Ask question

Ask question

All categories

s344n2d4d5 [400]

3 years ago

15

A competitive go-cart driver is traveling 32 m/s. He sees a caution flag go up, so he slows at a rate of -1.5 m/s^2 in 10.8 s. W

hat is his final velocity?

1 answer:

Svetllana [295]3 years ago

5 0

Answer:

15.8 m/s

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 32 m/s.

Acceleration (a) = – 1.5 m/s²

Time (t) = 10.8 s.

Final velocity (v) =?

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

Acceleration (a) = [final velocity (v) – initial velocity (u)] / time (t)

a = (v – u) /t

With the above formula, we can obtain the final velocity of go-cart driver as follow:

Initial velocity (u) = 32 m/s.

Acceleration (a) = – 1.5 m/s²

Time (t) = 10.8 s.

Final velocity (v) =?

a = (v – u) /t

– 1.5 = (v – 32) / 10.8

Cross multiply

(v – 32) = –1.5 × 10.8

v – 32 = – 16.2

Collect like terms

v = – 16.2 + 32

v = 15.8 m/s

Therefore, the final velocity of go-cart driver is 15.8 m/s.

You might be interested in

How long will it take to melt 3500 kg of steel in a 50 kW oven operating at 65% efficiency from room temperature to 1600°C?

zepelin [54]

Answer:

71195 seconds or 19.8 hours

Explanation:

Because the steel will be melted, beginning from room temperature, energy is needed to heat it from room temperature to 1600°C and another to melt it at 1600°C.

We take room temperature to be 27°C, the specific heat capacity of steel to be 420 J/kg/K and the specific latent heat of fusion of steel to be 440 J/kg.

The heat required to heat the steel from room temperature to 1600°C is given by

$H_S = mc_S\delta\theta$

$m$ is the mass, $c_S$ is the specific heat capacity and $\delta\theta$ is the change in temperature.

$H_S = 3500\times420\times(1600-27) = 2312310000$

The heat required to melt it at 1600°C is

$H_L = ml_S$

$m$ is the mass and $l_S$ is the specific latent heat of fusion of steel.

$H_L = 3500\times440 = 1540000$

Adding both, the heat required is $2312310000+1540000=2313850000\text{ J}$

Because the oven is 65% efficient, its output power = $65\%\times50000 = 32500 \text{ W}$

Now, energy = power * time

Time = Energy/power

$t = \dfrac{2313850000}{32500}=71195 \text{ seconds} = 19.8 \text{ seconds}$

4 0

3 years ago

This type of telescope uses a lens to collect<br> light.

bezimeni [28]

Optical Telescopes....................................

7 0

4 years ago

Https://student.masteryconnect

maria [59]

Answer:

a nucleus

Explanation:

ik.... ....... v

4 0

3 years ago

A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio

Gnesinka [82]

Answer:

a). $a_p=-2.39x10^{-12} rad/s^2$

b). $t=1016298.8 years$

c). $T_i=80.58x10^{-3}s$

Explanation:

a).

The acceleration for definition is the derive of the velocity so:

$a_p=\frac{dw}{dt}$

$w=\frac{2\pi}{t}$

$a_p=\frac{dw}{dt}=-\frac{2\pi}{t^2}*\frac{dT}{dt}$

$dT=0.0808s$

$dt=1 year*\frac{365d}{1year} \frac{24hr}{1d} \frac{60minute}{1hr} \frac{60s}{1minute}=31.536x10^{6}s$

Replacing

$a_p=-\frac{2\pi}{0.082s^2}*\frac{9.84x10^{-7}}{31.536x10^{6}s}= -2.39x10^{-12} rad/s^2$

b).

If the pulsar will continue to decelerate at this rate, it will stop rotating at time:

$t=\frac{w}{a_p}$

$w=\frac{2\pi }{t}=\frac{2\pi }{0.0820s}=76.62 rad/s$

$t=\frac{76.62 rad/s}{2.39x10^{-12}rad/s^2}= 3.2058x10^{13}s$

$t=1016298.8 years$

c).

582 years ago to 2019

1437

$T_i=0.0820-9.84x10^{-7}*1437)=80.58x10^{-3}s$

5 0

3 years ago

- Atoms of different elements:

TiliK225 [7]

(C) have the same heat capacity

3 0

3 years ago