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4 years ago

A go-kart accelerates at a rate

Physics

1 answer:

stira [4]4 years ago

4 0

Answer:

if it is at a speed of 12.5 m / s it would take 2 minutes

Explanation:

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What is homeostasis?

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4 years ago

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A car traveling with 500,000 J of kinetic energy is brought to a kinetic energy of

allsm [11]

Answer:

33,333.33 N

Explanation:

Given that :

Initial kinetic energy = 500,000 J

Final kinetic energy = 100,000 J

Using the relation :

Force * time = change in momentum (Newton's law)

Force (F) * 0.12 = (500,000 - 100,000)

0.12F = 400,000 J

Force = (400,000 J) / 0.12s

Force = 33333.333

Force = 33,333.33 N

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3 years ago

A light ray traveling from a higher index of refraction medium into a lower index of refraction medium (for example, from water

masya89 [10]

Answer:

option C

Explanation:

The correct answer is option C

When the ray of light pass from higher refractive index to lower refractive index then the light bend away from the normal.

when the refracted ray is parallel to the boundary of the medium, θ₂ = 90°

and the incident angle θ₁

so, the angle θc is known as the critical angle.

At critical angle the refracted ray becomes parallel to boundary surface.

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3 years ago

A 6.9-kg wheel with geometric radius m has radius of gyration computed about its mass center given by m. A massless bar at angle

Vilka [71]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The value is $\alpha =2.538 \ rad/s^2$

Explanation:

From the question we are told that

The mass of the wheel is m = 6.9 kg

The radius is $r = 0.69 \ m$

The radius of gyration is $k_G = 0.4\ m$

The angle is $\theta = 47^o$

The force which the massless bar is subjected to $F = 22.5 \ N$

Generally given that the wheels rolls without slipping on the flat stationary ground surface, it implies that point A is the center of rotation.

Generally the moment of inertia about A is mathematically represented as

$I_a = I_G + M* r^2$

Here $I_G$ is the moment of inertia about G with respect to the radius of gyration which is mathematically represented as

$I_G = M * k_G$

=> $I_a = k_G* M + M* r^2$

=> $I_a =0.4 * 6.9 + 6.9 * 0.69^2$

=> $I_a =6.045 \ kg \cdot m^2$

Generally the torque experienced by the wheel is mathematically represented as

$\tau = F * cos (47)$

=> $\tau = 22.5 * cos (47)$

=> $\tau = 15.34 \ kg \cdot m^2 \cdot s^{-2}$

Generally this torque is also mathematically represented as

$\tau = I_a * \alpha$

=> $15.34 = 6.045 * \alpha$

=> $\alpha =2.538 \ rad/s^2$

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3 years ago

A bicyclist moves along a straight line with an initial velocity vo and slows downs. Which of the following the best describes t

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