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3 years ago

Kayla drew a diagram to compare convex and concave lenses which labels belong in the areas marked XYZ

Physics

1 answer:

lesya692 [45]3 years ago

8 0

Answer:

X: Causes light rays to bend toward the principal axis

Y: Is capable of producing images that are smaller than objects

Z: Causes light rays to bend away from the principal axis

Explanation:

just took the test...

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The car travels 25 miles in the first 0.5 hours

Westkost [7]

Answer:

$\huge\boxed{\sf v = 50\ miles / hr}$

Explanation:

Given Data:

Distance = S = 25 miles

Time = t = 0.5 hours

Required:

Speed = v = ?

Formula:

v = S/t

Solution:

v = 25 / 0.5

v = 50 miles / hr

$\rule[225]{225}{2}$

Hope this helped!

<h3>~AnonymousHelper1807</h3>

4 0

3 years ago

To test the performance of its tires, a car

velikii [3]

Answer:

The coefficient of static friction between the tires and the road is 1.987

Explanation:

Given:

Radius of the track, r = 516 m

Tangential Acceleration $a_r$ = 3.89 m/s^2

Speed,v = 32.8 m/s

To Find:

The coefficient of static friction between the tires and the road = ?

Solution:

The radial Acceleration is given by,

$a_{R = \frac{v^2}{r}$

$a_{R = \frac{(32.8)^2}{516}$

$a_{R = \frac{(1075.84)}{516}$

$a_{R = 2.085 m/s^2$

Now the total acceleration is

$\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2$

=> $= \sqrt{ (a_r)^2+(a_R)^2}$

=> $\sqrt{ (3.89 )^2+( 2.085)^2}$

=> $\sqrt{ (15.1321)+(4.347)^2}$

=> $19.4791 m/s^2$

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of static friction is

$\mu =\frac{f}{W}$

From (1) and (2)

$\mu =\frac{ma}{mg}$

$\mu =\frac{a}{g}$

Substituting the values, we get

$\mu =\frac{19.4791}{9.8}$

$\mu =1.987$

8 0

3 years ago

Which is the product of cellular respiration? A. ATTP B. light C. oxygen D.sugar

MrMuchimi

The answer to the question is A

7 0

3 years ago

A block of wood is 4 cm wide, 5 cm long, and 10 cm high. It weighs 100 grams. Calculate its volume. Calculate its density. Will

77julia77 [94]

Answer:

$V=200cm^3\\\\\rho =0.500g/cm^3$

It will float.

Explanation:

Hello.

In this case, given the width, length and height, we can compute the volume as follows:

$V=W*L*H\\\\V=4cm*5cm*10cm\\\\V=200cm^3$

Moreover, since the density is computed via the division of the mass by the volume:

$\rho =\frac{m}{V}$

We obtain:

$\rho =\frac{100g}{200cm^3} \\\\\rho =0.500g/cm^3$

In such a way, since the solid has a lower density than the water, we infer it will float.

Best regards.

3 0

3 years ago

A 5-cm-high peg is placed in front of a concave mirror with a radius of curvature of 20 cm.

Andrei [34K]

Answer:

Explanation:

Using the magnification formula.

Magnification = Image distance(v)/object distance(u) = Image Height(H1)/Object Height(H2)

M = v/u = H1/H2

v/u = H1/H2...1

3) Given the radius of curvature of the concave lens R = 20cm

Focal length F = R/2

f = 20/2

f = 10cm

Object distance u = 5cm

Object height H2= 5cm

To get the image distance v, we will use the mirror formula

1/f = 1/u+1/v

1/v = 1/10-1/5

1/v = (1-2)/10

1/v =-1/10

v = -10cm

Using the magnification formula

(10)/5 = H1/5

10 = H1

H1 = 10cm

Image height of the peg is 10cm

4) If u = 15cm

1/v = 1/f-1/u

1/v = 1/10-1/15

1/v = 3-2/30

1/v = 1/30

v = 30cm

30/15 = H1/5

15H1 = 150

H1/= 10cm

5) if u = 20cm

1/v = 1/f-1/u

1/v = 1/10-1/20

1/v = 2-1/20

1/v = 1/20

v = 20cm

20/20 = H1/5

20H1 = 100

H1 = 5cm

6) If u = 30cm

1/v = 1/f-1/u

1/v = 1/10-1/30

1/v = 3-1/30

1/v = 2/30

v = 30/2 cm

v =>15cm

15/30 = Hi/5

30H1 = 75

H1 = 75/30

H1 = 2.5cm

4 0

3 years ago