All categories

3 years ago

Calculate the force it would take to accelerate a 250 kg motorcycle at a rate of 5m/s^2

Physics

1 answer:

bija089 [108]3 years ago

5 0

1250 N of force is required to accelerate a 250 kg motorcycle at a rate of 5 m/ $s^{2}$ .

<u>Explanation:</u>

As per the second law of motion stated by Newton, the net unbalanced external force acting on any object is directly proportional to the product of mass and acceleration experienced by the object.

As here the object is given as motorcycle of mass 250 kg. Also the acceleration exhibited by the motorcycle is stated as 5 m/ $s^{2}$ . So from the Newton's second law of motion,

Net force = Mass of the motorcycle× Acceleration of the motorcycle

Net force = 250 × 5 = 1250 N.

Thus, 1250 N of force is required to accelerate a 250 kg motorcycle at a rate of 5 m/ $s^{2}$ .

You might be interested in

An object begins with a speed of 20 meters per second and slows down to a speed of 10 meters per second over a time of 5 seconds

Andreyy89

Answer:

V = 6 m/s

Explanation:

Given that,

Initial speed of an object is 20 m/s

Final speed of an object is 10 m/s

Time, t = 5 s

We need to find the average speed of the object during these 5 seconds. Let it is equal to V. Here, time is same. The average speed is given by :

$V=\dfrac{x+y}{2}\\\\\text{Putting values}\\\\V=\dfrac{20+10}{5}\\\\V=6\ m/s$

So, the average speed of the object is 6 m/s.

3 0

3 years ago

An open organ pipe is 2.46 m long, and the speed of the air in the pipe is 345 m/s.

lukranit [14]

Answer:

Fundamental frequency is 70.12 m

Explanation:

For an open organ pipe, the fundamental frequency is given by :

$f=\dfrac{nv}{2l}$

n = 1 for fundamental frequency

v is speed of sound in air, v = 345 m/s

l is length of open organ pipe, l = 2.46 m

Substituting values in above formula. So,

$f=\dfrac{1\times 345}{2\times 2.46}\\\\f=70.12\ Hz$

So, the fundamental frequency of this pipe is 70.12 m.

6 0

3 years ago

a projectile leaves the ground with a velocity of 35 meters per second at an angle 32° what is the maximum height

Nina [5.8K]

To determine the maximum height for the projectile, we can use the timeless kinematics equation $V _{f} ^{2} -V_{i}^{2}=2a \Delta y$ and plug in what we know. For $V _{f} ^{2}$ , we know it is zero because at the highest location of flight, there is no vertical velocity. For $V _{i} ^{2}$ , we know it is $(35sin(32))^{2}$ since the vertical component is the sine of the velocity. We need to solve for $y$ , so it is left as-is. Since the only acceleration is $g$ , we can substitute -9.81 into it. Solving for the equation yields a solution of 17.55 m for $\Delta y$ .