The period for which an organism lives is called its ______

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

A varying force is given by F=Ae ^-kx, where x is the position;A and I are constants that have units of N and m^-1 , respectivel

Burka [1]

W = ∫ (x from 0.1 to +oo) F dx

= ∫ (x from 0.1 to +oo) A e^(-kx) dx

= A/k x [ - e^(-kx) ](between 0.1 and +oo)

= A/k x [ 0 + e^(-k * 0.1) ]

<span> = A/k x e^(-k/10) </span>

4 0

3 years ago

A van starts off 152 miles directly north from the city of Springfield. It travels due east at a speed of 25 miles per hour. Aft

erastovalidia [21]

Answer:

12.84 miles per hour

Explanation:

Given:

Vertical distance of starting point of van from Springfield (d) = 152 miles

Speed in east direction (s) = 25 mph

Distance traveled in east direction (e) = 91 miles

Let the direct distance from Springfield of the van be 'x' at any time 't'.

Now, from the question, it is clear that, the vertical distance of van is fixed at 152 miles and only the horizontal distance is changing with time.

Now, consider a right angled triangle SNE representing the given situation.

Point S represents Springfield, N represents the starting point of van and E represents the position of van at any time 't'.

SN = d = 152 miles (fixed)

Now, using the pythagorean theorem, we have:

$SE^2=SN^2+NE^2\\\\x^2=d^2+e^2\\\\x^2=(152)^2+e^2----(1)$

Now, differentiating both sides with respect to time 't', we get:

$2x\frac{dx}{dt}=0+2e\frac{de}{dt}\\\\\frac{dx}{dt}=\frac{e}{x}\frac{de}{dt}$

Now, we are given speed as 25 mph. So, $\frac{de}{dt}=25\ mph$

Also, when $e=91\ mi$ , we can find 'x' using equation (1). This gives,

$x^2=23104+(91)^2\\\\x=\sqrt{31385}=177.16\ mi$

Now, plug in the values of 'e' and 'x' and solve for $\frac{dx}{dt}$ . This gives,

$\frac{dx}{dt}=\frac{91}{177.16}\times 25\\\\\frac{dx}{dt}=12.84\ mph$

Therefore, the distance between the van and Springfield is changing at a rate of 12.84 miles per hour

6 0

3 years ago

A 6 kilogram block in outer space is moving at -100 m/s (to the left). It suddenly experiences three forces as shown below.

Alika [10]

Newton's second law and the kinematic relations allow to find the results for the questions about forces and the movement of the block are:

B) the force applied to maintain the system is equilibrium is: F = 0.39 N with an angle of tea = 180º

C) The maximum force is: F = 24 N

D) The time to stop the block is: t = 25 s

Newton's second law establishes a relationship between the net force, the mass, and the acceleration of the body. In the special case that the acceleration is zero it is called the equilibrium condition.

B) They indicate a diagram of forces on the block, let's look for the components of the force that the block maintains with zero acceleration, in the attached we have a free-body diagram including the force applied to keep the system in equilibrium.

x-axis

-10 + 12 sin 60 + Fₓ = 0

Fₓ = 10- 12 sin 60 = -0.39 N

y-axis

12 cos 60 - 6 + F_y = 0

F_y = 6 - 12 cos 60 = 0 N

We can give the result of the force in two ways:

Form of coordinates F = -0.39 i ^ N

Form of module and angle.

Let's use Pythagoras' theorem to find the modulus.

$F = \sqrt{F_x^2 + F_y^2 } \\F = \sqrt{0.39^2 +0^2}$

F = 0.39N

We use trigonometry for the angle.

$tan \theta = \frac{F_y}{F_x}$

tan θ= 0º

The component of the force is negative therefore this angle is in the second quadrant, to measure the angle from the positive side of the x axis in a counterclockwise direction.

θ = 180 + θ'

θ = 180 + 0

θ = 180º

C) if the three forces can be moved and the maximum force occurs when they are all linear.

10+ 6 + 6 + F = 0

F = -24 N

D) if we maintain this force and eliminate the other three, the block stops, let's look for its acceleration.

$a = \frac{F}{m}$

a = $\frac{24}{6}$

a = 4 m / s²

The acceleration is in the opposite direction of the initial velocity of the block v₀ = -100 m / s

If we use kinematic relations.

v = v₀ - a t

Final velocity when stopped is zero

t = $\frac{0-v_o}{a}$

t = 100/4

t = 25 s

In conclusion using Newton's second law and the kinematics relations we can find the results for the questions about the forces and the motion of the block are:

B) the force applied to maintain the system is equilibrium is: F = 0.39 N with an angle of tea = 180º

C) The maximum force is: F = 24 N

D) The time to stop the block is: t = 25 s

Learn more about Newton's second law here: brainly.com/question/25545050

3 0

3 years ago

What distance will a car cover while uniformly accelerating from 12m/s to 26m/in 14 seconds

VikaD [51]

Answer:

Distance = 266m

Explanation:

6 0

4 years ago

The period for which an organism lives is called its _______