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3 years ago

12

Two equal forces are applied to a door. The first force is applied at the midpoint of the door, the second force is applied at t

he doorknob. Both forces are applied perpendicular to the door. Which force exerts the greater torque?A) both exert equal non-zero torquesB) the first at the midpointC) both exerts zero torquesD) the second at the doorknobE) additional information is needed

1 answer:

Orlov [11]3 years ago

4 0

Answer:

D) the second at the doorknob

Explanation:

The torque exerted by a force is given by:

$\tau = Fdsin \theta$

where

F is the magnitude of the force

d is the distance between the point of application of the force and the centre of rotation

$\theta$ is the angle between the direction of the force and d

In this problem, we have:

- Two forces of equal magnitude F

- Both forces are perpendicular to the door, so $\theta=90^{\circ}, sin \theta=1$

- The first force is exerted at the midpoint of the door, while the 2nd force is applied at the doorknob. This means that d is the larger for the 2nd force

--> therefore, the 2nd force exerts a greater torque

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Send me the circuit diagram

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3 years ago

WILL GIVE BRAINLIEST AND 50 POINTS!

7nadin3 [17]

Answer:

1) Current decreases; 2) Inverse proportionally; 3) 1[A]

Explanation:

1)

As we can see as the resistance increases the current decreases, if we take two points as an example, when the resistance is equal to 50 [ohms] the current is equal to 1[amp] and when the resistance is equal to 200 [ohms] the current tends to have a value below 0.5 [amp]. Thus demonstrating the decrease in current.

2)

Inverse proportionally, by definition we know that the law of ohm determines the voltage according to resistance and amperage. This is the voltage will be equal to the product of the voltage by the resistance.

$V=I*R\\V = voltage [volts]\\I = current[amp]\\R = resistance [ohms]$

where:

$R =\frac{V}{I} \\or\\I=\frac{V}{R}$

And whenever we have in a fractional number the denominator the variable we are interested in, we can say that this is inversely proportional to the value we are interested in determining. In this case, we can see from the two previous expressions that both the current and the resistance appear in the denominator, therefore they are inversely proportional to each other.

3)

If we place ourselves on the graph on the resistance axis, we see that at 50 [ohm] will correspond a current value equal to 1 [A].

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3 years ago

Read 2 more answers

Explain how mirrors can produce images that are larger or smaller than life size, as well as upright or inverted

galina1969 [7]

Answer:

1) When $d_{o}$ < $d_{i}$ (hence $d_{o}$ < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

2) When $d_{o}$ > $d_{i}$ (and $d_{o}$ < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

3) When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror

Explanation:

The position of an object in front of a concave mirror of radius of curvature, R, determines the size and orientation of the image of the object as illustrated in the mirror equation

$\dfrac{1}{f}=\dfrac{1}{d_{o}} + \dfrac{1}{d_{i}}$

$Magnification, \, m = \dfrac{h_{i}}{h_{o}} = -\dfrac{d_{i}}{d_{o}}$

Where:

f = Focal length of the mirror = R/2

$d_{i}$ = Image distance from the mirror

$d_{o}$ = Object distance from the mirror

$h_{i}$ = Image height

$h_{o}$ = Object height

$d_{o}$ is positive for an object placed in front of the mirror and negative for an object placed behind the mirror

$d_{i}$ is positive for an image formed in front of the mirror and negative for an image formed behind the mirror

m is positive when the orientation of the image and the object is the same

m is negative when the orientation of the image and the object is inverted

f and R are positive in the situation where the center of curvature is located in front of the mirror (concave mirrors) and f and R are negative in the situation where the center of curvature is located behind the mirror (convex mirrors)

∴ When $d_{o}$ < $d_{i}$ (hence $d_{o}$ < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

When $d_{o}$ > $d_{i}$ (and $d_{o}$ < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror.

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4 years ago

Consider a concave spherical mirr or that has focal length f = +19.5 cm.

lidiya [134]

The distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.

<h3>What is concave mirror?</h3>

A concave mirror has a reflective surface that is curved inward and away from the light source.

Concave mirrors reflect light inward to one focal point and it usually form real and virtual images.

<h3>Object distance of the concave mirror</h3>

Apply mirrors formula as shown below;

1/f = 1/v + 1/u

where;

f is the focal length of the mirror
v is the object distance
u is the image distance

when image height = object height, magnification = 1

u/v = 1

v = u

Substitute the given parameters and solve for the distance of the object from the mirror's vertex

1/f = 1/v + 1/v

1/f = 2/v

v = 2f

v = 2(19.5 cm)

v = 39 cm

Thus, the distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.

Learn more about concave mirror here: brainly.com/question/27841226

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2 years ago

In one contest at the county fair, a spring-loaded plunger launches a ball at a speed of 3.2m/s from one corner of a smooth, fla

lara31 [8.8K]

Answer:

Explanation:

Given

Speed of ball $u=3.2\ m/s$

Plane is inclined at an angle $20^{\circ}$

To win the Game we need to hit the target at $x=2.4\ m$ away

Launch angle of ball $\theta$

Motion of ball can be considered in two planes i.e. Vertical to the plane and horizontal to the plane

So Net acceleration in vertical plane is $g\sin 20$

Range of Projectile is given by

$R=\frac{u^2\sin 2\theta }{g}$

for $R=2.4\ m$

$2.4=\frac{3.2^2\times sin 2\theta }{g\sin 20}$

$\sin 2\theta =\frac{2.4\times 9.8\times \sin 20}{3.2^2}$

$\sin 2\theta =0.7855$

$2\theta =51.77$

$\theta =25.88^{\circ}$

so ball must be launched at an angle of $25.88^{\circ}$

4 0

3 years ago