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Contact [7]
2 years ago
14

Speed of a train is 5m/s, find its value in km/hr.​

Physics
2 answers:
antiseptic1488 [7]2 years ago
4 0

Answer: 18km/hr

0.005km/s

1 sec --> 0.005

3600sec --> 0.005*3600 is 18

So, 18km/hr

Free_Kalibri [48]2 years ago
3 0

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

Here's the solution ~

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:5 \: m \: s {}^{ - 1}

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:\dfrac{5}{1000} km \:  {s}^{ - 1}

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:\dfrac{5}{1000} \times 3600 \:  km \:  {h}^{ - 1}

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \: \dfrac{5}{5}  \times 18 \: km \: h {}^{ - 1}

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \: 18 \: km \: h {}^{ - 1}

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Anyone know these questions?
salantis [7]
400m in 32sec: (400/32)>12.5meters per second>
(12.5)(60)(60)(1/1000)=45km per hour
Constant speed would mean that the two forces are equivalent
7 0
3 years ago
A listener increases his distance from a sound source by a factor of 4.49.
noname [10]

Answer: Δβ (dB) = -13.1dB

Explanation:

The intensity of sound is inversely proportional to the square of the distance between them.

I ∝ 1/r²

I₁/I₂= r₂²/r₁² .....1

When the listener increases his distance from the source by a factor of 4.49.

Then,

r₂/r₁= 4.49

From equation 1

I₁/I₂ = (4.49)²

I₁/I₂ = 20.16

I₂/I₁ = 1/20.16

The change in sound intensity in dB can be given as

Δβ (dB) = 10 log(I₂/l₁) = 10log(1/20.6) = -13.1dB

6 0
3 years ago
A bicycle with 0.80-m-diameter tires is coasting on a level road at 5.6 m/s . A small blue dot has been painted on the tread of
skelet666 [1.2K]

Answer:

a)

14 rad/s

b)

11.2 m/s

Explanation:

a)

d = diameter of tire = 0.80 m

r = radius of tire = (0.5) d = (0.5) (0.80) = 0.40 m

v = speed of bicycle = 5.6 m/s

w = angular speed of the tire

Speed of cycle is given as

v = r w

5.6 = (0.40) w

w = 14 rad/s

b)

v' = speed of blue dot

Speed blue of dot is given as

v' = v + rw

v' = 5.6 + (0.40) (14)

v' = 11.2 m/s

4 0
3 years ago
A small propeller airplane can comfortably achieve a high enough speed to take off on a runway that is 1/4 mile long. A large, f
QveST [7]

Answer:

1 mile

Explanation:

We can use the following equation of motion to solve for this problem:

v^2 - v_0^2 = 2a\Delta s

where v m/s is the final take-off velocity of the airplane, v_0 = 0 initial velocity of the can when it starts from rest, a is the acceleration of the airplanes, which are the same, and \Delta s is the distance traveled before takeoff, which is minimum runway length:

v^2 - 0^2 = 2a\Delta s

\Delta s = \frac{v^2}{2a}

From here we can calculate the distance ratio

\frac{\Delta s_1}{\Delta s_2} = \frac{v_1^2/2a_1}{v_2^2/2a_2}

\frac{\Delta s_1}{\Delta s_2} = \left(\frac{v_1}{v_2}\right)^2\frac{a_2}{a_1}

Since the 2nd airplane has the same acceleration but twice the velocity

\frac{\Delta s_1}{\Delta s_2} = 0.5^2* 1

\Delta s_2 = 4 \Delta s_1 = 4*(1/4) = 1 mile

So the minimum runway length is 1 mile

6 0
3 years ago
You grab the handle of the door in the picture, and pull as hard as you can. Unfortunately, this door should
chubhunter [2.5K]

Answer:

Explanation:

The solved solution is on the attach document

8 0
3 years ago
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