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2 years ago

Liquids usually have higher volume coefficients of expansion than solids do. true or false

Physics

2 answers:

Ulleksa [173]2 years ago

5 0

Answer: true

Explanation:

iren [92.7K]2 years ago

5 0

The answer is true! Hope this helps!! :)

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A motorcycle is speeding along at a velocity of120kph.Then, it changes to a velocity of 150 kph for 2 minutes.What is the motorc

Taya2010 [7]

Answer:

Explanation:

a=v-u/t

a=acceleration

v=final velocity

u=initial velocity

t=tme taken

we need to convert from kph to ms⁻¹

v= 150*1000/60*60= 41.67ms⁻¹

u= 120*1000/60*60= 33.33ms⁻¹

t= 2*60= 120s

a=41.67-33.33/120

a=8.34/120

a=0.0694ms⁻²

8 0

2 years ago

An air compressor compresses 6 L of air at 120 kPa and 22°C to 1000 kPa and 400°C. Determine the flow work, in kJ/kg, required b

Mariana [72]

Answer:

The work flow required by the compressor = 100.67Kj/kg

Explanation:

The solution to this question is obtained from the energy balance where the initial and final specific internal energies and enthalpies are taken from A-17 table from the given temperatures using interpolation .

The work flow can be determined using the equation:

M1h1 + W = Mh2

U1 + P1alph1 + ◇U + Workflow = U2 + P2alpha2

Workflow = P2alpha2 - P1alpha1

Workflow = (h2 -U2) - (h1 - U1)

Workflow = ( 684.344 - 491.153) - ( 322.483 - 229.964)

Workflow = ( 193.191 - 92.519)Kj/kg

Workflow = 100.672Kj/kg

6 0

3 years ago

A 50.0-kg block is being pulled up a 16.0Â° slope by a force of 250 n that is parallel to the slope. the coefficient of kinetic

Drupady [299]

When dealing with multiple forces acting on a body, it is advisable to draw a free-body diagram like that shown in the picture. There are four forces acting on the box: weight (W) pointing straight down, normal force perpendicular to the slope denoted as Fn, force used to push the box upwards along the slope and the frictional force acting opposite to the direction of motion of the box denoted as Ff. Frictional force is equal to coefficient of kinetic friction (μk) multiplied with Fn.

∑Fy = Fn - mgcos30° = 0
Fn = (50)(9.81)(cos 16) = 471.5 N

When in motion, the net force is equal to mass times acceleration according to Newton's 2nd Law of Motion:

Fnet = F - μk*Fn - mgsin30° = ma
250 - (0.2)(471.5 N) - (50)(sin 16°) = (50)(a)
a = 2.84 m/s²

8 0

3 years ago

A truck travels up a hill with a 5.7° incline. The truck has a constant speed of 22 m/s. What is the horizontal component of the

tester [92]

Answer:

The horizontal component of the velocity is 21.9 m/s.

Explanation:

Please see the attached figure for a better understanding of the problem.

Notice that the vector v and its x and y-components (vx and vy) form a right triangle. Then, we can use trigonometry to find the magnitude of vx, the horizontal component of the velocity.

To find vx, let´s use the following trigonometric rule of right triangles:

cos α = adjacent / hypotenuse

cos 5.7° = vx / 22 m/s

22 m/s · cos 5.7° = vx

vx = 21.9 m/s

The horizontal component of the velocity is 21.9 m/s.

8 0

2 years ago

High-speed stroboscopic photographs show that the head of a 210-g golf club is traveling at 56 m/s just before it strikes a 46-g

tamaranim1 [39]

Explanation:

It is given that,

Mass of golf club, m₁ = 210 g = 0.21 kg

Initial velocity of golf club, u₁ = 56 m/s

Mass of another golf ball which is at rest, m₂ = 46 g = 0.046 kg

After the collision, the club head travels (in the same direction) at 42 m/s. We need to find the speed of the golf ball just after impact. Let it is v.

Initial momentum of golf ball, $p_i=m_1u_1=0.21\ kg\times 56\ m/s=11.76\ kg-m/s$

After the collision, final momentum $p_f=0.21\ kg\times 42\ m/s+0.046v$

Using the conservation of momentum as :

$p_i=p_f$

$11.76\ kg-m/s=0.21\ kg\times 42\ m/s+0.046v$

v = 63.91 m/s

So, the speed of the golf ball just after impact is 63.91 m/s. Hence, this is the required solution.

3 0

3 years ago