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3 years ago

A bike travels at a constant speed of 4.0m/s for 5 seconds. How far it goes ?

Physics

2 answers:

Y_Kistochka [10]3 years ago

7 0

Answer:

Distance, d = 20 meters

Explanation:

It is given that,

Speed of the bike, v = 4 m/s

Time taken, t = 5 sec

We need to find the distance covered by the bike during this time. The distance covered by bike is calculated by the product of speed and time

d = v × t

d = 4 × 5

d = 20 meters

So, the bike will go to a distance of 20 meters. Hence, this is the required solution.

alex41 [277]3 years ago

3 0

The speed of 4m/s means that the bike goes 4 meters for each second of movement. Multiplying this by 5 seconds we have a total distance travelled of 20m.

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sdas [7]

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3 years ago

The denizens of a distant planet live under an atmosphere made predominantly of CO2 which is a dispersive medium while attending

LuckyWell [14K]

Since the frequency of sound in a medium is constant, therefore, the concert-goers would hear the low notes and high notes at the same time.

<h3>What is a dispersive medium?</h3>

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Since CO2 is a dispersive medium, it means sound waves passing through it would be dispersed based on wavelength.

The note of a sound depends on its frequency, the higher the frequency, the higher the note.

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Learn more about dispersion of sound at: brainly.com/question/781734

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2 years ago

On a snowy day, max (mass = 15 kg) pulls his little sister maya in a sled (combined mass = 20 kg) through the slippery snow. max

sesenic [268]

Work done by a given force is given by

$W = F.d$

here on sled two forces will do work

1. Applied force by Max

2. Frictional force due to ground

Now by force diagram of sled we can see the angle of force and displacement

work done by Max = $W_1 = Fdcos\theta$

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$W_1 = 57.96 J$

Now similarly work done by frictional force

$W_2 = Fdcos\theta$

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3 years ago

A rope passes over a fixed sheave with both ends hanging straight down. The coefficient of friction between the rope and sheave

Oliga [24]

Answer:3.51

Explanation:

Given

Coefficient of Friction $\mu =0.4$

Consider a small element at an angle \theta having an angle of $d\theta$

Normal Force $=T\times \frac{d\theta }{2}+(T+dT)\cdot \frac{d\theta }{2}$

$N=T\cdot d\theta$

Friction $f=\mu \times Normal\ Reaction$

$f=\mu \cdot N$

and $T+dT-T=f=\mu Td\theta$

$dT=\mu Td\theta$

$\frac{dT}{T}=\mu d\theta$

$\int_{T_2}^{T_1}\frac{dT}{T}=\int_{0}^{\pi }\mu d\theta$

$\frac{T_2}{T_1}=e^{\mu \pi}$

$\frac{T_2}{T_1}=e^{0.4\times \pi }$

$\frac{T_2}{T_1}==e^{1.256}$

$\frac{T_2}{T_1}=3.51$

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3 years ago

In the high jump, the kinetic energy of an athlete is transformed into gravitational potential energy without the aid of a pole.

Fiesta28 [93]

Answer:

6.0 m/s

Explanation:

According to the law of conservation of energy, the total mechanical energy (potential, PE, + kinetic, KE) of the athlete must be conserved.

Therefore, we can write:

$KE_i+PE_i =KE_f+PE_f$

$\frac{1}{2}mu^2+0=\frac{1}{2}mv^2+mgh$

where:

m is the mass of the athlete

u is the initial speed of the athlete (at the bottom)

0 is the initial potential energy of the athlete (at the bottom)

v = 0.80 m/s is the final speed of the athlete (at the top)

$g=9.8 m/s^2$ is the acceleration due to gravity

h = 1.80 m is the final height of the athlete (at the top)

Solving the equation for u, we find the initial speed at which the athlete must jump:

$u=\sqrt{v^2+2gh}=\sqrt{0.80^2+2(9.8)(1.80)}=6.0 m/s$

4 0

3 years ago