The unicorn suddenly ran through the school at 5.6 m/s and it has a mass of 450 kg. What is the unicorn’s kinetic energy?

if a torque of 55.0 N/m is required and the largest force that can be exerted by you is 135 N what is th e length of the lever a

Whitepunk [10]

Answer:

$r=0.41m$

Explanation:

Torque is defined as the cross product between the position vector ( the lever arm vector connecting the origin to the point of force application) and the force vector.

$\tau=r\times F$

Due to the definition of cross product, the magnitude of the torque is given by:

$\tau=rFsin\theta$

Where $\theta$ is the angle between the force and lever arm vectors. So, the length of the lever arm (r) is minimun when $sin\theta$ is equal to one, solving for r:

$r=\frac{\tau}{F}\\r=\frac{55\frac{N}{m}}{135N}\\r=0.41m$

7 0

3 years ago

A child of mass m is standing at the edge of a carousel. Both the carousel and the child are initially stationary. The carousel

suter [353]

Answer:

the angular velocity of the carousel after the child has started running =

$\frac{2F}{mR} \delta t$

Explanation:

Given that

the mass of the child = m

The radius of the disc = R

moment of inertia I = $\frac{1}{2} mR^2$

change in time = $\delta \ t$

By using the torque around the inertia ; we have:

T = I×∝

where

R×F = I × ∝

R×F = $\frac{1}{2} mR^2$ ∝

F = $\frac{1}{2} mR$ ∝

∝ = $\frac{2F}{mR}$ ( expression for angular angular acceleration)

The first equation of motion of rotating wheel can be expressed as :

$\omega = \omega_0 + \alpha \delta t$

where ;

∝ = $\frac{2F}{mR}$

Then;

$\omega = 0+ \frac{2F}{mR} \delta t$

$\omega = \frac{2F}{mR} \delta t$

∴ the angular velocity of the carousel after the child has started running =

$\frac{2F}{mR} \delta t$

7 0

3 years ago

The force generated by a single muscle fiber can be increased by increasing is called

mars1129 [50]

Answer:

The force generated by a single muscle fiber can be increased by increasing the frequency of action potentials

Explanation:

The force generated by a muscle fiber is the result of the shortening of the skeletal muscle, and this force is also know as muscle tension. The larger motor units shorten along with the smaller units to produce the muscle force. The time lapsed between the beginning of the action potential in the muscle and the beginning of the contraction is the latent period. Action potential is the result of the difference electrical potential as a result of passage of an impulse along the membrane of a muscle or nerve cell.

4 0

3 years ago

A spring hangs at rest from a support. If you suspend a 0.46 kg mass from the spring it elongates 0.79m.

Fudgin [204]

a. The restoring force in the spring has magnitude

F[spring] = k (0.79 m)

which counters the weight of the mass,

F[weight] = (0.46 kg) g = 4.508 N

so that by Newton's second law,

F[spring] - F[weight] = 0 ⇒ k = (4.508 N) / (0.79 m) ≈ 5.7 N/m

b. Using the same equation as before, we now have

F[weight] = (0.75 kg) g = 7.35 N

so that

(5.7 N/m) x - 7.35 N = 0 ⇒ x = (7.35 N) / (5.7 N/m) ≈ 1.3 m

4 0

2 years ago

Un automovil parte del reposo y acelera uniformemente hasta alcanzar una rapidez de 0,255km/h en un tiempo de 3/4 Minutos determ

Elden [556K]

Answer:

a = 1.5*10^-3 m/s^2

x = 0.033m = 3.3cm

Explanation:

To calculate the acceleration and the distance traveled by the car you use the following formulas:

$v=v_o+at$ (1)

$x=v_ot+\frac{1}{2}at^2$ (2)

v: final velocity = 0,255 km/h

vo: initial velocity = 0 m/s

t: time = 3/4 min

a: acceleration = ?

x: distance

In order to use the equations (1) and (2) you first convert the units of the final velocity to m/s, and the time to seconds.

$v=0,255\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}\\\\v=0.07m/s\\\\t=\frac{3}{4}min*\frac{60s}{1min}=45s$

Next, you solve the equation (1) for the acceleration a:

$a=\frac{v}{t}=\frac{0.07m/s}{45s}=1.5*10^{-3}\frac{m}{s^2}$

With this value of a you can calculate the distance traveled by the car, by using the equation (2):

$x=\frac{1}{2}(1.5*10^{-3}m/s^2)(45s)^2=0.033m=3.3cm$

hence, the acceleration of the car is 1.5*10^-3 m/s^2 and the distance traveled in 3/4 min is 0.033m

5 0

3 years ago