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2 years ago

Ill mark brainliest!!!

Physics

1 answer:

navik [9.2K]2 years ago

6 0

Answer: A

si, di, are expressed as negative and f is expressed as positive values.

Explanation:

When the object is located at a location in front of the focal point of a convex len, the image will always be located somewhere on the same side of the lens as the object. The image is located behind the object. In this case, the image will be an upright image and the image is enlarged

In this question, the relationship between the focal length, image distance di and object distance is

I/f = 1/di + 1/do that is

I/f - 1/di - 1/do = 0

The image size si is also negative since the image is a virtual image.

Therefore, si, di, are expressed as negative, f is expressed as positive values.

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A racetrack curve has radius 70.0 m and is banked at an angle of 12.0 ∘. The coefficient of static friction between the tires an

zubka84 [21]

Answer:

See attachment below

Explanation:

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3 years ago

A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provid

Sergeeva-Olga [200]

Answer:

$8.37*10^5$ rpm

Explanation:

Given that rotational kinetic energy = $4.66*10^9J$

Mass of the fly wheel (m) = 19.7 kg

Radius of the fly wheel (r) = 0.351 m

Moment of inertia (I) = $\frac{1}{2} mr ^2$

Rotational K.E is illustrated as $(K.E)_{rt} = \frac{1}{2} I \omega^2$

$\omega = \sqrt{\frac{2(K.E)_{rt}}{I} }$

$\omega = \sqrt{\frac{2(KE)_{rt}}{1/2 mr^2} }$

$\omega = \sqrt{\frac{4(K.E)_{rt}}{mr^2} }$

$\omega = \sqrt{\frac{4*4.66*10^9J}{19.7kg*(0.351)^2} }$

$\omega = 87636.04$

$\omega = 8.76*10^4 rad/s$

Since 1 rpm = $\frac{2 \pi}{60} rad/s$

$\omega = 8.76*10^4(\frac{60}{2 \pi})$

$\omega = 836518.38$

$\omega = 8.37 *10^5 rpm$

3 0

2 years ago

the cross section area of a hole is 725cm^2. Given that the area of a circle is A=3.14r^2 , find the radius of the hole.

asambeis [7]

$The\ area\ of\ a\ circle\ = \pi r^2 \\ 725\ cm^2 = 3.14r^2 \\
230.57 cm^2 = r^2 \\ \sqrt{230.57\ cm^2} = r \\ \boxed{r = 15.18\ cm}$

7 0

2 years ago

What is the greatest distance an image can be located behind a convex spherical mirror?

Afina-wow [57]

Answer:

Maximum distance of image from mirror is equal to focal length of the mirror

Explanation:

As we know by the equation of mirror we have

$\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}$

here we know for convex mirror

object position is always negative as it will be placed behind the mirror always

while the focal length of the convex mirror is always taken positive

So here we have

$\frac{1}{d_i} + \frac{1}{-d_o} = \frac{1}{f}$

$\frac{1}{d_i} = \frac{1}{d_o} + \frac{1}{f}$

so here maximum value of image distance is equal to focal length of the mirror

6 0

2 years ago

PLEASE HELP WILL GIVE BRAINLIEST AND EXTRA POINTS!! Super simple I’ve just been sick so I have no clue what’s going on.

Tju [1.3M]

Answer: 12) 1.07 m/s (right) 13) 4.05 m/s 14) 73 m/s 15) 10.9 m/s

Explanation:

12) Conservation of momentum. Momentum is the produce of mass and velocity.

13(2) + 15(-5) = 13(-5) + 15v

v = 1.06666... ≈ 1.07 m/s (right)

13) 18(9) + 22(0) = 18v + 22v

v = 18(9)/40 = 4.05 m/s

14) 0.65(35) + 0.08(0) = 0.65(26) + 0.08v

v = 73.125

15) This is a bit trickier. Let's ASSUME you jump off at 7 m/s relative to the truck. Doing this, we can assume that the reference frame is moving along with the truck at 10 m/s

the conservation of momentum equation becomes

600(0) + 80(0) = 600v + 80(-7)

v = 0.9333333... m/s

adding back the velocity of the reference frame means the truck is now traveling.

10.9333333... ≈ 10.9 m/s

4 0

2 years ago