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2 years ago

A car traveling at 10 m/s speeds up to 20 m/s in 3 seconds. determine the acceleration of the car

Physics

2 answers:

Sergeeva-Olga [200]2 years ago

7 0

Answer:

66m/s

Explanation:

because 10 meters/per second is going to be 22 mph/35 km, and 20m/s is 44 mph/70 km and in 3 seconds it would boost its mph/km(m/s) to 66 only because from 22 mph to 44 mph is 22 but 44/22 = 22, however you have top multiply that by 3 seconds which will coem out to 66 m/s, mph, km.

astra-53 [7]2 years ago

4 0

Answer: 3.33m/s²

Explanation:

Acceleration = rate of change in velocity with time

Acceleration = change in velocity / time

Initial velocity (u) = 10 m/s

Final velocity (v) = 20m/s

time (t) = 3s

a = V - U / t

a = (20 - 10) / 3

a = 10 / 3 = 3.33m/s²

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3 years ago

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2 years ago

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SCALCET8 3.9.018.MI. A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the

Firlakuza [10]

Answer:

The length of his shadow is decreasing at a rate of 1.13 m/s

Explanation:

The ray of light hitting the ground forms a right angled triangle of height H, which is the height of the building and width, D which is the distance of the tip of the shadow from the building.

Also, the height of the man, h which is parallel to H forms a right-angled triangle of width, L which is the length of the shadow.

By similar triangles,

H/D = h/L

L = hD/H

Also, when the man is 4 m from the building, the length of his shadow is L = D - 4

So, D - 4 = hD/H

H(D - 4) = hD

H = hD/(D - 4)

Since h = 2 m and D = 12 m,

H = 2 m × 12 m/(12 m - 4 m)

H = 24 m²/8 m

H = 3 m

Since L = hD/H

and h and H are constant, differentiating L with respect to time, we have

dL/dt = d(hD/H)/dt

dL/dt = h(dD/dt)/H

Now dD/dt = velocity(speed) of man = -1.7 m/s ( negative since he is moving towards the building in the negative x - direction)

Since h = 2 m and H = 3 m,

dL/dt = h(dD/dt)/H

dL/dt = 2 m(-1.7 m/s)/3 m

dL/dt = -3.4/3 m/s

dL/dt = -1.13 m/s

So, the length of his shadow is decreasing at a rate of 1.13 m/s

5 0

2 years ago

What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8 cm from equilibrium and then

Sedbober [7]

Answer:

$x=(0.088m)\cos(\sqrt{\frac{k}{m} } t)$

Explanation:

We first identify the elements of this simple harmonic motion:

The amplitude A is 8.8cm, because it's the maximum distance the mass can go away from the equilibrium point. In meters, it is equivalent to 0.088m.

The angular frequency ω can be calculated with the formula:

$\omega =\sqrt{\frac{k}{m}}$

Where k is the spring constant and m is the mass of the particle.

Now, since the spring starts stretched at its maximum, the appropriate function to use is the positive cosine in the equation of simple harmonic motion:

$x=A\cos(\omega t)$

Finally, the equation of the motion of the system is:

$x=(0.088m)\cos(\omega t)$

$x=(0.088m)\cos(\sqrt{\frac{k}{m} } t)$

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3 years ago