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3 years ago

A person speed walking down a street covers 20m in 5s. What is the speed of the individual? *

Physics

1 answer:

omeli [17]3 years ago

8 0

Answer:

4 m/s.

Explanation:

The following data were obtained from the question:

Distance travalled (d) = 20 m

Time (t) = 5 secs

Speed (S) =?

Speed is defined as the rate of change of distance moved with time. Mathematically, it is expressed as:

Speed (S) = Distance (d) /time (t)

S = d/t

With the above formula, we can easily calculate the speed of the individual as follow:

Distance travalled (d) = 20 m

Time (t) = 5 secs

Speed (S) =?

S = d/t

S = 20/5

S = 4 m/s

Therefore, the speed of the individual is 4 m/s

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A certain white dwarf star was once an average star like our Sun. But now it is in the last stage of its evolution and is the si

solmaris [256]

Answer:

4.384 * 10^13

Explanation:

Given the expression :

[(6.67 * 10^-11) * (1.99 * 10^30)] ÷ [(1.74*10^3)*(1.74*10^3)]

Applying the laws of indices

[(6.67 * 1.99) *10^(-11 + 30)] ÷ [(1.74 * 1.74) * 10^3+3]

13.2733 * 10^19 ÷ 3.0276 * 10^6

(13.2733 / 3.0276) * 10^(19 - 6)

4.3840996 * 10^13

= 4.384 * 10^13

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3 years ago

An instrument is thrown upward with a speed of 15 m/s on the surface of planet X where the acceleration due to gravity is 2.5 m/

Katen [24]

<h2>Answer: 12 s</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told <u>the instrument is thrown upward</u> from the surface, we will only use the equations related to the Y axis.

In this sense, the main movement equation in the Y axis is:

$y-y_{o}=V_{o}.t-\frac{1}{2}g.t^{2}$ (1)

Where:

$y$ is the instrument's final position

$y_{o}=0$ is the instrument's initial position

$V_{o}=15m/s$ is the instrument's initial velocity

$t$ is the time the parabolic movement lasts

$g=2.5\frac{m}{s^{2}}$ is the acceleration due to gravity at the surface of planet X.

As we know $y_{o}=0$ and $y=0$ when the object hits the ground, equation (1) is rewritten as:

$0=V_{o}.t-\frac{1}{2}g.t^{2}$ (2)

Finding $t$ :

$0=t(V_{o}-\frac{1}{2}g.t^{2})$ (3)

$t=\frac{2V_{o}}{g}$ (4)

$t=\frac{2(15m/s)}{2.5\frac{m}{s^{2}}}$ (5)

Finally:

$t=12s$

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3 years ago

2) A motorcycle is moving at a constant speed of 40 km/h. How long does it take the motorcycle to

docker41 [41]

Answer:

2 hours

Explanation:

The motorcycle travels 40 km per hour.

80km / 40km/h = 2 hours.

7 0

3 years ago

Read 2 more answers

It takes 23 hours 56 minutes and 4 seconds for the earth to make one revolution (mean sidereal day). What is the angular speed o

Dominik [7]

Answer:

Explanation:

Check attachment for solution

3 0

3 years ago

A 65 kg bungee jumper leaps from a bridge. She is tied to a bungee cord that is 12 m long when unstretched, and falls a total of

Nimfa-mama [501]

Answer:

(a) k = 30.33 N/m

(b) a = 9.8 m/s²

Explanation:

First, we need to find the force acting on the bungee jumper. Since, this is a free fall motion. Therefore, the force must be equal to the weight of jumper:

F = W = mg

F = (65 kg)(9.8 m/s²)

F = 637 N

(a)

Now applying Hooke's Law:

F = k Δx

where,

k = spring constant = ?

Δx = change in length of bungee cord = 33 m - 12 m = 21 m

Therefore,

637 N = k(21 m)

k = 637 N/21 m

<u></u>

(b)

Since, this is free fall motion. Thus, the maximum acceleration will be the acceleration due to gravity.

a = g

7 0

3 years ago