Ask question

Ask question

All categories

3 years ago

8

g Initially, the motorcycle travels along a straight road with a speed of 35 m/s (this is almost 80 mph). The maximum decelerati

on of the motorcycle is -1.2 m/s2. If the biker is skilled and uses maximum deceleration, how long will it take them to come to a complete stop

1 answer:

astraxan [27]3 years ago

4 0

Given:

Initial speed of the motorcycle (u) = 35 m/s

Final speed of the motorcycle (v) = 0 m/s (Complete Stop)

Maximum deceleration of the motorcycle (a) = -1.2 m/s²

Required Equation:

$\boxed{\bf{ v = u + at}}$

Answer:

By substituting values in the equation, we get:

$\rm \longrightarrow 0 = 35 + ( - 1.2)t \\ \\ \rm \longrightarrow 0 = 35 - 1.2t \\ \\ \rm \longrightarrow 35 - 1.2t = 0 \\ \\ \rm \longrightarrow 35- 35 - 1.2t = 0 - 35 \\ \\ \rm \longrightarrow - 1.2t = - 35 \\ \\ \rm \longrightarrow \dfrac{ - 1.2t}{ - 1.2} = \dfrac{ - 35}{ - 1.2} \\ \\ \rm \longrightarrow t = 29.167 \: s$

$\therefore$ Time taken by motorcycle to come to a complete stop (t) = 29.167 s

You might be interested in

ANSWER ASAP PLS

posledela

Answer:

<h3>JAWAB SECEPATNYA pliss</h3><h3 /><h3>Anda memiliki rangkaian paralel 10 volt, dengan 2 resistor di atasnya. Berapakah tegangan pada</h3><h3>resistor pertama? Di seberang kedua?</h3><h3 /><h3>(saya akan menandai tercerdas tolong bantu)</h3>

Explanation:

Hukum Ohm

= tegangan

= kuat arus

= ketahanan

Kalau kamu mau mencari tegangan listrik, kamu gunakan rumus V = I.R. Kalau ternyata kamu perlu mencari kuat arus listrik, maka gunakan rumus I = V/R. Nah, kalau yang kamu cari adalah hambatan listrik, maka gunakan rumus R = V/I.

4 0

2 years ago

Draw a schematic diagram of a circuit consisting of a battery of five 2 V cells, a 5 ohm resistor, a 10 ohm resistor, a 15 ohm r

MakcuM [25]

Answer:

Current = 0.33 A

Explanation:

For diagram refer the attachment.

It is given that five cells of 2V are connected in series, so total voltage of the battery:

$\dashrightarrow \: \: \sf V = 2 \times 5 = 10 V$

Three resistor of 5 $\Omega$ , 10 $\Omega$ , 15 $\Omega$ are connected in Series, so the net resistance:

$\dashrightarrow \: \: \sf R_{n} = R_{1} + R_{2} + R_{3}$

$\dashrightarrow \: \: \sf R = 5 + 10 + 15$

${ \pink{\dashrightarrow \sf \: \: { \underbrace{R = 30 \: \Omega}}}}$

According to ohm's law:

$\dashrightarrow \sf\: \: V = IR$

$\dashrightarrow \sf \: \: I = \dfrac{V}{R}$

On substituting resultant voltage (V) as 10 V and resultant resistant, as 30 ${\pmb{\sf{\Omega}}}$ we get:

$\dashrightarrow \sf \: \: I = \dfrac{10V}{30\Omega}$

${ \pink{\dashrightarrow \sf \: \: { \underbrace{I = 0.33 A}}}}$

$\therefore$ The electric current passing through the above circuit when the key is closed will be <u>0.33 A</u>

4 0

2 years ago

Read 2 more answers

The frequency of a wave is 560 Hz. What is it’s period

Crazy boy [7]

The answer would be, "1/560 seconds".

4 0

3 years ago

A high powered rifle can shoot a bullet at a speed of 1500 mi/hr. On the moon, with almost no atmosphere and an acceleration due

Amiraneli [1.4K]

Answer:

Explanation:

On the Moon :----

1500 x 1.6 = 2400 m /s is initial velocity of bullet .

g = 1.6 m /s²

v = u - gt

0 = 2400 - 1.6 t

t = 1500 s

This is time of ascent

Time of decent will also be the same

Total time of flight = 2 x 1500 = 3000 s

On the Earth : ---

v = u - a₁ t

0 = u - a₁ x 18

u = 18a₁

v² = u² - 2 x a₁ x 2743.2

0 = (18a₁ )² - 2 x a₁ x 2743.2

a₁ = 16.93

For downward return

s = ut + 1/2 a₂ x t²

2743.2 = 0 + .5 x a₂ x 31²

a₂ = 5.7 m /s²

If d be the deceleration produced by air

g + d = 16.93 ( during upward journey )

g - d = 5.7

g = (16.93 + 5.7) / 2

= 11.315 m / s

d = 5.6 m /s²

So air is creating a deceleration of 5.6 m /s².

6 0

3 years ago

A flywheel turns through 40 rev as it slows from an angular speed of 1.5 rad/s to a stop.

Lynna [10]

Answer:

-0.0047 rad/s²

335.103 seconds

99.18 seconds

Explanation:

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity = 1.5 ra/s

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation = 40 rev

t = Time taken

Equation of rotational motion

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-1.5^2}{2\times 2\pi \times 40}\\\Rightarrow \alpha=-0.0047\ rad/s^2$

Acceleration while slowing down is -0.0047 rad/s²

$t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-1.5}{-0.0047}\\\Rightarrow t=335.103\ s$

Time taken to slow down is 335.103 seconds

$\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow 20\times 2\pi=1.5\times t+\frac{1}{2}\times -0.0047\times t^2\\\Rightarrow 0.00235t^2-1.5t+125.66=0$

Solving the equation

$t=\frac{-\left(-1.5\right)+\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}, \frac{-\left(-1.5\right)-\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}\\\Rightarrow t=539.11, 99.18\ s$

The time required for it to complete the first 20 is 99.18 seconds as 539.11>335.103

4 0

3 years ago