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3 years ago

A bus slams on its breaks and goes from 30 km/hr to 15km/hr in 4 seconds. What is its acceleration?

1 answer:

5 0

Acceleration = (change in speed) / (time for the change).

Change in speed = (15 - 30) = -15 km/hr
Time for the change = 4 sec

Acceleration = (-15 km/hr) / (4 sec) = -3.75 km/hr per second .

Is that a lot ? Not much ?

Let's convert it to a unit that we can think about:

(-15 km/hr) x (1,000 m/km) x (1 hr / 3,600 sec) =

(-15 x 1,000) / (3,600) = -(4 and 1/6) m/sec .

So the acceleration of the bus is -(4 and 1/6) m/sec² .

The negative sign means that it slowed down.

(4 and 1/6) m/sec² is about 42% of the acceleration of gravity ...
the acceleration the bus would have if it drove off of a cliff.

When the car or the bus you're riding in slows down at that rate,
you feel 42% of your weight pulling you forward against your
seat belt. That's quite a drastic acceleration !

You might be interested in

How does an object's acceleration change if the net

topjm [15]

C.the acceleration is doubled

5 0

3 years ago

A copper rod of cross-sectional area 11.6 cm2 has one end immersed in boiling water and the other in an ice-water mixture, which

julia-pushkina [17]

Answer:

0.686 g of ice melts each second.

Solution:

As per the question:

Cross-sectional Area of the Copper Rod, A = $11.6\ cm^{2} = 11.6\times 10^{- 4}\ m^{2}$

Length of the rod, L = 19.6 cm = 0.196 m

Thermal conductivity of Copper, K = $390\ W/m.^{\circ}C$

Conduction of heat from the rod per second is given by:

$q = \frac{KA\Delta T}{L}$

where

$\Delta T = 100^{\circ} - 0^{\circ} = 100^{\circ}C$ = temperature difference between the two ends of the rod.

Thus

$q = \frac{390\times 11.6\times 10^{- 4}\times 100}{0.196} = 228.48\ J/s$

Now,

To calculate the mass, M of the ice melted per sec:

$M = \frac{q}{L_{w}}$

where

$L_{w}$ = Latent heat of fusion of water = 333 kJ/kg

$M = \frac{228.48}{333\times 10^{3}} = 6.86\times 10^{- 4}\ kg = 0.686\ g$

5 0

3 years ago

A solid sphere, a solid disk, and a thin hoop are all released from rest at the top of the incline (h0 = 20.0 cm).

Ede4ka [16]

Answer:

a. The object with the smallest rotational inertia, the thin hoop

b. The object with the smallest rotational inertia, the thin hoop

c. The rotational speed of the sphere is 55.8 rad/s and Its translational speed is 1.67 m/s

Explanation:

a. Without doing any calculations, decide which object would be spinning the fastest when it gets to the bottom. Explain.

Since the thin has the smallest rotational inertia. This is because, since kinetic energy of a rotating object K = 1/2Iω² where I = rotational inertia and ω = angular speed.

ω = √2K/I

ω ∝ 1/√I

since their kinetic energy is the same, so, the thin hoop which has the smallest rotational inertia spins fastest at the bottom.

b. Again, without doing any calculations, decide which object would get to the bottom first.

Since the acceleration of a rolling object a = gsinФ/(1 + I/MR²), and all three objects have the same kinetic energy, the object with the smallest rotational inertia has the largest acceleration.

This is because a ∝ 1/(1 + I/MR²) and the object with the smallest rotational inertia has the smallest ratio for I/MR² and conversely small 1 + I/MR² and thus largest acceleration.

So, the object with the smallest rotational inertia gets to the bottom first.

c. Assuming all objects are rolling without slipping, have a mass of 2.00 kg and a radius of 3.00 cm, find the rotational and translational speed at the bottom of the incline of any one of these three objects.

We know the kinetic energy of a rolling object K = 1/2Iω² + 1/2mv² where I = rotational inertia and ω = angular speed, m = mass and v = velocity of center of mass = rω where r = radius of object

The kinetic energy K = potential energy lost = mgh where h = 20.0 cm = 0.20 m and g = acceleration due to gravity = 9.8 m/s²

So, mgh = 1/2Iω² + 1/2mv² = 1/2Iω² + 1/2mr²ω²

Let I = moment of inertia of sphere = 2mr²/5 where r = radius of sphere = 3.00 cm = 0.03 m and m = mass of sphere = 2.00 kg

So, mgh = 1/2Iω² + 1/2mr²ω²

mgh = 1/2(2mr²/5 )ω² + 1/2mr²ω²

mgh = mr²ω²/5 + 1/2mr²ω²

mgh = 7mr²ω²/10

gh = 7r²ω²/10

ω² = 10gh/7r²

ω = √(10gh/7) ÷ r

substituting the values of the variables, we have

ω = √(10 × 9.8 m/s² × 0.20 m/7) ÷ 0.03 m

= 1.673 m/s ÷ 0.03 m

= 55.77 rad/s

≅ 55.8 rad/s

So, its rotational speed is 55.8 rad/s

Its translational speed v = rω

= 0.03 m × 55.8 rad/s

= 1.67 m/s

So, its rotational speed is of the sphere is 55.8 rad/s and Its translational speed is 1.67 m/s

6 0

2 years ago

What happens to the mechanical energy of an apple as it falls from a tree?

Vikentia [17]

Answer:

When the apple falls from the tree, it has some gravitational potential energy due to its height. Now, When it starts falling, the Gravitation Potential energy will starts converting into the Kinetic Energy. When the apple is about to strikes the ground, the Gravitational Potential energy have been converted into the Kinetic Energy.

Explanation:

4 0

2 years ago

A 1200 kg car accelerates from 13 m/s to 17 m/s. find the change in momentum of the car.

mixas84 [53]

P=4800kgm/s
As
p=mΔv
where p is momentum, m is mass and v is velocity
Given values is
m =1200kg
Δv= 17m/s-13m/s=4m/s
Now
p=mΔv
p=(1200kg)*(4m/s)
p=4800kgm/s

8 0

2 years ago