All categories

2 years ago

A bicycle is heading West. It goes 5000m in 500s what's its velocity

1 answer:

8 0

There are several information's of immense importance already given in the question. Based on the given information's the answer to the question can easily be determined.
Distance covered by the bicycle = 5000 meter
Time taken by the bicycle to reach the distance = 500 second.
Velocity of the bicycle = Distance / Time taken
= 5000/500 meter/second
= 50 meter/second
So the velocity of the bicycle is 50 meter per second. I hope the procedure is clear enough for you to understand. In future you can always use this procedure for solving similar problems.

You might be interested in

A bird sits on top of a 639 m tall tower. If it's gravitational potential energy up there is 2033 J, what is its mass?

iris [78.8K]

The mass of the bird is 0.32 kg.

Explanation:

Gravitational potential energy, the energy exhibited by an object at rest due to the influence of gravitational force. So the increase in distance of object from the surface of earth leads to increase in the gravitational potential energy. Thus,

$\text {Gravitational potential energy}=m \times \text { Acceleration } \times \text { Distance of bird from bottom }$

So, as the gravitational potential energy is given as 2033 J and the position of bird placed on the tall tower is 639 m away from the bottom, then the mass (m) of the bird can be found as below.

$m o f \text { bird }=\frac{\text {Gravitational potential energy}}{a \times \text {Distance}}=\frac{2033}{9.8 \times 639}=\frac{2033}{6262.2}$

So, finally we get the bird's mass as,

m of bird = 0.32 kg

7 0

3 years ago

A mass on a spring with k=88.7 N/m oscillates 15 times in 9.24s. what is the objects mass? unit=kg?

sweet [91]

The mass on the spring is 0.86 kg

Explanation:

The period of a mass-spring system is given by the equation

$T=2\pi \sqrt{\frac{m}{k}}$

where

m is the mass

k is the spring constant

In this problem, we have:

k = 88.7 N/m is the spring constant

The system makes 15 oscillations in 9.24 s: therefore, the period of the system is

$T=\frac{9.24}{15}=0.62 s$

Now we can re-arrange the first equation to solve for the mass:

$m=k(\frac{T}{2\pi})^2=(88.7)(\frac{0.62}{2\pi})^2=0.86 kg$

Learn more about period:

brainly.com/question/5438962

#LearnwithBrainly

3 0

2 years ago

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

vlada-n [284]

Answer:

$\frac{1}{10}$ M

Explanation:

To apply the concept of angular momentum conservation, there should be no external torque before and after

As the asteroid is travelling directly towards the center of the Earth, after impact ,it does not impose any torque on earth's rotation, So angular momentum of earth is conserved

⇒ $I_{1} \times W_{1} =I_{2} \times W_{2}$

$I_{1}$ is the moment of interia of earth before impact
$W_{1}$ is the angular velocity of earth about an axis passing through the center of earth before impact
$I_{2}$ is moment of interia of earth and asteroid system
$W_{2}$ is the angular velocity of earth and asteroid system about the same axis

let $W_{1}=W$

since $\text{Time period of rotation}∝$ $\frac{1}{\text{Angular velocity}}$

⇒ if time period is to increase by 25%, which is $\frac{5}{4}$ times, the angular velocity decreases 25% which is $\frac{4}{5}$ times

therefore $W_{1}$ $=$ $\frac{4}{5} \times W_{1}$

$I_{1}=\frac{2}{5} \times M\times R^{2}$ (moment of inertia of solid sphere)

where M is mass of earth

R is radius of earth

$I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}$

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where $M_{1}$ is mass of asteroid