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Amiraneli [1.4K]
3 years ago
8

9. What happens to the particles of a medium when the temperature is high?

Physics
1 answer:
Karo-lina-s [1.5K]3 years ago
4 0

Answer:

IT IS D

Any help you can ask me

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When you measure the boiling point of mercury, you are investigating a ___. a.chemical change b.chemical property c.physical cha
Amanda [17]
D. physical property

the bonds between molecules of mercury are breaking so it's physical and it's not changing the chemical composition of the substance
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What is a risky behaviour​
Sindrei [870]

Explanation:

Doing homework is risky behaviour broo

7 0
3 years ago
Read 2 more answers
1. A student lifts a box of books that weighs 185 N. The box is
aksik [14]

1)  148 J

When lifting an object, the work done on the object is equal to its change in gravitational potential energy. Mathematically:

W = \Delta U = (mg) \Delta h

where

mg is the weight of the object

\Delta h is the change in height

For the box in this problem,

mg = 185 N

\Delta h = 0.800 m

Substituting into the equation, we find:

W=(185)(0.800)=148 J

2) (a) 28875 J

The work done by a force applied parallel to the direction of motion of the object is given by

W=Fd

where

F is the magnitude of the force

d is the displacement

In this problem,

F = 825 N is the force applied by the two students together

d = 35 m is the displacement of the car

Substituting,

W=(825)(35)=28875 J

2) (b) 57750 J

As seen previously, the equation that gives the work done by the force is

W=Fd

We see that the work done is proportional to the magnitude of the force: therefore, if the force is doubled, then the work done is also doubled.

The work done previously was

W = 28875 J

Now the force is doubled, so the new work done will be

W' = 2(28875)=57750 J

3) 4.4 J

In this case, the force acting on the ball is the force of gravity, whose magnitude is:

F = mg

where

m = 0.180 kg is the mass of the ball

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

Solving the equation,

F=(0.180)(9.8)=1.76 N

Now we find the work done by gravity using the same formula applied before:

W=Fd

where d = 2.5 m is the displacement of the ball. We can apply this version of the formula since the force is parallel to the displacement. Substituting,

W=(1.76)(2.5)=4.4 J

4) 595.2 kg

In this case, we have the work done on the box:

W = 7.0 kJ = 7000 J

And we also know the change in height of the box:

\Delta h = 1.2 m

As we stated in part a), the work done on the box is equal to its change in gravitational potential energy:

W=mg \Delta h

Solving for m, we find

m=\frac{W}{g \Delta h}

And substituting the numerical values, we find the mass of the box:

m=\frac{7000}{(9.8)(1.2)}=595.2 kg

5) They do the same work

In fact, the net work done by each person on the box is equal to the change in gravitational potential energy of the box:

W=mg \Delta h

Where \Delta h is the difference in height between the final position and the initial position of the box.

This means that the work done on the box depends only on its initial and final position, not on the path taken. The two men carry the box along different paths, however the reach at the end the same position, and they started from the same position: this means that the value of \Delta h is the same for both of them, so the work they have done is exactly the same.

5 0
4 years ago
A projectile is fired with initial speed vo at an angle of 45o above the horizontal. Assume no air resistance.
katrin2010 [14]

Answer:

The correct answer is a

Explanation:

At projectile launch speeds are

X axis     vₓ = v₀ = cte

Y axis     v_{y} = v_{oy} –gt

The moment is defined as

         p = mv

For the x axis

         pₓ = mvₓ = m v₀ₓ

As the speed is constant the moment is constant

For the y axis

        p_{y} = m v_{y} = m (v_{oy} –gt) = m v_{oy} - m (gt)

Speed ​​changes over time, so the moment also changes over time

Let's examine the answer

i   True

ii False.  The moment changes with time

The correct answer is a

7 0
3 years ago
The magnitude of the magnetic flux through the surface of a circular plate is 5.90 10-5 T · m2 when it is placed in a region of
valkas [14]

Answer:

The strength of the magnetic field is 3.5 x 10⁻³ T

Explanation:

Given;

magnitude of the magnetic flux , Φ = 5.90 x 10⁻⁵ T·m²

angle of inclination of the field, θ = 42.0°

radius of the circular plate, r = 8.50 cm = 0.085 m

Generally magnetic flux in a uniform magnetic field is given as;

Φ = BACosθ

where;

B is the strength of the magnetic field

A is the area of the circular plate

Area of the circular plate:

A = πr²

A = π (0.085)² = 0.0227 m²

The strength of the magnetic field:

B = Φ / ACosθ

B = ( 5.90 x 10⁻⁵) / ( 0.0227 x Cos42)

B = 3.5 x 10⁻³ T

Therefore, the strength of the magnetic field is 3.5 x 10⁻³ T

3 0
3 years ago
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