A Subtle Relationship

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

Art [367]

Answer:

1.02 m/s²

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 0 m/s

Final velocity (v) = 6.6 m/s

Time (t) = 6.5 s

Acceleration (a) =.?

Acceleration can simply be defined as the change of velocity with time. Mathematically, it can be expressed as:

a = (v – u) / t

Where:

a is the acceleration.

v is the final velocity.

u is the initial velocity.

t is the time.

With the above formula, we can obtain the acceleration of the car as follow:

Initial velocity (u) = 0 m/s

Final velocity (v) = 6.6 m/s

Time (t) = 6.5 s

Acceleration (a) =.?

a = (v – u) / t

a = (6.6 – 0) / 6.5

a = 6.6 / 6.5

a = 1.02 m/s²

Therefore, the acceleration of the car is 1.02 m/s²

3 0

3 years ago

What is the ideal banking angle for a gentle turn of 1.20-km radius on a highway with a 105 km/h speed limit (about 65 mi/h), as

Mnenie [13.5K]

Answer:

4.14°

Explanation:

given:

r = 1.2 km

v = 105 km/h

1) <em>convert your given </em>

a) r = 1.2 km to m = 1200m

b) v = 105 km/h to m/s = 29.2 m/s

2) <em>plug into your ideal banking angle equation</em>

$tan^-1$ ( $\frac{v^2}{rg}$ ) = $\frac{29.2^2}{(1200)(9.8)}$ = 4.14°

8 0

2 years ago

A proton is projected in the positive x direction into a region of a uniform electric field E S 5 126.00 3 105 2 i^ N/C at t 5 0

Dafna11 [192]

Answer:

(a). The magnitude of the acceleration of the proton is $5.74\times10^{13}\ m/s^2$

(b). The initial peed of the protion is $2.83\times10^{6}\ m/s$

(c). The time is $0.493\times10^{-7}\ sec$ .

Explanation:

Given that,

Electric field $E=-6.00\times10^{5}i\ N/C$

Time = 5.0 sec

Distance 7.00 cm

(a). We need to calculate the acceleration

Using formula of force

$F=F_{e}$

$ma=qE$

$a=\dfrac{Eq}{m}$

Where, E = electric field

m = mass of proton

q = charge of proton

Put the value into the formula

$a=\dfrac{-6.00\times10^{5}\times1.6\times10^{-19}}{1.67\times10^{-27}}$

$a=-5.74\times10^{13}\ m/s62$

The magnitude of the acceleration of the proton is $5.74\times10^{13}\ m/s^2$

(b). We need to calculate the initial peed

Using equation of motion

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

Where, s = distance

Put the value into the formula

$0-u^2=2\times(-5.74\times10^{13})\times7.00\times10^{-2}$

$u=\sqrt{2\times5.74\times10^{13}\times7.00\times10^{-2}}$

$u=2834783.94$

$u=2.83\times10^{6}\ m/s$

The initial peed of the protion is $2.83\times10^{6}\ m/s$

(c). We need to calculate the time interval over which the proton comes to rest

Using formula

$t=\dfrac{u}{a}$

Where, u = initial velocity

a = acceleration

Put the value into the formula

$t=\dfrac{2.83\times10^{6}}{5.74\times10^{13}}$

$t=0.493\times10^{-7}\ sec$

Hence, (a). The magnitude of the acceleration of the proton is $5.74\times10^{13}\ m/s^2$

(b). The initial peed of the protion is $2.83\times10^{6}\ m/s$

(c). The time is $0.493\times10^{-7}\ sec$ .

5 0

3 years ago

A cylinder of gas is at room temperature (20°C). The air conditioner breaks down, and the temperature rises

noname [10]

Answer: we divide this equation by pV and use {C}_{p}={C}_{V}+ . ... The temperature, pressure, and volume of the resulting gas-air mixture

Explanation:

4 0

3 years ago

Seen through a traditional optical telescope, the space between stars and galaxies is pitch black. But, with a radio telescope,

Gemiola [76]

That's the Cosmic Microwave Background (CMB). (D)

It doesn't "provide evidence for the ... Theory". The Big Bang theory is the best explanation (so far) for the CMB and a lot of other observations.

Evidence isn't used to prove theories. Theories are attempts to explain evidence.

8 0

3 years ago