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3 years ago

How long will it take for a man to walk a distance of 2 km if his walking speed is 2 m/s?

Physics

2 answers:

laila [671]3 years ago

8 0

Answer:

1000s

Explanation:

because as 2km means 2000m then if the distance is divided my speed the answe comes 1000s

Alexxx [7]3 years ago

3 0

Answer:

1000 seconds

Explanation:

2 m/s is equivalent to 0.002 km/s. If you want to find how long it will take to go 2 km, you just take 2 and divide by your rate, which is 0.002. 2/0.002=1000

You might be interested in

A cyclist moves at a constant speed of 5 m/s if the cyclist does not accelerate during the next 20 seconds he will travel at?

Verizon [17]

5 m/s because the speed is constant

8 0

4 years ago

the mass of a high speed train is 4.5 ×× 105 kgkg , and it is traveling forward at a velocity of 8.3 ×× 101 m/sm/s . given that

svlad2 [7]

1. The value of m is 3.735

2. The value of n is 7

<h3>What is momentum? </h3>

Momentum is defined as the product of mass and velocity. It is expressed as

Momentum = mass × velocity

<h3>How to determine the value of m and n</h3>

We can obtain the value of m and n by simply obtaining the momentum in scientific notation. This is illustrated below:

Mass of train = 4.5×10⁵ Kg
Velocity of train = 8.3×10¹ m/s
Momentum =?

Momentum = mass × velocity

Momentum = 4.5×10⁵ × 8.3×10¹

Momentum = 3.735×10⁷ Kg⋅m/s

Thus, the value of m and n are: 3.735 and 7

Learn more about momentum:

brainly.com/question/250648

#SPJ1

4 0

1 year ago

A 2.0-kg projectile is fired with initial velocity components v0x = 30 m/s and v0y = 40 m/s from a point on the Earth's surface.

Alexxandr [17]

Answer:

Kinetic energy of the projectile at the vertex of the trajectory: $900\; {\rm J}$ .

Work done when firing this projectile: $2500\; {\rm J}$ .

Explanation:

Since the drag on this projectile is negligible, the horizontal velocity $v_{x}$ of this projectile would stay the same (at $30\; {\rm m\cdot s^{-1}}$ ) throughout the flight.

The vertical velocity $v_{y}$ of this projectile would be $0\; {\rm m\cdot s^{-1}}$ at the vertex (highest point) of its trajectory. (Otherwise, if $v_{y} > 0$ , this projectile would continue moving up and reach an even higher point. If $v_{y} < 0$ , the projectile would be moving downwards, meaning that its previous location was higher than the current one.)

Overall, the velocity of this projectile would be $v = 30\; {\rm m\cdot s^{-1}}\!$ when it is at the top of the trajectory. The kinetic energy $\text{KE}$ of this projectile (mass $m = 2.0\; {\rm kg}$ ) at the vertex of its trajectory would be:

$\begin{aligned} \text{KE} &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2} \times 2.0\; {\rm kg} \times (30\; {\rm m\cdot s^{-1}})^{2} \\ &= 900\; {\rm J} \end{aligned}$ .

Apply the Pythagorean Theorem to find the initial speed of this projectile:

$\begin{aligned}v &= \sqrt{(v_{x})^{2} + (v_{y})^{2}} \\ &= \left(\sqrt{900 + 1600}\right)\; {\rm m\cdot s^{-1}} \\ &= 50\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Hence, the initial kinetic energy $\text{KE}$ of this projectile would be:

$\begin{aligned} \text{KE} &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2} \times 2.0\; {\rm kg} \times (50\; {\rm m\cdot s^{-1}})^{2} \\ &=2500\; {\rm J} \end{aligned}$ .

All that energy was from the work done in launching this projectile. Hence, the (useful) work done in launching this projectile would be $2500\; {\rm J}$ .

7 0

2 years ago

What changes occurred when you heated sugar ?<br><br><br> Please help <br><br> Test

nirvana33 [79]

When sugar is first heated, it begins to melt. As it melts its color starts to change from white to golden brown, and then to dark brown (if you continue heating it). If you apply heat for even longer, it becomes black and gives off unpleasant fumes.

8 0

3 years ago

You were driving your car to UTD at a speed of 35 miles per hour. You stopped at the FloydCampbell intersection with the signal

ser-zykov [4K]

Answer:

green light have high energy

Explanation:

We have given the wavelength of the red light $\lambda =6.45\times 10^{-5}cm=6.45\times 10^{-7}m$

Speed of the light $c=3\times 106{8}m/sec$

The energy of the signal is given by $E=h\nu =h\frac{c}{\lambda }=\frac{6.67\times 10^{-34}\times 3\times 10^{8}}{6.45\times 10^{-7}}=3.1023\times 10^{-15}j$

The frequency of the green light is given by:

$f=5.80\times 10^{14}s^{-1}$

So energy $E=h\nu =6.67\times 10^{-34}\times 5.80\times 10^{14}=3.8686\times 10^{-19}j$

So green light have high energy

8 0

3 years ago