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3 years ago

What is the acceleration of a racing car if it’s speed is increased uniformly form 44m/s to 66 m/s over an 11 sec. period.

Physics

1 answer:

jonny [76]3 years ago

7 0

Answer:

$a=2\ m/s^2$

Explanation:

Given that,

Initial velocity, u = 44 m/s

Final velocity, v = 66 m/s

Time, t = 11 s

Acceleration = rate of change of velocity

$a=\dfrac{v-u}{t}\\\\a=\dfrac{66-44}{11}\\\\a=2\ m/s^2$

So, the acceleration of the car is $2\ m/s^2$ .

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1_1H + 1_1H →2_1H + e^+ + v + energy
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3 years ago

Read 2 more answers

A velocity vs. time graph starts at 0 and ends at 10 m/s, stretching over a time- span of 15 s with constant acceleration. What

Alex777 [14]

Answer:

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3 years ago

A parallel-plate capacitor is made from two aluminum-foil sheets, each 5.9 cm wide and 5.6 m long. Between the sheets is a Teflo

Zarrin [17]

Answer:

1.98 x 10⁻⁷ F

Explanation:

w = width of the sheet = 5.9 cm = 0.059 m

L = length of the sheet = 5.6 m

Area of the sheet is given as

A = L w = (5.6) (0.059) = 0.3304 m²

d = distance between the sheets = 3.1 x 10⁻⁵ m

k = dielectric constant of teflon = 2.1

Capacitance is given as

$C = \frac{k\epsilon _{o}A}{d}$

$C = \frac{(2.1)(8.85\times 10^{-12})(0.3304)}{3.1\times 10^{-5}}$

C = 1.98 x 10⁻⁷ F

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3 years ago

A 21.6−g sample of an alloy at 93.00°C is placed into 50.0 g of water at 22.00°C in an insulated coffee-cup calorimeter with a h

IrinaK [193]

Answer : The specific heat capacity of the alloy $1.422J/g^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$C_1$ = specific heat of alloy = ?

$C_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of alloy = 21.6 g

$m_2$ = mass of water = 50.0 g

$T_f$ = final temperature of system = $31.10^oC$

$T_1$ = initial temperature of alloy = $93.00^oC$

$T_2$ = initial temperature of water = $22.00^oC$

Now put all the given values in the above formula, we get

$21.6g\times c_1\times (31.10-93.00)^oC=-50.0g\times 4.18J/g^oC\times (31.10-22.00)^oC$

$c_1=1.422J/g^oC$

Therefore, the specific heat capacity of the alloy $1.422J/g^oC$

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3 years ago

A 350-g mass is attached to a spring whose spring constant is 64 N/m. Its maximum acceleration is 5.3 m/s2. What is the frequenc

max2010maxim [7]

The frequency of oscillation is 2.153 Hz

What is the frequency of spring?

Spring Frequency is the natural frequency of spring with a weight at the lower end. Spring is fixed from the upper end and the lower end is free.

For the mass-spring system in this problem,

The Frequency of spring is calculated with the equation:

$f = \frac{1}{2\pi } \sqrt{\frac{k}{m} }$

Where,

f = frequency of spring

k = spring constant = 64 N/m

m = mass attached to spring = 350g = 0.350 kg

a = maximum acceleration = 5.3 m/s^2

Substituting the values in the equation,

$f = \frac{1}{2\pi } \sqrt{\frac{64}{0.350} }$

$f = \frac{1}{2\pi } ( 13.522)$

$f = 2.1535 Hz$

Hence,

The frequency of oscillation is 2.153 Hz

Learn more about frequency here:

<u>brainly.com/question/13978015</u>

#SPJ4

6 0

1 year ago