A 55 kg boy is riding a skateboard at a speed of 1.0 m/s. What is his<br> kinetic energy?

Which four equations can be used to solve for acceleration

emmainna [20.7K]

The four equations for acceleration are obtained from the three equations of motion and from second law of motion.

Explanation:

Acceleration is defined as the rate of change of velocity with respect to time. So the change in velocity with respect to time can be determined using the three equations of motions.

So from the first equation of motion, v = u + at , we can determine the value of acceleration if time taken, final and initial velocity is known. The equation can be re-written as $a = \frac{v-u}{t}$

Similarly, from the second equation of motion, s = ut + 1/2 at², we can determine the equation for acceleration as $a = 2*\frac{s-ut}{t^{2} }$

So this is second equation for acceleration.

Then from the third equation of motion, $v^{2}- u^{2} = 2* a *s$

the acceleration equation is determined as $a = \frac{v^{2}-u^{2} }{2s}$

In addition to these three equation, another equation is present to determine the acceleration with respect to force from the Newton's second law of motion. F = Mass × acceleration. From this, acceleration = Force/mass.

So, these are the four equations for acceleration.

8 0

3 years ago

When responding to sound, the human eardrum vibrates about its equilibrium position. suppose an eardrum is vibrating with an amp

vodomira [7]

745.92 hz (b)
this is hte answer only because i seen it on the question lol

5 0

3 years ago

The parallel axis theorem relates Icm, the moment of inertia of an object about an axis passing through its center of mass, to I

erma4kov [3.2K]

Answer:

Part a)

$I_{end} = \frac{mL^2}{3}$

Part b)

$I_{edge} = \frac{2ma^2}{3}$

Explanation:

As we know that by parallel axis theorem we will have

$I_p = I_{cm} + Md^2$

Part a)

here we know that for a stick the moment of inertia for an axis passing through its COM is given as

$I = \frac{mL^2}{12}$

now if we need to find the inertia from its end then we will have

$I_{end} = I_{cm} + Md^2$

$I_{end} = \frac{mL^2}{12} + m(\frac{L}{2})^2$

$I_{end} = \frac{mL^2}{3}$

Part b)

here we know that for a cube the moment of inertia for an axis passing through its COM is given as

$I = \frac{ma^2}{6}$

now if we need to find the inertia about an axis passing through its edge

$I_{edge} = I_{cm} + Md^2$

$I_{edge} = \frac{ma^2}{6} + m(\frac{a}{\sqrt2})^2$

$I_{edge} = \frac{2ma^2}{3}$

7 0

3 years ago

A car travels at a constant speed around a circular track whose radiu is 2.6 km. The goes once arond the track in 360s . What is

AveGali [126]

Answer:

Centripetal acceleration = 0.79 m/s²

Explanation:

<u>Given the following data;</u>

Radius, r = 2.6 km

Time = 360 seconds

<em><u>Conversion:</u></em>

2.6 km to meters = 2.6 * 1000 = 2600 meters

To find the magnitude of centripetal acceleration;

First of all, we would determine the circular speed of the car using the formula;

$Circular \; speed (V) = \frac {2 \pi r}{t}$

Where;

r represents the radius and t is the time.

Substituting into the formula, we have;

$Circular \; speed (V) = \frac {2*3.142*2600}{360}$

$Circular \; speed (V) = \frac {16338.4}{360}$

Circular speed, V = 45.38 m/s

Next, we find the centripetal acceleration;

Mathematically, centripetal acceleration is given by the formula;

$Centripetal \; acceleration = \frac {V^{2}}{r}$

Where;

V is the circular speed (velocity) of an object.
r is the radius of circular path.

Substituting into the formula, we have;

$Centripetal \; acceleration = \frac {45.38^{2}}{2.6}$

$Centripetal \; acceleration = \frac {2059.34}{2600}$

<em>Centripetal acceleration = 0.79 m/s²</em>

3 0

3 years ago

HURRY I NEED IT NOW What is the density at 20oC of 12.0 milliliters of a liquid that has a mass of 4.05 grams?

zzz [600]

Density is the mass per unit volume of any object. It is calculated by dividing the mass of an object by its volume. This is:

ρ = m/V

ρ = 4.05 g / 12 mL

ρ = 0.3375 g/mL

OPTION A

4 0

3 years ago

A 55 kg boy is riding a skateboard at a speed of 1.0 m/s. What is his kinetic energy?