y_f^2 = 0 + 2* 9.8* 40.82\n\nv_y_f^2 = 800.07\n\nv_y_f = √(800.07) \n\nv_y_f = 28.28 \ m/s$

The horizontal velocity doesn't change.

the final horizontal velocity, $v_x_f$ = initial horizontal velocity = 10 m/s

The resultant of the final velocity of the stone before it hits the ground;

$v _f= √(v_x_f^2 + v_y_f^2) \n\nv_f = √(10^2 + 28.28^2) \n\nv_f= 29.99 \ m/s \approx 30 \ m/s$

Thus, the magnitude of the velocity of the stone just before it hits the ground is 30 m/s.

Learn more here:brainly.com/question/13533552

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A positive magnification means the image is inverted compared to the object​

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A positive magnification means the image is inverted compared to the object