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wlad13 [49]
3 years ago
12

a Ferrari with an initial velocity of 10 m/s, comes to a complete stop in 5 seconds, what will its acceleration be?

Physics
2 answers:
Aleks04 [339]3 years ago
8 0

Acceleration = (change in speed) / (time for the change)

Change in speed = (speed at the end) minus (speed at the beginning)

Change in speed = (zero) minus (10 m/s) = negative 10 m/s

Acceleration = (-10 m/s) / (5 sec)

Acceleration of the Ferrari = -2 m/s²

Democratically, it doesn't have to be a Ferrari. If my 16-yr-old Dodge is going 10 m/s and stops in 5 sec, its acceleration would be the same number.

11111nata11111 [884]3 years ago
6 0

10 m/s divided by 5 s

-2m/s/s

minus - deceleration

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shtirl [24]

Answer:

3g/cm³

Explanation:

<em>Use the formula:</em>

density = mass ÷ volume

<em>Substitute (plug in) the values:</em>

density = 3 ÷ 1 = 3g/cm³

4 0
2 years ago
A car horn emits a frequency of 400 Hz. A car traveling at 20.0 m/s sounds the horn as it approaches a stationary pedestrian. Wh
Temka [501]

Answer:

The observed frequency by the pedestrian is 424 Hz.

Explanation:

Given;

frequency of the source, Fs = 400 Hz

speed of the car as it approaches the stationary observer, Vs = 20 m/s

Based on Doppler effect, as the car the approaches the stationary observer, the observed frequency will be higher than the transmitted (source) frequency because of decrease in distance between the car and the observer.

The observed frequency is calculated as;

F_s = F_o [\frac{v}{v_s + v} ] \\\\

where;

F₀ is the observed frequency

v is the speed of sound in air = 340 m/s

F_s = F_o [\frac{v}{v_s + v} ] \\\\400 = F_o [\frac{340}{20 + 340} ] \\\\400 = F_o (0.9444) \\\\F_o = \frac{400}{0.9444} \\\\F_o = 423.55 \ Hz \\

F₀ ≅ 424 Hz.

Therefore, the observed frequency by the pedestrian is 424 Hz.

8 0
3 years ago
"My distance to the center of the earth is about 4000 miles when I am on the surface. If I go to a height of 8000 miles above th
lord [1]

Given,

Distance from the surface to the center of the earth, d=4000 miles

Distance from the center to you at a height of 8000 miles, a= 8000+4000=12000 miles

The gravitational force acting on a person at the surface is equal to his weight.

From Newton's Universal Law of Gravitation, the gravitational force is

F=\frac{G\times M\times m}{r^2}

Where G is the gravitational constant, M is the mass of the earth, m is the mass of the object/person, r is the distance between the center of the earth and the object/person

At the surface, this force is equal to the weight of the person, W=mg

i.e.

F_s=\frac{G\times M\times m}{d^2}=W

On substituting the of d,

W=\frac{\text{GMm}}{4000^2}

At a height of 8000 miles from the surface, the gravitational force is equal to,

F_a=\frac{GMm}{12000^2}

On dividing the above two equations,

\frac{F_a}{W}=\frac{4000^2^{}}{12000^2}=\frac{1}{9}

Therefore,

F_a=\frac{1}{9}W

Therefore at a height of 8000 miles above the surface of the earth, the force of gravity becomes 1/9 time your weight.

5 0
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Pani-rosa [81]

Hey how's your day going

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Ice is less dense than water.

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When water freezes the molecules inside completely stop moving (They still vibrate but don't change their position much). In doing so, they spread out a touch which makes it less dense than liquid water. So ice floats

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A train increase its speed steadily from 10ms to 20ms in 1 minutes
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Answer:

Explanation:  nigerian what

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