All categories

2 years ago

The crankshaft in a race car goes from rest to 3000 rpm in 2.0 s.

Physics

2 answers:

Marta_Voda [28]2 years ago

5 0

Answer:

a. 157 rad/s²

b. 50 revolutions

Explanation:

Initial angular velocity, u = 0

Final v = 3000 rpm = 3000 × 2π/60 rad/s= 314 rad/s

time, t = 2.0 s

a. angular acceleration is given by first equation of rotational motion:

α = (v-u)/t = (314 rad/s-0)/ 2.0 s = 157 rad/s²

b. number of revolutions made before reaching final angular velocity of 3000 rpm. Time taken = 2.0 s.

Use second equation of rotational motion:

θ = u t + 0.5 α t²

⇒ θ = 0 + 0.5 × 157 rad/s² × (2.0 s.)² = 314 rad

⇒n ( number of revolutions) = θ / 2π = 314/ 2π = 50 revolutions

a_sh-v [17]2 years ago

3 0

The equations are analogous to that for linear movement:
acceleration = (final velocity - initial velocity) / time
acceleration = (3000 rpm - 0 rpm) / 2.0 s
a) acceleration = 1500 rpm/s or 25 rp(s^2)
For the displacement
displacement = initial velocity*time + 0.5*acceleration*time^2
displacement = (0)*(2 s) + (0.5)(25 rps^2)*(2 s)^2
b) displacement = 50 revolutions

You might be interested in

If a cell phone is dropped from a very tall building, how far has the phone fallen after 2.7 seconds, neglecting air resistance?

Zielflug [23.3K]

The free fall of the phone is an uniformly accelerated motion toward the ground, with constant acceleration equal to
$g=9.81 m/s^2$

So, assuming the downward direction as positive direction of the motion, since the phone starts from rest the distance covered by the phone after a time t is given by
$y(t) = \frac{1}{2}gt^2$
And if we substitute t=2.7 s, we find the distance covered:
$y(t)= \frac{1}{2}(9.81 m/s^2)(2.7 s)^2=35.8 m$

6 0

2 years ago

You are driving on the highway, and you come to a steep downhill section. As you roll down the hill, you take your foot off the

Vinvika [58]

Answer:

Weight of the car, normal force, drag force

Explanation:

The forces acting on the car are:

The normal force which acts perpendicularly to the downhill plane
The weight of the car which acts vertically downwards
The drag force due to air resistance which acts in opposition to the motion of the car

Friction is ignored, so the force due to friction is assumed negligible

6 0

2 years ago

Here's a question from ~ [ AIEEE 2002 ]

lyudmila [28]

r=150m
coefficient of friction= $\mu$ =0.6

As car is avoid skidding

$\\ \sf\hookrightarrow \dfrac{mv^2}{r}=\mu mg$

Cancel m

$\\ \sf\hookrightarrow \dfrac{v^2}{r}=\mu g$

$\\ \sf\hookrightarrow v^2=\mu rg$

$\\ \sf\hookrightarrow v^2=0.6(10)(150)$

$\\ \sf\hookrightarrow v^2=60(150)$

$\\ \sf\hookrightarrow v^2=900$

$\\ \sf\hookrightarrow v=30ms^{-1}$

Done

7 0

2 years ago

Read 2 more answers

Based on Newton’s 3rd Law of motion, if a baseball player hits a ball with the bat, the action force is the impact of the bat ag

kicyunya [14]

The forums is a is equal to f divided by m

6 0

2 years ago

Some1 help!!

Evgesh-ka [11]

There are four atoms that are found in the products.

When a molecule of I2 reacts with NaCl, we must first note that no reaction will take place. This is because Iodine is lower than chlorine in the series hence I2 can not displace chlorine from its salt.

Since we are still going to end up with the substances that we started with, we will have a total of four atoms.

Learn more: brainly.com/question/1388366

3 0

2 years ago