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adoni [48]
3 years ago
7

can you help me create a sketch of two different objects with one that has a greater density than the other?

Physics
1 answer:
sergey [27]3 years ago
6 0

I don't know how good you are at sketching ... I'm terrible. 
But you can put the point across in a dramatic way if you
can sketch a bowling ball and a basketball ... you'll need
to clearly identify them with the markings you sketch on
each ball. 

They're the same shape and nearly the same size, but
there's a huge difference in their densities.

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A man wishes to pull a crate 15m across a rough floor by exerting a force of 100 N. The
Darina [25.2K]

Answer:

option (E) is correct.

Explanation:

Work done is defined as the product of force and the distance in the direction of force.

force, f = 100 N

Coefficient of friction, = 0.25

distance = 15 m

So, net force F = f - friction force

F = 100 - 0.25 x m g

Work = (100 - 0.25 mg) x d cosθ

For minimum work, the angle should be maximum.

So, the value of θ is 76°.

thus, option (E) is correct.

3 0
3 years ago
An oceanographer is studying how the ion concentration in seawater depends on depth. She makes a measurement by lowering into th
Black_prince [1.1K]

Answer:

a)  R = ρ₀ L /π(r_b² - R_a²) , b)  ρ₀ = V / I    π (r_b² - R_a²) / L

Explanation:

a) The resistance of a material is given by

          R = ρ l / A

where ρ is the resistivity, l is the length and A is the area

the length is l = L and the resistivity is ρ = ρ₀

the area is the area of ​​the cylindrical shell

           A = π r_b² - π r_a²

           A = π (r_b² - r_a²)

we substitute

         R = ρ₀ L /π(r_b² - R_a²)

b) The potential difference is related to current and resistance by ohm's law

         V = i R

         

we subsist the expression of resistance

          V = I ρ₀ L /π (r_b² - R_a²)

           ρ₀ = V / I    π (r_b² - R_a²) / L

6 0
3 years ago
What is the direction of the normal contact force of the road on the wheels?
Gekata [30.6K]

Answer:

The direction of the contact forces acting on a body is not necessarily perpendicular to the contact surface. The resolution of contact forces in two components i.e. perpendicular to contact surface and along surface. Perpendicular component is normal force and parallel component is friction.

Explanation:

3 0
2 years ago
An aurora occurs when _____.
drek231 [11]
The correct answer is A. Charged particles from the sun exite the atmosphere near the poles to create auroras.
6 0
4 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
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