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3 years ago

Starting from rest and moving in a straight line, a cheetah

Physics

1 answer:

UNO [17]3 years ago

3 0

The average acceleration of the cheetah is 7.75 m/s²

The given parameters;

velocity of the cheetah, v = 31 m/s

time of motion, t = 4 seconds

The average acceleration of the cheetah is calculated as follows;

$Average \ acceleration = \frac{\Delta velocity}{\Delta time } \\\\Average \ acceleration = \frac{v-u}{t}$

where;

v is the final velocity

u is the initial velocity

t is the time of motion

Assuming the cheetah started from rest, the initial velocity, u = 0

The average acceleration is calculated as follows;

$Average \ acceleration = \frac{v-u}{t} = \frac{31-0}{4} = 7.75 \ m/s^2$

Thus, the average acceleration of the cheetah is 7.75 m/s²

Learn more here: brainly.com/question/17280180

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A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the

Margaret [11]

Hi there!

$\boxed{\omega = 0.38 rad/sec}$

We can use the conservation of angular momentum to solve.

$\large\boxed{L_i = L_f}$

Recall the equation for angular momentum:

$L = I\omega$

We can begin by writing out the scenario as a conservation of angular momentum:

$I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)$

$I_m$ = moment of inertia of the merry-go-round (kgm²)

$\omega_m$ = angular velocity of merry go round (rad/sec)

$\omega_f$ = final angular velocity of COMBINED objects (rad/sec)

$I_b$ = moment of inertia of boy (kgm²)

$\omega_b$ = angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

$I = MR^2$

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

$\omega = \frac{v}{r}$

$L_b = MR^2(\frac{v}{R}) = MRv$

Plug in the given values:

$L_b = (20)(3)(5) = 300 kgm^2/s$

Now, we must solve for the boy's moment of inertia:

$I = MR^2\\I = 20(3^2) = 180 kgm^2$

Use the above equation for conservation of momentum:

$600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}$

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3 years ago

,how do charged objects react???quiet urgent

son4ous [18]

Any charged object can<span> exert the force upon other objects ... i think tell me if im right</span>

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The magnitude of electrical force between a pair of charged particles is ____ as much when the particles are moved half as far a

Gnesinka [82]

The magnitude of the electrical force between a pair of charged particles is 4 Times as much when the particles are moved half as far apart.

This can be easily understood by Columb's law,

$F_{new} = \frac{kQ_{1}Q_{2}}{r^{2}}$

which state's that the amount of electrical force experienced by two charged particles is inversely proportional to the square of the distance between them.

∴ $\frac{F_{new} }{F_{old} } = \frac{Distance_{new}^{2} }{Distance_{old}^{2} }$

Now, we know the new distance is half the original distance,

$F_{new} = \frac{kQ_{1}Q_{2}}{\frac{r}{2}^{2} } \\F_{new} = 4\frac{kQ_{1}Q_{2}}{r^{2}}$

$F_{new} = 4F_{old}$

The electrical force of attraction or electrostatic force of attraction between two charged particles refers to the amount of attractive or repulsive force that exists between the two charges. This can be calculated by Columb's Law.

A charged particle in physics is a particle that has an electric charge. It might be an ion, such as a molecule or atom having an excess or shortage of electrons in comparison to protons. The same charge is thought to be shared by an electron, a proton, or another primary particle.

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1 year ago

A baseball bat changes the momentum of a ball with an impulse of 13.8 Nᐧs. What is the average force that the bat exerts on the

vladimir2022 [97]

Answer:

13800 N

Explanation:

Impulse is the product of average force and time expressed as I=Ft where I is the impulse which results into change in momentum, F is the average force and t is the time of impact. Making F the subject of formula then

$F=\frac {I}{t}$

Substituting I with 13.8 N.s and time, t witg 0.001 s then the average force is calculated as

$F=\frac {13.8 N.s}{0.001}=13800N$

Therefore, the average force is equivalent to 13800 N

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3 years ago

When is there the least amount of heat transfer within a liquid

Annette [7]

In the beginning, the melting stage they call it.

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