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3 years ago

Which equation describes the fastest runner?

Physics

2 answers:

fomenos3 years ago

8 0

The equation distance $\bold{=2 \times \text { time }}$ describes the fastest runner.

Answer: Option D

Explanation:

The fastest runner is determined by how quick a person can reach the destination (i.e) in a less amount of time compared to others. Consider it took 10 seconds to the reach the destination x. Substitute time = 10 seconds,

$0.5 \times 10=5$

$0.33 \times 10=3.3$

$0 \times 10=0$

$2 \times 10=20$

From the four options, the highest distance travelled for constant time is distance $\bold{=2 \times \text {time}}$ .

irina [24]3 years ago

7 0

I think the answer will be A

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10. At point A, a runner is jogging 5 m/s. Forty seconds later, at point B

kondaur [170]

Answer:

Deceleration of 0.075m/s²

Explanation:

Given parameters:

Initial velocity = 5m/s

Time = 40s

Final velocity = 2m/s

Unknown:

The jogger's acceleration = ?

Solution:

To solve this problem, we use;

a = $\frac{v - u}{t}$

a is the acceleration

v is the final velocity

u is the initial velocity

t is the time taken

a = $\frac{2 - 5}{40}$ = - 0.075m/s²

5 0

3 years ago

Any help is appreciated

sdas [7]

image distance,di=10 cm

object distance,do=20cm

magnification, m=di/do

=10/20

=0.5

since the image is virtual, magnification is negative.

therefore m=-0.5

8 0

3 years ago

How much voltage (in terms of the power source voltage bV) will the capacitor have when it has started at zero volts potential d

Archy [21]

Answer:

The voltage is $V = 0.993V_b$

Explanation:

From the question we are told that

The time that has passed is $t = \frac{\tau}{2}$

Here $\tau$ is know as the time constant

The voltage of the power source is $V_b$

Generally the voltage equation for charging a capacitor is mathematically represented as

$V = V_b [1 - e^{- \frac{t}{\tau} }]$

=> $V = V_b [1 - e^{- \frac{\frac{\tau}{2}}{\tau} }]$

=> $V = V_b [1 - e^{- \frac{\tau}{2\tau} }]$

=> $V = V_b [1 - e^{- \frac{1}{2} }]$

=> $V = 0.993V_b$

5 0

3 years ago

What total mass must be converted into energy

Eduardwww [97]

This question apparently wants you to get comfortable
with E = m c² . But I must say, this question is a lame
way to do it.

c = 3 x 10⁸ m/s
E = m c²

 1.03 x 10⁻¹³ joule = (m) (3 x 10⁸ m/s)²

Divide each side by (3 x 10⁸ m/s)²:

 Mass = (1.03 x 10⁻¹³ joule) / (9 x 10¹⁶ m²/s²)

 = (1.03 / 9) x (10⁻¹³ ⁻ ¹⁶) (kg)

 = 1.144 x 10⁻³⁰ kg . (choice-1)

This is roughly the mass of (1 and 1/4) electrons, so it seems
that it could never happen in nature. The question is just an
exercise in arithmetic, and not a particularly interesting one.
______________________________________

Something like this could have been much more impressive:

The Braidwood Nuclear Power Generating Station in northeastern
Ilinois USA serves Chicago and northern Illinois with electricity.
The station has two pressurized water reactors, which can generate
a net total of 2,242 megawatts at full capacity, making it the largest
nuclear plant in the state.
If the Braidwood plant were able to completely convert mass
to energy, how much mass would it need to convert in order
to provide the total electrical energy that it generates in a year,
operating at full capacity ?

Energy = (2,242 x 10⁶ joule/sec) x (86,400 sec/day) x (365 da/yr)

 = (2,242 x 10⁶ x 86,400 x 365) joules

 = 7.0704 x 10¹⁶ joules .

How much converted mass is that ?

 E = m c²

Divide each side by c² : Mass = E / c² .
c = 3 x 10⁸ m/s

 Mass = (7.0704 x 10¹⁶ joules) / (9 x 10¹⁶ m²/s²)

 = 0.786 kilogram ! ! !

THAT should impress us ! If I've done the arithmetic correctly,
then roughly (1 pound 11.7 ounces) of mass, if completely
converted to energy, would provide all the energy generated
by the largest nuclear power plant in Illinois, operating at max
capacity for a year !

7 0

3 years ago

Read 2 more answers

FREE BRAINIEST IF YOU ANSWER THIS

Inessa [10]

The watt is a rate, similar to something like speed (miles per hour) and other time-interval related measurements.

Specifically, watt means Joules per Second. We are given that the electrical engine has 400 watts, meaning it can make 400 joules per second. If we need 300 kJ, or 3000 Joules, then we can write an equation to solve the time it would take to reach this amount of joules:

w * t = E

w: Watts

t: Time

E: Energy required

(Watts times time is equal to the energy required)

Input our values:

400 * t = 3000

(We need to write 3000 joules instead of 300 kilojoules, since Watts is in joules per second. It's important to make sure your units are consistent in your equations)

Divide both sides by 400 to isolate t:

$\frac{400t}{400}$ = $\frac{3000}{400}$

t = 7.5 (s)

It will take 7.5 seconds for the 400 W engine to produce 300 kJ of work.

If you have any questions on how I got to the answer, just ask!

- breezyツ

6 0

3 years ago