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Darina [25.2K]
3 years ago
10

How can you tell when an object changes direction on a displacement vs. time graph

Physics
1 answer:
adoni [48]3 years ago
7 0

Answer:

You can see it looking at the slope of the curve.

Explanation:

The slope of the curve its displacement / time, so its the celerity of the object. Now, when the object change direction of its movement, its must change the direction on which its moving, by definition. So, if the object is first getting away from its starting position, you will see its displacement grow, and the slope will be positive (or negative, it depends of the direction you put an positive displacement) Now, when its change position, the displacement will start to shrink, and the slope of the curve will be negative (or positive). To find out the exact moment in which the object changes direction, you can just look at when the slope its equal to zero. If the slope goes from positive to negative, or viceversa, it must pass by zero, the exact momento the object change direction.

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Calculate the Schwarzschild radius (in kilometers) for each of the following.1.) A 1 ×108MSun black hole in the center of a quas
Westkost [7]

Answer:

(I). The Schwarzschild radius is 2.94\times10^{8}\ km

(II). The Schwarzschild radius is 17.7 km.

(III). The Schwarzschild radius is 1.1\times10^{-7}\ km

(IV). The Schwarzschild radius is 7.4\times10^{-29}\ km

Explanation:

Given that,

Mass of black hole m= 1\times10^{8} M_{sun}

(I). We need to calculate the Schwarzschild radius

Using formula of radius

R_{g}=\dfrac{2MG}{c^2}

Where, G = gravitational constant

M = mass

c = speed of light

Put the value into the formula

R_{g}=\dfrac{2\times6.67\times10^{-11}\times1\times10^{8}\times1.989\times10^{30}}{(3\times10^{8})^2}

R_{g}=2.94\times10^{8}\ km

(II). Mass of block hole m= 6 M_{sun}

We need to calculate the Schwarzschild radius

Using formula of radius

R_{g}=\dfrac{2MG}{c^2}

Put the value into the formula

R_{g}=\dfrac{2\times6.67\times10^{-11}\times6\times1.989\times10^{30}}{(3\times10^{8})^2}

R_{g}=17.7\ km

(III). Mass of block hole m= mass of moon

We need to calculate the Schwarzschild radius

Using formula of radius

R_{g}=\dfrac{2MG}{c^2}

Put the value into the formula

R_{g}=\dfrac{2\times6.67\times10^{-11}\times7.35\times10^{22}}{(3\times10^{8})^2}

R_{g}=1.1\times10^{-7}\ km

(IV). Mass = 50 kg

We need to calculate the Schwarzschild radius

Using formula of radius

R_{g}=\dfrac{2MG}{c^2}

Put the value into the formula

R_{g}=\dfrac{2\times6.67\times10^{-11}\times50}{(3\times10^{8})^2}

R_{g}=7.4\times10^{-29}\ km

Hence, (I). The Schwarzschild radius is 2.94\times10^{8}\ km

(II). The Schwarzschild radius is 17.7 km.

(III). The Schwarzschild radius is 1.1\times10^{-7}\ km

(IV). The Schwarzschild radius is 7.4\times10^{-29}\ km

8 0
3 years ago
The seven sciences the study of kinesiology includes are: exercise physiology, exercise anatomy, biomechanics, exercise psycholo
Alex

Answer:

True

Explanation:

5 0
3 years ago
The electric potential at the surface of a charged conductor _______.
frez [133]

Answer:

The electric potential at the surface of a charged conductor<u> is always such that the potential is zero at all points inside the conductor.</u>

Explanation:

Each point on the surface of a balanced charged conductor has the same electrical potential.

The surface on any charged conductor in electrostatic equilibrium is an equipotential surface. Since the electric field is equal to zero inside the conductor, the electric potential at any point inside and on the surface is equivalent to its value.

6 0
3 years ago
For a caffeinated drink with a caffeine mass percent of 0.65% and a density of 1.00 g/mL, how many mL of the drink would be requ
slava [35]

Explanation:

First we will convert the given mass from lb to kg as follows.

        157 lb = 157 lb \times \frac{1 kg}{2.2046 lb}

                   = 71.215 kg

Now, mass of caffeine required for a person of that mass at the LD50 is as follows.

         180 \frac{mg}{kg} \times 71.215 kg

         = 12818.7 mg

Convert the % of (w/w) into % (w/v) as follows.

      0.65% (w/w) = \frac{0.65 g}{100 g}

                           = \frac{0.65 g}{(\frac{100 g}{1.0 g/ml})}

                           = \frac{0.65 g}{100 ml}

Therefore, calculate the volume which contains the amount of caffeine as follows.

   12818.7 mg = 12.8187 g = \frac{12.8187 g}{\frac{0.65 g}{100 ml}}

                       = 1972 ml

Thus, we can conclude that 1972 ml of the drink would be required to reach an LD50 of 180 mg/kg body mass if the person weighed 157 lb.

5 0
3 years ago
Why would an object float in water, but sink in rubbing alcohol?
Romashka [77]

Because water is more dense than the object but rubbing alcohol is less dense than the object

6 0
3 years ago
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