What is the weight of a 20 kg box on the Earth?<br> A) 2 N <br> B) 20 N<br> C) 200 N

A system absorbs 194 kj of heat and the surroundings do 120 kj of work on the system. internal eneergy change

notsponge [240]

We can solve the problem by using the first law of thermodynamics, which states that:
$\Delta U = Q-W$
where
$\Delta U$ is the change in internal energy of the system
Q is the heat absorbed by the system
W is the work done by the system

In our problem, the heat absorbed by the system is Q=+194 kJ, while the work done is W=-120 kJ, where the negative sign means the work is done by the surroundings on the system. Therefore, the variation of internal energy is
$\Delta U= Q-W=+194 kJ - (-120 kJ)=+314 kJ$

6 0

3 years ago

A substance with a define shape and volume is a

Vlad1618 [11]

Solid is the answer.

8 0

3 years ago

Bus starts from rest if the acceleration of the bus is 0.5 MS squared what will be the velocity at the end of two minutes and wh

Nutka1998 [239]

Explanation:

Given that,

Initial speed of the bus, u = 0

Acceleration of the bus, a = 0.5 m/s²

Let v is the velocity at the end of 2 minutes. The change in velocity divided by time equals acceleration.

So,

$a=\dfrac{v-u}{t}\\\\v=u+at\\\\v=0+0.5\times 120\\\\v=60\ m/s$

Let d is the distance cover during that time. So,

$v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{(60)^2}{2\times 0.5}\\\\d=3600\ m$

So, the final speed is 60 m/s and the distance covered during that time is 3600 m.

4 0

3 years ago

A car's bumper is designed to withstand a 5.04 km/h (1.4-m/s) collision with an immovable object without damage to the body of t

emmasim [6.3K]

Answer:

the magnitude of the average force on the bumper is 3189.8 N

Explanation:

Given the data in the question;

In terms of force and displacement, work done is;

W = $F^>$ × $x^>$

W = $Fxcos\theta$ ------- let this be equation 1

where F is force applied, x is displacement and θ is angle between force and displacement.

Now, since the displacement of the bumper and force acting on it is in the same direction,

hence, θ = 0°

we substitute into equation 1

W = $Fxcos($ 0° $)$

W = $Fx$ ------- let this be equation 2

Now, using work energy theorem,

total work done on the system is equal to the change in kinetic energy of the system.

$W_{net$ = ΔKE

= $\frac{1}{2}$ mv² - $\frac{1}{2}$ mu² --------- let this be equation 3

where m is mass of object, v is final velocity, u is initial velocity.

from equation 2 and 3

$Fx$ = $\frac{1}{2}$ mv² - $\frac{1}{2}$ mu²

we make F, the subject of formula

F = $\frac{m}{2x}$ ( v² - u² )

given that mass of car m = 830 kg, x = 0.255 m, v = 0 m/s, and u = 1.4 m/s

so we substitute

F = $\frac{830}{(2)(0.255)}$ ( (0)² - (1.4)² )

F = 1627.45098 ( 0 - 1.96 )

F = 1627.45098 ( - 1.96 )

F = -3189.8 N

The negative sign indicates that the direction of the force was in opposite compare to the direction of the velocity of the car.

Therefore, the magnitude of the average force on the bumper is 3189.8 N

6 0

2 years ago

A thin rod (length = 1.09 m) is oriented vertically, with its bottom end attached to the floor by means of a frictionless hinge.

Murljashka [212]

Answer:

a) w = 4.24 rad / s , b) α = 8.99 rad / s²

Explanation:

a) For this exercise we use the conservation of kinetic energy,

Initial. Vertical bar

Emo = U = m g h

Final. Just before touching the floor

Emf = K = ½ I w2

As there is no friction the mechanical energy is conserved

Emo = emf

mgh = ½ m w²

The moment of inertial of a point mass is

I = m L²

m g h = ½ (m L²) w²

w = √ 2gh / L²

The initial height h when the bar is vertical is equal to the length of the bar

h = L

w = √ 2g / L

Let's calculate

w = RA (2 9.8 / 1.09)

w = 4.24 rad / s

b) Let's use Newton's equation for rotational motion

τ = I α

F L = (m L²) α

The force applied is the weight of the object, which is at a distance L from the point of gro

mg L = m L² α

α = g / L

α = 9.8 / 1.09

α = 8.99 rad / s²

5 0

3 years ago

What is the weight of a 20 kg box on the Earth? A) 2 N B) 20 N C) 200 N