Looters break a statue into pieces. how do you expect the weathering of pieces of rock to change

A car with speed v and an identical car with speed 2v both travel the same circular section of an unbanked road. If the friction

yawa3891 [41]

Answer:

$F'=\dfrac{F}{4}$

Explanation:

Let m is the mass of both cars. The first car is moving with speed v and the other car is moving with speed 2v. The only force acting on both cars is the centripetal force.

For faster car on the road,

$F=\dfrac{mv^2}{r}$

v = 2v

$F=\dfrac{m(2v)^2}{r}$

$F=4\dfrac{m(v)^2}{r}$ ..........(1)

For the slower car on the road,

$F'=\dfrac{mv^2}{r}$ ............(2)

Equation (1) becomes,

$F=4F'$

$F'=\dfrac{F}{4}$

So, the frictional force required to keep the slower car on the road without skidding is one fourth of the faster car.

8 0

4 years ago

A 12000 kg train engine moving at 2.2 m/s hits and locks into 3 boxcars with a total mass of 25000 kg sitting still. If the coll

PilotLPTM [1.2K]

Answer:

V = 0.714m/s

Explanation:

Full solution calculation can be found in the attachment below.

From the principle of conservation of linear momentum, the sum of momentum before collision equals the sum of momentum after collision.

Before collision only the train had momentum. After the collision the train and the boxcars stick together and move as one body. The initial momentum of the train is now shared with the boxcars as they move together as one body. The both move with a common velocity v.

See the attachment below for the solution calculation.

5 0

3 years ago

Read 2 more answers

A centrifuge rotor (hollow disk) is rotating at 10,000 rpm is shut off and brought to rest by a constant frictional torque of 1.

Y_Kistochka [10]

Answer:

The no of revolutions rotor turn before coming to rest is 1,601.1943 and time taken is equal to 19.21 seconds

Explanation:

here we know that torque = I×α

α= angular acceleration

I = moment of inertia of hollow disc = m× $k^{2}$

given that m=4.37kg

k=0.0710m

torque=1.2Nm

$w_{o}=10000\times \frac{\pi}{30} rad /s$

$\alpha =\frac{torque}{I}$

from the above equation we can calculate the angular acceleration of the hollow disc .

since $w^{2}-w_o^{2} =2\alpha$ $\theta$

from this above equation $\theta=\frac{w^{2}-w_{o}^{2}}{2\alpha }$

no of revolutions = $\frac{\theta}{2\pi}$ = 1,601.1943.

Now to calculate time we know that time = $\frac{2\theta}{w+w_{o}}$

so upon calculating we will be getting t=19.21 seconds

5 0

3 years ago

A car is traveling at 120 km/h (75 mph). When applied the braking system can stop the car with a deceleration rate of 9.0 m/s2.

Bumek [7]

Answer:

the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15

Explanation:

Given that;

speed of car V = 120 km/h = 33.3333 m/s

Reaction time of an alert driver = 0.8 sec

Reaction time of an alert driver = 3 sec

extra time taken by sleepy driver over an alert driver = 3 - 0.8 = 2.2 sec

now, extra distance that car will travel in case of sleepy driver will be'

S_d = V × 2.2 sec

S_d = 33.3333 m/s × 2.2 sec

S_d = 73.3333 m

hence, number of car of additional car length n will be;

n = S_n / car length

n = 73.3333 m / 5m

n = 14.666 ≈ 15

Therefore, the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15

8 0

3 years ago

A 190 g glider on a horizontal, frictionless air track is attached to a fixed ideal spring with force constant 160 N/m. At the i

laiz [17]

(a) Let x be the maximum elongation of the spring. At this point, the glider would have zero velocity and thus zero kinetic energy. The total work W done by the spring on the glider to get it from the given point (4.00 cm from equilibrium) to x is

W = - (1/2 kx ² - 1/2 k (0.0400 m)²)

(note that x > 4.00 cm, and the restoring force of the spring opposes its elongation, so the total work is negative)

By the work-energy theorem, the total work is equal to the change in the glider's kinetic energy as it moves from 4.00 cm from equilibrium to x, so

W = ∆K = 0 - 1/2 m (0.835 m/s)²

Solve for x :

- (1/2 (160 N/m) x ² - 1/2 (160 N/m) (0.0400 m)²) = -1/2 (0.190 kg) (0.835 m/s)²

==> x ≈ 0.0493 m ≈ 4.93 cm

(b) The glider attains its maximum speed at the equilibrium point. The work done by the spring as it is stretched away from equilibrium to the 4.00 cm position is

W = - 1/2 k (0.0400 m)²

If v is the glider's maximum speed, then by the work-energy theorem,

W = ∆K = 1/2 m (0.835 m/s)² - 1/2 mv ²

Solve for v :

- 1/2 (160 N/m) (0.0400 m)² = 1/2 (0.190 kg) (0.835 m/s)² - 1/2 (0.190 kg) v ²

==> v ≈ 1.43 m/s

(c) The angular frequency of the glider's oscillation is

√(k/m) = √((160 N/m) / (0.190 kg)) ≈ 29.0 Hz

3 0

3 years ago