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gavmur [86]
3 years ago
15

Suppose a car travels at a constant speed of 12.0 m/s. How far would it move in 3600.0 s?

Physics
1 answer:
beks73 [17]3 years ago
7 0

Answer:

43200 m

Explanation:

speed = 12.0 m/s

time = 3600.0 s

distance = speed * time

distance = 12.0 m/s * 3600.0 s

distance = 43200 m

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MATCH THE NUMBER WITH THE APPROPRIATE VARIABLE <br> HELPP!!!!! its for my final
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Use the diagram below to answer the following question:
d1i1m1o1n [39]

Answer:

3.0 cm

Explanation:

We can solve this problem by using the mirror equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length of the mirror

p is the distance of the object from the mirror

q is the distance of the image from the mirror

In this problem we have:

f = 1.5 cm is the focal length of the mirror (positive for a concave mirror)

p = 3.0 cm is the distance of the object from the mirror

Therefore, the distance of the image is:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{1.5}-\frac{1}{3.0}=\frac{1}{3.0}\\\rightarrow q=3.0 cm

And the positive sign means that the image is real.

(The second part of the exercise is just the description of the image of the first exercise).

5 0
3 years ago
The Southwest Indian Ridge, shown in red, moves at an average rate of 20 mm/year, making it among the ultraslow spreading ridges
blagie [28]

v = average speed of movement of the Southwest Indian Ridge = 20 mm/year

d = distance moved by the Southwest Indian Ridge = 100 mm

t = number of years required to move distance "d"

distance traveled is given as

d = v t

inserting the above values in the formula

100 mm = (20 mm/year) t

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7 0
3 years ago
Read 2 more answers
A cannon, located 60.0 m from the base of a vertical 25.0-m-tall cliff, shoots a 15-kg shell at 43.0o above the horizontal towar
Artist 52 [7]

Answer:

a)   v₀ = 32.64 m / s , b)  x = 59.68 m

Explanation:

a) This is a projectile launching exercise, we the distance and height of the cliff

         x = v₀ₓ t

         y = v_{oy} t - ½ g t²

We look for the components of speed with trigonometry

         sin 43 = v_{oy} / v₀

         cos 43 = v₀ₓ / v₀

         v_{oy} = v₀ sin 43

         v₀ₓ = v₀ cos 43

Let's look for time in the first equation and substitute in the second

         t = x / v₀ cos 43

         y = v₀ sin 43 (x / v₀ cos 43) - ½ g (x / v₀ cos 43)²

          y = x tan 43 - ½ g x² / v₀² cos² 43

          1 / v₀² = (x tan 43 - y) 2 cos² 43 / g x²

           v₀² = g x² / [(x tan 43 –y) 2 cos² 43]

Let's calculate

          v₀² = 9.8 60 2 / [(60 tan 43 - 25) 2 cos 43]

          v₀ = √ (35280 / 33.11)

          v₀ = 32.64 m / s

.b) we use the vertical distance equation with the speed found

         y = v_{oy} t - ½ g t²

         .y = v₀ sin43 t - ½ g t²

        25 = 32.64 sin 43 t - ½ 9.8 t²

        4.9 t² - 22.26 t + 25 = 0

         t² - 4.54 t + 5.10 = 0

We solve the second degree equation

         t = (4.54 ±√(4.54 2 - 4 5.1)) / 2

         t = (4.54 ± 0.46) / 2

         t₁ = 2.50 s

         t₂ = 2.04 s

The shortest time is when the cliff passes and the longest when it reaches the floor, with this time we look for the horizontal distance traveled

         x = v₀ₓ t

         x = v₀ cos 43 t

         x = 32.64 cos 43  2.50

         x = 59.68 m

8 0
3 years ago
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