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3 years ago

Which statement describes an intensive property of matter

Physics

1 answer:

Fantom [35]3 years ago

4 0

Answer:

where are the statements so i can see which one is correct

Explanation:

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I am Pushing down at a 20

s344n2d4d5 [400]

Answer:

hivhhhh

Explanation:

iiiiuuuujrurj

7 0

2 years ago

A bus travels from el Paso, texas, to chihuahua, mexico, in 5 hr with an average velocity if 75km/hr to the south. How far is Ch

katen-ka-za [31]

375km

Explanation:

Given parameters:

Time of travel = 5hr

Average velocity = 75km/hr south

Unknown:

Displacement between Chihuahua and El paso = ?

Solution:

Velocity is the rate of change of displacement of a body with time. It is a vector quantity that has both magnitude and direction;

Velocity = $\frac{displacement}{time}$

Since the unknown is displacement;

displacement = velocity x time

input the values;

displacement = 75km/hr x 5hr = 375km

learn more:

Velocity brainly.com/question/10883914

#learnwithBrainly

4 0

3 years ago

A single-turn circular loop of wire of radius 5.0 cm lies in a plane perpendicular to a spatially uniform magnetic field. During

Artemon [7]

Answer:

Induced EMF, $\epsilon=0.0143\ volts$

Explanation:

Given that,

Radius of the circular loop, r = 5 cm = 0.05 m

Time, t = 0.0548 s

Initial magnetic field, $B_i=200\ mT=0.2\ T$

Final magnetic field, $B_f=300\ mT=0.3\ T$

The expression for the induced emf is given by :

$\epsilon=\dfrac{d\phi}{dt}$

$\phi$ = magnetic flux

$\epsilon=\dfrac{d(BA)}{dt}$

$\epsilon=A\dfrac{d(B)}{dt}$

$\epsilon=A\dfrac{B_f-B_i}{t}$

$\epsilon=\pi (0.05)^2\times \dfrac{0.3-0.2}{0.0548}$

$\epsilon=0.0143\ volts$

So, the induced emf in the loop is 0.0143 volts. Hence, this is the required solution.

5 0

3 years ago

What is a homogenous mixture

steposvetlana [31]

A homogenous mixture is a mixture that is consistent throughout. If I pulled a portion of this mixture out twice, like two cup fulls, you wouldn’t be able to notice a difference between the two.

4 0

3 years ago

Suppose a plane accelerates from rest for 32.3, achieving a takeoff speed of 47.1 m/s after traveling a distance of 607 m down t

V125BC [204]

Answer:

217.28 m/s

Explanation:

u = 0, t 32.3 s, v = 47.1 m/s, s = 607 m

Let a be the acceleration.

Use third equation of motion.

v^2 = u^2 + 2 a s

47.1 x 47.1 = 0 + 2 a x 607

a = 1.83 m/s^2

For small plane

a = 1.83 m/s^2 , v = 28.2 m/s, u = 0, Let teh distance be s.

Use third equation of motion

28.2^2 = 0 + 2 x 1.83 x s

s = 217.28 m/s

6 0

3 years ago