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2 years ago

A car accelerates uniformly from rest to a speed of 6.6 m/s in 6.5 s. Find the distance the car travels during this time.

Physics

2 answers:

andrew11 [14]2 years ago

6 0

It traveled 39.6 meters

stiks02 [169]2 years ago

3 0

Answer:

Distance covered by the car, s = 21.56 meters

Explanation:

Given that,

Initially, the car is at rest, u = 0

It reaches a speed, final speed, v = 6.6 m/s

Time, t = 6.5 s

We need to find the distance covered by the car during this time. Firstly, we can find the acceleration of the car using equation of motion as :

$v=u+at$

$6.6=0+a\times 6.5$

$a=1.01\ m/s^2$

Now using third equation of motion to find distance (s) as :

$s=\dfrac{v^2-u^2}{2a}$

$s=\dfrac{(6.6)^2}{2\times 1.01}$

s = 21.56 meters

So, the distance traveled by the car during 6.5 seconds is 21.56 meters. Hence, this is the required solution.

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Calculate the energy in the form of heat (in kJ) required to change 75.0 g of liquid water at 27.0 °C to ice at –20.0 °C. Assume

REY [17]

Explanation:

The given data is as follows.

mass, m = 75 g

$T_{1} = 0^{o}C$

$T_{2} = 27^{o}C$

Specific heat of water = 4.18

First, we will calculate the heat required for water is as follows.

q = $m C \times (T_{1} - T_{2})$

= $75 g \times 4.18 J/g^{o}C \times (0 - 27)^{o}C$

= 8464.5 J/mol

= 8.46 kJ ......... (1)

Also, it is given that $T_{3} = -20^{o}C$ = (20 + 273) K = 293 K and specific heat of ice is 2.108 kJ/kg K.

Now, we will calculate the heat of fusion as follows.

q = $mC \times (T_{3} - T_{1})$

= $0.075 kg \times 2.108 kJ/kg K \times (-293 - 0) K$

= -46.32 kJ ......... (2)

Now, adding both equations (1) and (2) as follows.

8.46 kJ - 46.32 kJ

= -37.86 kJ

Therefore, we can conclude that energy in the form of heat (in kJ) required to change 75.0 g of liquid water at $27.0^{o}C$ to ice at $-20.0^{o}C$ is -37.86 kJ.

4 0

2 years ago

Inna Hurry is traveling at 6.8 m/s, when she realizes she is late for an appointment. She accelerates at 4.5 m/s^2 for 3.2 s. Wh

Alborosie

Answer:

1) v = 21.2 m/s

2) S = 63.33 m

3) s = 61.257 m

4) Deceleration, a = -4.32 m/s²

Explanation:

1) Given,

The initial velocity of Inna, u = 6.8 m/s

The acceleration of Inna, a = 4.5 m/s²

The time of travel, t = 3.2 s

Using the first equation of motion, the final velocity is

v = u + at

= 6.8 + 4.5 x 3.2

= 21.2 m/s

The final velocity of Inna is, v = 21.2 m/s

2) Given,

The initial velocity of Lisa, u = 12 m/s

The final velocity of Lisa, v = 26 m/s

The acceleration of Lisa, a = 4.2 m/s²

Using the III equations of motion, the displacement is

v² = u² +2aS

S = (v² - u²) / 2a

= (26² -12²) / 2 x 4.2

= 63.33 m

The distance Lisa traveled, S = 63.33 m

3) Given,

The initial velocity of Ed, u = 38.2 m/s

The deceleration of Ed, d = - 8.6 m/s²

The time of travel, t = 2.1 s

Using the II equations of motion, the displacement is

s = ut + 1/2 at²

=38.2 x 2.1 + 0.5 x(-8.6) x 2.1²

= 61.257 m

Therefore, the distance traveled by Ed, s = 61.257 m

4) Given,

The initial velocity of the car, u = 24.2 m/s

The final velocity of the car, v = 11.9 m/s

The time taken by the car is, t = 2.85 s

Using the first equations of motion,

v = u + at

∴ a = (v - u) / t

= (11.9 - 24.2) / 2.85

= -4.32 m/s²

Hence, the deceleration of the car, a = = -4.32 m/s²

5 0

2 years ago

Read 2 more answers

A toroidal solenoid has an inner radius of 12.0 cm and an outer radius of 15.0 cm . It carries a current of 1.50 A . Part A How

tensa zangetsu [6.8K]

Answer:

The number of turns is $N = 1750 \ turns$

Explanation:

From the question we are told that

The inner radius is $r_i = 12.0 \ cm = 0.12 \ m$

The outer radius is $r_o = 15.0 \ cm = 0.15 \ m$

The current it carries is $I = 1.50 \ A$

The magnetic field is $B = 3.75 mT = 3.75 *10^{-3} \ T$

The distance from the center is $d = 14.0 \ cm = 0.14 \ m$

Generally the number of turns is mathematically represented as

$N = \frac{2 * \pi * d * B}{ \mu_o * r_o }$

Generally $\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi * 10^{-7} \ N/A^2$

$N = \frac{2 * 3.142 * 0.14 * 3.75 *10^{-3} }{ 4\pi * 10^{-7} * 0.15 }$

$N = 1750 \ turns$

5 0

2 years ago

A car travels 10 m/s east. Another car travels 10 m/s north. The relative speed of the first car with respect to the second is:

Thepotemich [5.8K]

Answer:

d. less than 20m/s

Explanation:

To the 2nd car, the first car is travelling 10m/s east and 10m/s south. So the total velocity of the first car with respect to the 2nd car is

[tex]\sqrt{10^2 + 10^2} =10\sqrt{2}=14.14m/s

As 14.14m/s is less than 20m/s. d is the correct selection for this question.

3 0

2 years ago

A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.

MakcuM [25]

Answer:

b. 9.5°C

Explanation:

$m_i$ = Mass of ice = 50 g

$T_i$ = Initial temperature of water and Aluminum = 30°C

$L_f$ = Latent heat of fusion = $3.33\times 10^5\ J/kg^{\circ}C$

$m_w$ = Mass of water = 200 g

$c_w$ = Specific heat of water = 4186 J/kg⋅°C

$m_{Al}$ = Mass of Aluminum = 80 g

$c_{Al}$ = Specific heat of Aluminum = 900 J/kg⋅°C

The equation of the system's heat exchange is given by

$m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C$

The final equilibrium temperature is 9.50022°C

4 0

2 years ago