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ICE Princess25 [194]
4 years ago
8

The auto in the sketch moves forward as the brakes are applied. A bystander says that during the interval of braking, the auto's

velocity and acceleration are in opposite directions. Do you agree or disagree?
Physics
1 answer:
Ivan4 years ago
3 0

Answer:

The statement is true: velocity and acceleration have opposite directions in the interval of braking.

Explanation:

Let's say we have a velocity v>0.

The acceleration a is the rate of change of the velocity v. This means that if v is <em>increasing during</em> time, then a must be positive. But if v is <em>decreasing over</em> time, then a will be negative (even though the velocity is positive).

Mathematically:

a=\frac{dv}{dt}

v decreases ⇒\frac{dv}{dt}

⇒a.

Example:

v(t)=e^{-t}>0 \\\\\frac{dv}{dt}=-te^{-t}

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Limewater can be used to detect carbon dioxide. If carbon dioxide is bubbled through limewater then it turns from clear to cloudy/milky in colour. This is why limewater used in a simple respirometer can show that more carbon dioxide is present in exhaled air compared to inhaled air.

Explanation:

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675 km equals how much cm ?
irga5000 [103]
KHDMDCM.
Now go from Kilometer to Centimeter: 5.
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7 0
4 years ago
The left end of a long glass rod 8.00 cm in diameter, with an index of refraction 1.60, is ground to a concave hemispherical sur
sp2606 [1]

Answer:

a) q = -9.23 cm, b)  h’= 0.577 mm , c) image is right and virtual

Explanation:

This is an optical exercise, where the constructor equation should be used

        1 / f = 1 / p + 1 / q

Where f is the focal length, p the distance to the object and q the distance to the image

A) The cocal distance is framed with the relationship

       1 / f = (n₂-1) (1 /R₁ -1 /R₂)

In this case we have a rod whereby the first surface is flat R1 =∞ and the second surface R2 = -4 cm, the sign is for being concave

       1 / f = (1.60 -1) (1 /∞ - 1 / (-4))

       1 / f = 0.6 / 4 = 0.15

        f = 6.67 cm

We have the distance to the object p = 24.0 cm, let's calculate

       1 / q = 1 / f - 1 / p

       1 / q = 1 / 6.67 - 1/24

       1 / q = 0.15 - 0.04167 = 0.10833

       q = -9.23 cm

distance to the negative image is before the lens

B) the magnification of the lenses is given by

       M = h ’/ h = - q / p

        h’= - q / p h

        h’= - (-9.23) / 24.0 0.150

        h’= 0.05759 cm

        h’= 0.577 mm

C) the object is after the focal length, therefore, the image is right and virtual

6 0
3 years ago
Can someone help me with these questions please. I will mark brainliest
VladimirAG [237]

Answer:

3: I can´t see the text/image, but it depend on the mass and the force applied to the ball, if both are too high, it will be harder to make a home run. (Second law)

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Explanation:

3 0
3 years ago
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