An engine develops 500 Nm of torque at 3900 rev/min. The power developed by the engine is ...

A boxer punches a sheet of paper in midair and brings it from rest up to a speed of 30 m/s in 0.060 s .

zimovet [89]

Answer:

Force exerted, F = 1.5 N

Explanation:

It is given that, a boxer punches a sheet of paper in midair and brings it from rest up to a speed of 30 m/s in 0.060 s.

i.e. u = 0

v = 30 m/s

Time taken, t = 0.06 s

Mass of the paper, m = 0.003 kg

We need to find the force the boxer exert on it. The force can be calculated using second law of motion as :

$F=m\times a$

$F=m\times (\dfrac{v-u}{t})$

$F=0.003\times (\dfrac{30}{0.06})$

F = 1.5 N

So, the force the boxer exert on the paper is 1.5 N. Hence, this is the required solution.

6 0

3 years ago

To introduce you to the concept of escape velocity for a rocket. the escape velocity is defined to be the minimum speed with whi

Mama L [17]

A projectile fired upward from the Earth's surface will usually slow down, come momentarily to rest, and return to Earth. For a certain initial speed, however it will move upward forever, with its speed gradually decreasing to zero just as its distance from Earth approaches infinity. The initial speed for this case is called escape velocity. You can find the escape velocity v for the Earth or any other planet from which a projectile might be launched using conservation of energy. The projectile of mass m leaves the surface of the body of mass M and radius R with a kinetic energy Ki = mv²/2 and potential energy Ui = -GMm/R. When the projectile reaches infinity, it has zero potential energy and zero kinetic energy since we are seeking the minimum speed for escape. Thus Uf = 0 and Kf = 0. And from conservation of energy,
Ki + Ui = Kf + Uf
mv²/2 -GMm/R = 0
∴ v = √(2GM/R)

This is the expression for escape velocity.

3 0

3 years ago

Read 2 more answers

A 350-g mass is attached to a spring whose spring constant is 64 N/m. Its maximum acceleration is 5.3 m/s2. What is the frequenc

max2010maxim [7]

The frequency of oscillation is 2.153 Hz

What is the frequency of spring?

Spring Frequency is the natural frequency of spring with a weight at the lower end. Spring is fixed from the upper end and the lower end is free.

For the mass-spring system in this problem,

The Frequency of spring is calculated with the equation:

$f = \frac{1}{2\pi } \sqrt{\frac{k}{m} }$

Where,

f = frequency of spring

k = spring constant = 64 N/m

m = mass attached to spring = 350g = 0.350 kg

a = maximum acceleration = 5.3 m/s^2

Substituting the values in the equation,

$f = \frac{1}{2\pi } \sqrt{\frac{64}{0.350} }$

$f = \frac{1}{2\pi } ( 13.522)$

$f = 2.1535 Hz$

Hence,

The frequency of oscillation is 2.153 Hz

Learn more about frequency here:

<u>brainly.com/question/13978015</u>

#SPJ4

6 0

1 year ago

During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be lau

Luden [163]

10.8 seconds is the correct answ

3 0

3 years ago

he atomic radii of Li and O2- ions are 0.068 and 0.140 nm, respectively. (a) Calculate the force of attraction in newtons betwee

Crazy boy [7]

Answer:

a

$F = -1.07 *10^{-8} \ N$

b

$F_r = 1.07 *10^{-8} \ N$

Explanation:

Generally the force of attraction between this two irons is mathematically represented as

$F = \frac{k * [Q_{Li} ] * [Q_{O} ] }{ r^2}$

Here k is known as the proportionality constant with value $k = 2.31 * 10^ {-28} J \cdot m$

substituting -2 for $Q_{O}$ i.e the charge on oxygen , +1 for $Q_{Li}$ i.e the charge on Lithium and $[0.140 + 0.068 ] nm= 0.208 nm = 0.208*10^{-9}$ for r

So

$F = \frac{ 2.31 * 10^ {-28}* +1 * -2 }{ ( 0.208*10^{-9} )^2 }$

$F = -1.07 *10^{-8} \ N$

Generally the force of repulsion will be the magnitude but different direction to the force o attraction

So Force of repulsionn is

$F_r = 1.07 *10^{-8} \ N$

6 0

3 years ago