Please help on this one? The mass on the spring in this diagram is in harmonic motion between x = -5cm and x=5cm

Calculate the kinetic energy in joules of an automobile weighing 2135 lb and traveling at 55 mph. (1 mile = 1.6093 km, 1 lb = 45

victus00 [196]

<span>Let's convert the speed to m/s: speed = (55 mph) (1609.3 m / mile) (1 hour / 3600 seconds) speed = 24.59 m/s Let's convert the mass to kilograms: mass = (2135 lb) (0.45359 kg / lb) mass = 968.4 kg We can find the kinetic energy KE: KE = (1/2) m v^2 KE = (1/2) (968.4 kg) (24.59 m/s)^2 KE = 292780 joules The kinetic energy of the automobile is 292780 joules.</span>

4 0

3 years ago

A truck accidentally rolls down a driveway for 8.0\,\text m8.0m8, point, 0, start text, m, end text while a person pushes agains

gavmur [86]

Answer:

Explanation:

According to work energy theorem

change in kinetic energy of truck = work done against it

work done against it = force x displacement

= - 850 x 8 = 6800 J

change in kinetic energy of truck = - 6800 J .

energy will be reduced by 6800 J

5 0

3 years ago

9. (02.04 LC)

ycow [4]

Answer:

Molecules speed up

Explanation:

This is caused because of the temperature increasing. The temperature increase is telling us that the thermal energy of the reaction is increasing. When the energy is increased molecules increase their speed, because they have more energy in them

7 0

3 years ago

If you place 1 C of positive charge on Earth and 1 C of negative charge on the moon, 384,500 km away, how much force would the p

ivolga24 [154]

Answer:

6.1 x 10^-8 newtons

Explanation:

F = 8.98 *109 *1*1/3845000002

4 0

2 years ago

A worker lifts a 20.0-kg bucket of concrete from the ground up to the top of a 25.0-m tall building. The bucket is initially at

yuradex [85]

Answer:

Minimum work = 5060 J

Explanation:

Given:

Mass of the bucket (m) = 20.0 kg

Initial speed of the bucket (u) = 0 m/s

Final speed of the bucket (v) = 4.0 m/s

Displacement of the bucket (h) = 25.0 m

Let 'W' be the work done by the worker in lifting the bucket.

So, we know from work-energy theorem that, work done by a force is equal to the change in the mechanical energy of the system.

Change in mechanical energy is equal to the sum of change in potential energy and kinetic energy. Therefore,

$\Delta E=\Delta U+\Delta K\\\\\Delta E= mgh+\frac{1}{2}m(v^2-u^2)$

Therefore, the work done by the worker in lifting the bucket is given as:

$W=\Delta E\\\\W=mgh+\frac{1}{2}m(v^2-u^2)$

Now, plug in the values given and solve for 'W'. This gives,

$W=(20\ kg)(9.8\ m/s^2)(25\ m)+\frac{1}{2}(20\ kg)(4^2-0^2)\ m^2/s^2\\\\W=4900\ J +160\ J\\\\W=5060\ J$

Therefore, the minimum work that the worker did in lifting the bucket is 5060 J.

7 0

3 years ago