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3 years ago

A dog runs after a car. The car is travelling at an average speed of 5 m/s.

Physics

2 answers:

n200080 [17]3 years ago

3 0

Answer:

<h2>The answer is 4 s</h2>

Explanation:

The time taken can be found by using the formula

$t = \frac{d}{v} \\$

where

d is the distance

v is the velocity

From the question we have

$t = \frac{20}{5} \\$

We have the final answer as

Hope this helps you

Lina20 [59]3 years ago

3 0

Answer:

4 secs

Explanation:

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Two airplanes leave an airport at the same time. The velocity of the first airplane is 730 m/h at a heading of 65.3 ◦ . The velo

yanalaym [24]

Answer:

Plane will 741.6959 m apart after 1.7 hour

Explanation:

We have given time = 1.7 hr

So if we draw the vectors of a 2d graph we see that the difference in angles is = $102^{\circ}-65.3^{\circ}=36.7^{\circ}$

Speed of first plane = 730 m/h

So distance traveled by first plane = 730×1.7 = 1241 m

Speed of second plane = 590 m/hr

So distance traveled by second plane = 590×1.7 = 1003 m

We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 58.6.

Using the law of cosine, $r^2$ representing the distance between the planes, we see that:

$r^2=1241^2+1003^2-2\times 1003\times 1241cos(36.7)=550112.8295$

r = 741.6959 m

3 0

3 years ago

A reconnaissance plane flies 560 km away from its base at 602 m/s, then flies back to its base at 903 m/s.

IrinaVladis [17]

Answer:

Approximately $722\; \rm m\cdot s^{-1}$ .

Explanation:

The average speed of a vehicle is calculated as:

$\displaystyle \text{average speed} = \frac{\text{total distance}}{\text{total time}}$ .

In this question, the total distance is $2 \times 560\; \rm km = 1120\; \rm km$ .

The unit of the speeds in this question is meters per second, while the unit of distance is kilometers. Convert the unit of distance to meters:

$560 \; \rm km = 560 \times 10^{3} \; \rm m = 5.6 \times 10^{5}\; \rm m$ .

$1120 \; \rm km = 1120 \times 10^{3} \; \rm m = 1.12 \times 10^{6}\; \rm m$ .

Time required for the first part of this trip:

$\displaystyle \frac{5.60 \times 10^{5}\; \rm m}{602\; \rm m\cdot s^{-1}} \approx 930\; \rm s$ .

Time required for the second part of this trip:

$\displaystyle \frac{5.60 \times 10^{5}\; \rm m}{903\; \rm m\cdot s^{-1}} \approx 620\; \rm s$ .

The time required for the entire trip would be approximately $930 + 620 = 1550\; \rm s$ .

Calculate the average speed of this plane:

$\begin{aligned} \text{average speed} &= \frac{\text{total distance}}{\text{total time}} \\ &\approx \frac{1.12\times 10^{6}\; \rm m}{1550\; \rm s} \approx 722\; \rm m \cdot s^{-1}\end{aligned}$ .

6 0

2 years ago

Which waves have wavelengths longer than those of visible light? Give an example of how each kind of wave is used.

Kazeer [188]

1. Radio Waves
ex. Wi-Fi
2. Microwaves
ex. Mobile Phones
3. Infrared Radiation
ex. Heat Lamps

6 0

2 years ago

Read 2 more answers

A 2-kg wheel rolls down the road with a linear speed of 15m/s. Find its trwansitional and rotational kinetic energies.

Elza [17]

Answer:

The translational kinetic energy is 225 J

The rotational kinetic energy is 225 J

Explanation:

Given;

mass of the wheel, m = 2-kg

linear speed of the wheel, v = 15 m/s

Transnational kinetic energy is calculated as;

E = ¹/₂MV²

where;

M is mass of the moving object

V is the velocity of the object

E = ¹/₂ x 2 x (15)²

E = 225 J

Rotational kinetic energy is calculated as;

E = ¹/₂Iω²

where;

I is moment of inertia

ω is angular velocity

$E = \frac{1}{2} I \omega^2\\\\E = \frac{1}{2} *mr^2*(\frac{v}{r})^2\\\\E = \frac{1}{2} *mr^2*\frac{v^2}{r^2} \\\\E = \frac{1}{2}mv^2$

E = ¹/₂ x 2 x (15)²

E = 225 J

Thus, the translational kinetic energy is equal to rotational kinetic energy

6 0

3 years ago

A 70 kg base runner begins his slide into second base while moving at a speed of 4.35 m/s. He slides so that his speed is zero j

sladkih [1.3K]

Answer

given,

Mass of the runner, M = 70 Kg

speed of the runner on the second base = 4.35 m/s

speed at the base = 0 m/s

Acceleration due to gravity,g = 9.8 m/s²

a) magnitude of mechanical energy lost

Mechanical energy lost is equal top gain in kinetic energy

$ME_{Lost}=\dfrac{1}{2}mv^2$

$ME_{Lost}=\dfrac{1}{2}\times 70\times 4.35^2$

$ME_{Lost}=662.29\ J$

b) Work done = Force x displacement

W = F. x

F = μ mg

W = μ mg . x

Work done is equal to 662.29 J

$x=\dfrac{W}{\mu m g}$

using the coefficient of the friction,μ = 0.7

$x=\dfrac{662.29}{ 0.7\times 70\times 9.8}$

x = 1.38 m

Hence, the runner will slide to 1.38 m.

3 0

3 years ago