Ask question

Ask question

All categories

4 years ago

5

A 2kg box is pushed along a flat frictionless surface with an applied force of 53.91newton j+19.62 j were j is horizontal and j

is vertical if the box moves 10m horizontal in 90 seconds what is the work and power that is required to complete the movement

1 answer:

tamaranim1 [39]4 years ago

5 0

Since the box doesn't move vertically at all, no work is done by the vertical component of the force.

If 53.91 Newtons is the horizontal component of the force (very unclear in the question), then the work done is

<u>Work = (force) x (distance)</u>

Work = (53.91 N) x (10 m)

<em>Work = 539.1 Joules</em>

<u>Power = (work done) / (time to do the work)</u>

Power = (539.1 joules) / (90 sec)

Power = (539.1/90) (joule/sec)

<em>Power = 5.99 watts</em>

=====================================

Note: If the mass of the box is only 2 kg, and you push it along the surface with a constant force of 53⁺ Newtons, and the surface really is frictionless, then that box is gonna cover a WHOOOOOLE LOT more than 10 meters in 90 seconds. I get 109,168 meters !

You might be interested in

What is the frequency of a wave that has a wavelength of 0.39 m and a speed

gogolik [260]

Answer:

<h3>The answer is option B</h3>

Explanation:

The frequency of a wave can be found by using the formula

$f = \frac{c}{ \lambda} \\$

where

c is the velocity

From the question

wavelength = 0.39 m

c = 86 m/s

We have

$f = \frac{86}{0.39} \\ = 220.512820...$

We have the final answer as

<h3>200 Hz</h3>

Hope this helps you

7 0

3 years ago

After a long day you go home in your friend's really fancy sports car which has a sun-roof on the top and a spoiler (a little wi

mixas84 [53]

Answer:

B: air pressure inside the car drops suddenly

Explanation:

Air in the car drops suddenly because the roof region has lowered pressure than the atmospheric pressure usually varying with the speed of the car

7 0

3 years ago

A block oscillating up and down on a spring comes to the top of its path, where it is momentarily at rest. At the instant it is

True [87]

Opposite to the direction of the velocity which led it to its current position.

Explanation:

The direction of momentum when a vertically oscillating block comes to the rest momentarily will be opposite to the direction of the velocity that it has just followed to reach reach its current position.

The direction of change in momentum at the bottom will be upwards and at the top will be downwards.

The change in momentum is mathematically defined as:

$\Delta P=m.v_f-m.v_i$

where:

$m=$ mass of the block

$v_f=$ final velocity of the block

$v_i=$ initial velocity of the block

When the block comes to rest it is due to the result of continuously decreasing velocity.

7 0

3 years ago

Cells store energy in the form of what

lukranit [14]

Cells store energy either as carbohydrates or lipids (fats and oils).

Carbohydrates provide for fast, short term energy. They are a key part of cellular respiration.

Lipids provide for long term energy storage because storing lipids is much more effective than storing carbs. One gram of lipid can store twice as much as one gram of carbs. So to access this energy storage, cells must conduct hydrolysis (inserting a hydrogen and a hydroxide group to split off the energy).

Hope this helps!

3 0

3 years ago

wo parallel-plate capacitors C1 and C2 are connected in parallel to a 12.0 V battery. Both capacitors have the same plate area o

tresset_1 [31]

Complete Question

Two parallel-plate capacitors C1 and C2 are connected in parallel to a 12.0 V battery. Both capacitors have the same plate area of 5.30 cm2 and plate separation of 2.65 mm. However, the first capacitor C1 is filled with air, while the second capacitor C2 is filled with a dielectric that has a dielectric constant of 2.10.

(a) What is the charge stored on each capacitor

(b) What is the total charge stored in the parallel combination?

Answer:

a

i $Q_1 = 2.124 *10^{-11} \ C$

ii $Q_2 = 4.4604 *10^{-11} \ C$

b

$Q_{eq} = 6.5844 *10^{-11} \ C$

Explanation:

From the question we are told that

The voltage of the battery is $V = 12.0 \ V$

The plate area of each capacitor is $A = 5.30 \ cm^2 = 5.30 *10^{-4} \ m^2$

The separation between the plates is $d = 2.65 \ mm = 2.65 *10^{-3} \ m$

The permittivity of free space has a value $\epsilon_o = 8.85 *10^{-12} \ F/m$

The dielectric constant of the other material is $z = 2.10$

The capacitance of the first capacitor is mathematically represented as

$C_1 = \frac{\epsilon * A }{d }$

substituting values

$C_1 = \frac{8.85 *10^{-12 } * 5.30 *10^{-4} }{2.65 *10^{-3} }$

$C_1 = 1.77 *10^{-12} \ F$

The charge stored in the first capacitor is

$Q_1 = C_1 * V$

substituting values

$Q_1 = 1.77 *10^{-12} * 12$

$Q_1 = 2.124 *10^{-11} \ C$

The capacitance of the second capacitor is mathematically represented as

$C_2 = \frac{ z * \epsilon * A }{d }$

substituting values

$C_1 = \frac{ 2.10 *8.85 *10^{-12 } * 5.30 *10^{-4} }{2.65 *10^{-3} }$

$C_1 = 3.717 *10^{-12} \ F$

The charge stored in the second capacitor is

$Q_2 = C_2 * V$

substituting values

$Q_2 = 3.717*10^{-12} * 12$

$Q_2 = 4.4604 *10^{-11} \ C$

Now the total charge stored in the parallel combination is mathematically represented as

$Q_{eq} = Q_1 + Q_2$

substituting values

$Q_{eq} = 4.4604 *10^{-11} + 2.124*10^{-11}$

$Q_{eq} = 6.5844 *10^{-11} \ C$

7 0

4 years ago