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2 years ago

the force of friction between a 1000 kg car and the road is 10000 N, what is the fastest acceleration the car can achieve?

Physics

2 answers:

nordsb [41]2 years ago

7 0

Answer:

1000

Explanation:

ololo11 [35]2 years ago

6 0

Answer:

10 m/s²

Explanation:

Newton's second law:

∑F = ma

10000 N = (1000 kg) a

a = 10 m/s²

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Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 9.0 kg gibbon

aleksklad [387]

Answer:

230 N

Explanation:

At the lowest position , the velocity is maximum hence at this point, maximum support force T is given by the branch.

The swinging motion of the ape on a vertical circular path , will require

a centripetal force in upward direction . This is related to weight as follows

T - mg = m v² / R

R is radius of circular path . m is mass of the ape and velocity is 3.2 m/s

T = mg - mv² / R

T = 8.5 X 9.8 + 8.5 X 3.2² / .60 { R is length of hand of ape. }

T = 83.3 + 145.06

= 228.36

= 230 N ( approximately )

5 0

2 years ago

A small lab cart and one of larger mass collide and rebound off each other. Which of them has the greater average force on it du

Elza [17]

When a small cart collide with a large mass then during collision they must be in contact with each other for some interval of time

During this contact interval we can say they will exert normal force on each other

This normal force is always equal and opposite on two balls which means this force will follow Newton's III law

It will be same in magnitude but opposite in the direction

So here correct answer would be

They both experience the same magnitude of the collision force.

4 0

2 years ago

A thin flake of mica (n=1.58) is used to cover one slit of a double-slit interference arrangement. The central point on the wiew

Llana [10]

To develop this problem it is necessary to apply the optical concepts related to the phase difference between two or more materials.

By definition we know that the phase between two light waves that are traveling on different materials (in this case also two) is given by the equation

$\Phi = 2\pi(\frac{L}{\lambda}(n_1-n_2))$

Where

L = Thickness

n = Index of refraction of each material

$\lambda =$ Wavelength

Our values are given as

$\Phi = 7(2\pi)$

$L=t$

$n_1 = 1.58$

$n_2 = 1$

$\lambda = 550nm$

Replacing our values at the previous equation we have

$\Phi = 2\pi(\frac{L}{\lambda}(n_1-n_2))$

$7(2\pi) = 2\pi(\frac{t}{\lambda}(1.58-1))$

$t = \frac{7*550}{1.58-1}$

$t = 6637.931nm \approx 6.64\mu m$

Therefore the thickness of the mica is 6.64μm

3 0

2 years ago

A woman with a mass of 52.0 kg is standing on the rim of a large disk that is rotating at an angular velocity of 0.470 rev/s abo

Charra [1.4K]

Explanation:

It is given that,

Mass of the woman, m₁ = 52 kg

Angular velocity, $\omega=0.47\ rev/s=2.95\ rad/s$

Mass of disk, m₂ = 118 kg

Radius of the disk, r = 3.9 m

The moment of inertia of woman which is standing at the rim of a large disk is :

$I={m_1r^2}$

$I={52\times 3.9^2}$

I₁ = 790.92 kg-m²

The moment of inertia of of the disk about an axis through its center is given by :

$I_2=\dfrac{m_2r^2}{2}$

$I_2=\dfrac{118\times (3.9)^2}{2}$

I₂ =897.39 kg-m²

Total moment of inertia of the system is given by :

$I=I_1+I_2$

$I=790.92+897.39$

I = 1688.31 kg-m²

The angular momentum of the system is :

$L=I\times \omega$

$L=1688.31 \times 2.95$

$L=4980.5\ kg-m^2/s$

So, the total angular momentum of the system is 4980.5 kg-m²/s. Hence, this is the required solution.

8 0

2 years ago

You stand on a frictionless platform that is rotating with an angular speed of 5.1 rev/s. Your arms are outstretched, and you ho

timurjin [86]

Answer:

$\omega'=19.419\ rev.s^{-1}$

Explanation:

Given:

angular speed of rotation of friction-less platform, $\omega=5.1\ rev.s^{-1}$

moment of inertia with extended weight, $I=9.9\ kg.m^2$

moment of inertia with contracted weight, $I'=2.6\ kg.m^2$

Now we use the law of conservation of angular momentum:

$I.\omega=I'.\omega'$

$9.9\times 5.1=2.6\times \omega'$

$\omega'=19.419\ rev.s^{-1}$

The angular speed becomes faster as the mass is contracted radially near to the axis of rotation.

5 0

3 years ago

the force of friction between a 1000 kg car and the road is 10000 N, what is the fastest acceleration the car can achieve?​

the force of friction between a 1000 kg car and the road is 10000 N, what is the fastest acceleration the car can achieve?