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Dmitry_Shevchenko [17]

3 years ago

10

Calculate the translational speed of a cylinder when it reaches the foot of an incline 7.35 m high. assume it starts from rest a

nd rolls without slipping.

1 answer:

allochka39001 [22]3 years ago

6 0

The total kinetic energy will be equivalent to the potential energy;
The total kinetic energy is given by the sum of transitional kinetic energy and the rotational kinetic energy.
Rotational kinetic energy
I= 1/2mR², but rotational energy =1/2 I w²
Thus rotational energy is equal to 1/4 mR²w², which is equivalent to
= 1/4 mv² since v= wR
Thus;
Total kinetic energy = 1/2 mv² + 1/4 mv²
hence;
Mgh = 3/4 mv²
gh = 3/4 v² ( assuming g = 9.81 m/s²)
72.1035 = 3/4 v²
v = 9.805
= 9.805 m/s

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A 20 metric ton train moves toward the south at 50 m/s. At what speed must it travel to have four times its original momentum

OverLord2011 [107]

Answer:

200 m/s

Explanation:

as momentum is a product of mass and speed, and mass is not changing, four times the speed will result in four times the momentum.

p = mv

4p = m(4v)

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2 years ago

Influenced by the gravitational pull of a distant star, the velocity of an asteroid changes from from +19.3 km/s to −18.8 km/s o

muminat

As per above given data

initial velocity = 19.3 km/s

final velocity = - 18.8 km/s

now in order to find the change in velocity

$\Delta v = v_f - v_i$

$\Delta v = -18.8 - 19.3$

$\Delta v = -38.1 km/s$

$\Delta v = -3.81 * 10^4 m/s$

Part b)

Now we need to find acceleration

acceleration is given by formula

$a = \frac{\Delta v}{\Delta t}$

given that

$\Delta v =- 3.81 * 10^4 m/s$

$\Delta t = 2.07 years = 6.53 * 10^7 s$

now the acceleration is given as

$a = \frac{-3.81 * 10^4}{6.53 * 10^7}$

$a = - 5.84 * 10^{-4}m/s^2$

so above is the acceleration

4 0

3 years ago

How does the mass of jupiter compare to the rest of the planets in the solar system?

katen-ka-za [31]

It is the most massive planet in the solar system.

3 0

3 years ago

A child whose weight is 276 N slides down a 5.90 m playground slide that makes an angle of 34.0° with the horizontal. The coeffi

garik1379 [7]

Answer:

Explanation:

Frictional force acting on the child = μ mg cosθ

, μ is coefficient of kinetic friction , m is mass of child θ is inclination

work done by frictional force

μ mg cosθ x d , d is displacement on inclined plane

work done = .13 x 276 x cos34 x 5.9

= 175.5 J

This work will be converted into heat energy.

b ) Initial energy of child = mgh + 1/2 m v ² , h is height , v is initial velocity

= 276 x 5.9 sin34 + 1/2 x 276 / 9.8 x .518² [ mass m = 276 / g ]

= 910.59 + 3.77

= 914.36 J

loss of energy due to friction = 175.5

Net energy at the bottom

= 738.86 J

If v be the velocity at the bottom

1/2 m v² = 738 .86

.5 x (276 / 9.8) x v² = 738.86

v² = 52.47

v = 7.24 m /s .

4 0

3 years ago

Read 2 more answers

A certain freely falling object, released from rest, requires 1.95 s to travel the last 23.5 m before it hits the ground. (a) Fi

Ratling [72]

Answer:

(a). The velocity of the object is -2.496 m/s.

(b). The total distance of the object travels during the fall is 23.80 m.

Explanation:

Given that,

Time = 1.95 s

Distance = 23.5 m

(a). We need to calculate the velocity

Using equation of motion

$s = ut+\dfrac{1}{2}gt^2$

Put the value into the formula

$-23.5=u\times1.95+\dfrac{1}{2}\times(-9.8)\times(1.95)^2$

$u=\dfrac{-23.5+4.9\times(1.95)^2}{1.95}$

$u=-2.496\ m/s$

(b). We need to calculate the total distance the object travels during the fall

Using equation of motion

$v = u+gt$

Put the value in the equation

$-2.496=0-9.8\times t$

$t =\dfrac{2.496}{9.8}$

$t=0.254\ sec$

The total time is

$t'=t+1.95$

$t'=0.254+1.95$

$t'=2.204\ sec$

We need to calculate the distance

Using equation of motion

$s = ut+\dfrac{1}{2}gt^2$

Put the value into the formula

$s=0+\dfrac{1}{2}\times9.8\times(2.204)^2$

$s=23.80\ m$

Hence, (a). The velocity of the object is -2.496 m/s.

(b). The total distance of the object travels during the fall is 23.80 m.

4 0

3 years ago