A spring is 4 cm long. A student hangs a weight of 2N. It is now 5.5 cm long.

Three moles of a monatomic ideal gas are heated at a constant volume of 1.20 m3. The amount of heat added is 5.22x10^3 J.(a) Wha

k0ka [10]

Answer:

A) 140 k

b ) 5.22 *10^3 J

c) 2910 Pa

Explanation:

Volume of Monatomic ideal gas = 1.20 m^3

heat added ( Q ) = 5.22*10^3 J

number of moles (n) = 3

A ) calculate the change in temp of the gas

since the volume of gas is constant no work is said to be done

heat capacity of an Ideal monoatomic gas ( Q ) = n.(3/2).RΔT

make ΔT subject of the equation

ΔT = Q / n.(3/2).R

= (5.22*10^3 ) / 3( 3/2 ) * (8.3144 J/mol.k )

= 140 K

B) Calculate the change in its internal energy

ΔU = Q this is because no work is done

therefore the change in internal energy = 5.22 * 10^3 J

C ) calculate the change in pressure

applying ideal gas equation

P = nRT/V

therefore ; Δ P = ( n*R*ΔT/V )

= ( 3 * 8.3144 * 140 ) / 1.20

= 2910 Pa

3 0

3 years ago

PART 2 OF ENERGY AND FORCES UNIT TEST

katrin2010 [14]

Answer:

1. at least two charged interacting parts

2. from the electric fields of charged subatomic particles

3 an arrow released from the bow

4Electrical fields of charged particles interact, bonding those with opposite charges.

5 the interaction of the electric fields of protons and electrons

6 The energy stored in the system increases.

7 Kinetic energy increases because the magnets move in the direction of the field.

8 Iron pieces accelerate toward the magnet, and the energy stored in the system decreases.

9

The energy stored in the field decreases because the magnet moves in the direction of the field.

10 The energy stored increases and then decreases.

11 The wire was not connected to the source.

12 The electromagnet will become more powerful.

the rest are written, hope this helps (:

4 0

3 years ago

A 150kg motorcycle starts from rest and accelerates at a constant rate along a distance of 350m. The applied force is 250N and t

notsponge [240]

A) The net force on the motorbike is 205.9 N

B) The acceleration of the motorbike is $1.37 m/s^2$

C) The final speed is 5.2 m/s

D) The elapsed time is 3.80 s

Explanation:

A)

There are two forces acting on the motorbike:

- The applied force, F = 250 N, forward

- The frictional force, $F_f$ , backward

The frictional force can be written as

$F_f = \mu mg$

where

$\mu=0.03$ is the coefficient of kinetic friction

$m=150 kg$ is the mass of the motorbike

$g=9.8 m/s^2$ is the acceleration of gravity

Therefore the net force is given by

$\sum F = F - F_f = F - \mu mg$

And substituting, we find

$\sum F=250 - (0.03)(150)(9.8)=205.9 N$

2)

The acceleration of the motorbike can be found by using Newton's second law, which states that the net force is equal to the product between mass and acceleration:

$\sum F = ma$

where

m is the mass

a is the acceleration

In this problem, we have

$\sum F = 205.9 N$ is the net force

m = 150 kg is the mass

Solving for a, we find the acceleration:

$a=\frac{\sum F}{m}=\frac{205.9}{150}=1.37 m/s^2$

C)

Since the motion of the motorbike is a uniformly accelerated motion, we can use the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For this motorbike, we have:

u = 0 (it starts from rest)

$a=1.37 m/s^2$

s = 350 m

Solving for v,

$v=\sqrt{u^2+2as}=\sqrt{0+2(1.37)(9.8)}=5.2 m/s$

4)

For this part of the problem, we can use the following suvat equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the elapsed time

Here we have:

v = 5.2 m/s

u = 0

$a=1.37 m/s^2$

Solving for t, we find

$t=\frac{v-u}{a}=\frac{5.2-0}{1.37}=3.80 s$

Learn more about Newton's laws of motion and accelerated motion:

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#LearnwithBrainly

6 0

3 years ago

Will mark as brainliest . pls help

ipn [44]

Answer:

answer D

Explanation:

circuit breaker is useful when you do something with your domestic electric circuit..

3 0

3 years ago

Speakers A and B are vibrating in phase. They are directly facing each other, are 6.69 m apart, and are each playing a 75.0-Hz t

maw [93]

Answer:

3.117 m

Explanation:

Given that:

the distance of separation between speaker A and speaker B (L) = 6.69 m

Frequency (F) = 750 -Hz tone

Velocity of speed of sound = 343 m/s

The distance from Speaker A to the first point (L₁) on the line can be calculated by using the formula:

$L_1=\frac{L-A}{2}$

where A = $\frac{Velocity ofthe sound (V)}{Frequency (F)}$

we have:

$L_1=\frac{L-\frac{V}{F} }{2}$

$L_1=\frac{6.69-\frac{343}{750} }{2}$

$L_1=\frac{6.69-0.457 }{2}$

$L_1=\frac{6.233 }{2}$

$L_1= 3.1165 m$

$L_1=3.117 m$

∴ the distance from speaker A to the first point on the line between the speakers where constructive interference occurs = 3.117 m

3 0

3 years ago