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Liula [17]
3 years ago
7

How do I find magnitude of acceleration?

Physics
1 answer:
Leno4ka [110]3 years ago
6 0
Divide the change in speed by the time for the change.
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A flat polished area is turned through 15 degrees. If the angle of incidence is 20 degrees, what is the angle between the incide
Ilya [14]

Answer:

<h2>50°</h2>

Explanation:

Angle of rotation of the flat polished surface \theta = 15°

angle of incidence i = 20°

Since the polished surface is turned at an angle of 15°, the angle of reflection

r = 2\theta (Note that the angle of rotation only have effect of the angle of reflection)

r = 2*15 = 30°

The angle between the reflected ray and the incident ray will be equal to the sum of the angle of incidence and the angle of reflection i.e i+r

The angle between the reflected ray and the incident ray = 20°+ 30° = 50°

7 0
4 years ago
Which temperature is lower ? -44°c -55°c​
Mice21 [21]

Answer:

-55 degrees c

Explanation:

because negative numbers work in reverse and that makes -55 actually lower.

6 0
3 years ago
A 40-kg worker climbs a ladder upwards for 15m. What work was done during their climb upwards?
Bingel [31]

Answer:

Explanation:

The work increased the potential energy

W = PE = mgh = 40(9.8)(15) = 5880 J(oules)

4 0
2 years ago
A 4500 kg car accelerates from rest to 45.0
Llana [10]

The car undergoes an acceleration <em>a</em> such that

(45.0 km/h)² - 0² = 2 <em>a</em> (90 m)

90 m = 0.09 km, so

(45.0 km/h)² - 0² = 2 <em>a</em> (0.09 km)

Solve for <em>a</em> :

<em>a</em> = (45.0 km/h)² / (2 (0.09 km)) = 11,250 km/h²

Ignoring friction, the net force acting on the car points in the direction of its movement (it's also pulled down by gravity, but the ground pushes back up). Newton's second law then says that the net force <em>F</em> is equal to the mass <em>m</em> times the acceleration <em>a</em>, so that

<em>F</em> = (4500 kg) (11,250 km/h²)

Recall that Newtons (N) are measured as

1 N = 1 kg • m/s²

so we should convert everything accordingly:

11,250 km/h² = (11,250 km/h²) (1000 m/km) (1/3600 h/s)² ≈ 0.868 m/s²

Then the force is

<em>F</em> = (4500 kg) (0.868 m/s²) = 3906.25 N ≈ 3900 N

8 0
4 years ago
An insect 1.1 mm tall is placed 1.0 mm beyond the focal point of the objective lens of a compound microscope. The objective lens
Pavlova-9 [17]

Answer:

Explanation:

For image formation in objective lens

object distance u = 14 +1 = 15 mm

focal length f = 14 mm .

image distance v = ?

lens formula

\frac{1}{v} -\frac{1}{u} =\frac{1}{f}

Putting the values

\frac{1}{v} +\frac{1}{15} =\frac{1}{14}

v = 210 mm .

B )

magnification = v / u

= 210 / 15

= 14

size of image = 14 x 1.1 mm

= 15.4 mm

= 15 mm approx

C )

For final image to be at infinity , image produced by objective lens must fall at the focal point of eye piece . so objective lens's distance from the image formed by objective must be equal to focal length of eye piece that is 21 mm .

21 mm is the answer .

D )

overall magnification =

\frac{210}{15} \times \frac{D}{f_e}

D = 25 cm , f_e = focal length of eye piece

= 14 x 250 / 21

= 166.67

= 170 ( in two significant figures )

7 0
3 years ago
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