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3 years ago

What is the average acceleration of a car that starts from

Physics

2 answers:

MAXImum [283]3 years ago

4 0

Answer:

1.5m/s^2

Explanation:

In the picture above.

mr Goodwill [35]3 years ago

4 0

Answer:

The car is moving at an average acceleration of 1.5 m/s2

Explanation:

According to the question,

The car is moving in a straight road

The speed of the car = 18 m/s

Time is given (t) = 12 s

Acceleration is the rate of change of speed with respect to time. It is computed by the formula;

$\text { Acceleration }=\frac{\text {Speed }}{\text {Time}}$

Acceleration = (18 m/s)/(12 s)= 1.5 m/s2

$\text { Acceleration }=\frac{18 \mathrm{m} / \mathrm{s}}{12 \mathrm{s}}=1.5 \mathrm{m} / \mathrm{s}^{2}$

Therefore, the car is moving at an average acceleration of 1.5 m/s2

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How many minutes in 1 hour

fgiga [73]

Answer:

Explanation:

Every 60 minutes is an hour

8 0

3 years ago

Read 2 more answers

The suspension system of a 1700 kg automobile "sags" 7.7 cm when the chassis is placed on it. Also, the oscillation amplitude de

spin [16.1K]

Answer:

the spring constant k = $5.409*10^4 \ N/m$

the value for the damping constant $\\ \\b = 1.518 *10^3 \ kg/s$

Explanation:

From Hooke's Law

$F = kx\\\\k =\frac{F}{x}\\\\where \ F = mg\\\\k = \frac{mg}{x}\\\\given \ that:\\\\mass \ of \ each \ wheel = 425 \ kg\\\\x = 7.7cm = 0.077 m\\\\g = 9.8 \ m/s^2\\\\Then;\\\\k = \frac{425 \ kg * 9.8 \ m/s^2}{0.077 \ m}\\\\k = 5.409*10^4 \ N/m$

Thus; the spring constant k = $5.409*10^4 \ N/m$

The amplitude is decreasing 37% during one period of the motion

$e^{\frac{-bT}{2m}}= \frac{37}{100}\\\\e^{\frac{-bT}{2m}}= 0.37\\\\\frac{-bT}{2m} = In(0.37)\\\\\frac{-bT}{2m} = -0.9943\\\\b = \frac{2m(0.9943)}{T}\\\\b = \frac{2m(0.9943)}{\frac{2 \pi}{\omega}}\\\\b = \frac{m(0.9943) \ ( \omega) )}{ \pi}$

$b = \frac{m(0.9943)(\sqrt{\frac{k}{m})}}{\pi}\\\\b = \frac{425*(0.9943)(\sqrt{\frac{5.409*10^4}{425}) } }{3.14}\\\\b = 1518.24 \ kg/s\\\\b = 1.518 *10^3 \ kg/s$

Therefore; the value for the damping constant $\\ \\b = 1.518 *10^3 \ kg/s$

5 0

3 years ago

A baseball is moving at a speed of 2.2m/s when it strikes the catchers glove. The paddding of the glove is compressed by 24mm be

Andre45 [30]

Average acceleration of the baseball: $-101 m/s^2$

Explanation:

Since the motion of the baseball is a uniformly accelerated motion, we can use the following suvat equation:

$v^2-u^2=2as$

where

u is the initial velocity

v is the final velocity

a is the acceleration

s is the displacement of the object

For the baseball in this problem, we have:

u = 2.2 m/s is the initial velocity

v = 0 is the final velocity (it comes to a stop)

s = 24 mm = 0.024 m is the displacement of the ball while decelerating

Therefore, we can solve for a to find the acceleration:

$a=\frac{v^2-u^2}{2s}=\frac{0-2.2^2}{2(0.024)}=-101 m/s^2$

where the negative sign means the baseball is slowing down.

Learn more about acceleration:

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brainly.com/question/2562700

#LearnwithBrainly

4 0

3 years ago

A resistor has four colored stripes in the following order: orange, orange, brown and silver. What is the resistance of the resi

zubka84 [21]

Answer:

Resistance =330 Ω

Tolerance = 33 Ω

Explanation:

see attached resistor color code table

The first stripe is orange, which means the leftmost digit is a 3.

The second stripe is orange , which means the next digit is a 3.

The third stripe is brown. Since brown is 1, it means add one zero to the right of the first two digits.

The resistance is:

orange-orange-brown= 330 Ω

The tolerance is:

The fourth color band indicates the resistor's tolerance. Tolerance is the percentage of error in the resistor's resistance.

silver is 10%

A 330 Ω resistor has a silver tolerance band.

Tolerance = value of resistor x value of tolerance band

= 330 Ω x 10% = 33 Ω

330 Ω stated resistance +/- 33 Ω tolerance means that the resistor could range in actual value from as much as 363 Ω to as little as 297 Ω.

7 0

3 years ago

A 180 lb crate is on the ground, and a strong rope is attached. You need to move it across the basement floor, which has a coeff

jekas [21]

Answer:

$F = 505.13 N$

Yes it is better to pull the rope rather than push it

Explanation:

Let the force is applied at an angle of 60 degree

so we will have net vertical force on the crate is given as

$F_n + Fsin60 = mg$

here we know

$m = 180 lb$

$m = 81.65 kg$

$F_n = 81.65(9.81) - Fsin60$

$F_n = 801 - 0.866 F$

now friction force on the crate is given as

$F_x = \mu F_n$

$Fcos60 = 0.7(801 - 0.866 F)$

$0.5F + 0.61F = 560.7$

$F = 505.13 N$

Yes it is better to pull the rope rather than push it

6 0

3 years ago