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zhenek [66]
3 years ago
5

Which is an example of the force of attraction between two objects that have mass?

Physics
1 answer:
Mariulka [41]3 years ago
4 0

Answer:

Gravity is an example of the force of attraction between two objects that have mass.

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A girl pushes a 1.04 kg book across a table with a horizontal applied force 10 points
mr Goodwill [35]

Answer:

Approximately 11.0\; \rm m \cdot s^{-1}. (Assuming that g = 9.81 \; \rm N \cdot kg^{-1}, and that the tabletop is level.)

Explanation:

Weight of the book:

W = m \cdot g = 1.04 \; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \approx 10.202\; \rm N.

If the tabletop is level, the normal force on the book will be equal (in magnitude) to weight of the book. Hence, F(\text{normal force}) \approx 10.202\; \rm N.

As a side note, the F_N and W on this book are not equal- these two forces are equal in size but point in the opposite directions.

When the book is moving, the friction F(\text{kinetic friction}) on it will be equal to

  • \mu_{\rm k}, the coefficient of kinetic friction, times
  • F(\text{normal force}), the normal force that's acting on it.

That is:

\begin{aligned}& F(\text{kinetic friction}) \\ &= \mu_{\rm k}\cdot F(\text{normal force})\\ &\approx 0.35 \times 10.202\; \rm N \approx 3.5708\; \rm N\end{aligned}.

Friction acts in the opposite direction of the object's motion. The friction here should act in the opposite direction of that 15.0\; \rm N applied force. The net force on the book shall be:

\begin{aligned}& F(\text{net force})  \\ &= 15.0 \; \rm N - F(\text{kinetic friction}) \\& \approx 15.0 - 3.5708\; \rm N \approx 11.429\; \rm N\end{aligned}.

Apply Newton's Second Law to find the acceleration of this book:

\displaystyle a = \frac{F(\text{net force})}{m} \approx \frac{11.429\; \rm N}{1.04\; \rm kg} \approx 11.0\; \rm m \cdot s^{-2}.

6 0
2 years ago
You find it takes 200 N of horizontal force tomove an unloaded pickup truck along a level road at a speed of2.4 m/s. You then lo
IgorC [24]

ANSWER:

F(h)= 230 N is the horizontal force you will need to move the pickup along the same road at the same speed.

STEP-BY-STEP EXPLANATION:

F(h) is Horizontal Force = 200 N

V is Speed = 2.4 m/s

The total weight increase by 42%

coefficient of rolling friction decrease by 19%

Since the velocity is constant so acceleration is zero; a=0

Now the horizontal force required to move the pickup is equal to the frictional force.

F(h) = F(f)

F(h) = mg* u

m is mass

g is gravitational acceleration = 9.8 m/s^2

200 = mg*u

Since weight increases by 42% and friction coefficient decreases by 19%

New weight = 1+0.42 = 1.42 = (1.42*m*g)

New friction coefficient = μ = 1 - 0.19 = 0.81 = 0.81 u

F(h) = (0.81μ) (1.42 m g)

       = (0.81) (1.42) (μ m g)

       = (0.81) (1.42) (200)

       = 230 N

4 0
3 years ago
The graph below is called a heating curve. It shows how water changes from one state of matter to another based on temperature a
Neporo4naja [7]

Explanation :

The heating curve shows how water changes from one state of matter to another based on temperature and the addition or removal of heat over time.

Initially, ice is heated until its temperature reaches 0^0\ C and changes to liquid state.

From the attached graph it is clear that until 100^0\ C the temperature will rise steadily. Here, the liquid begins to vaporize. Vaporization is the state of matter at which liquid state changes to the gaseous state.

So, E is the point which shows the gaseous state.

4 0
3 years ago
Read 2 more answers
What will the stopping distance be for a 2,000-kg car if -2,000 N of force are applied when the car is traveling 20 m/s?
astraxan [27]

Answer is B- 200 m

Given:

m (mass of the car) = 2000 Kg

F = -2000 N

u(initial velocity)= 20 m/s.

v(final velocity)= 0.

Now we know that

<u>F= ma</u>

Where F is the force exerted on the object

m is the mass of the object

a is the acceleration of the object

Substituting the given values

-2000 = 2000 × a

a = -1 m/s∧2

Consider the equation

<u>v=u +at</u>

where v is the initial velocity

u is the initial velocity

a is the acceleration

t is the time

0= 20 -t

t=20 secs


s = ut +1/2(at∧2)

where s is the displacement of the object

u is the initial velocity

t is the time

v is the final velocity

a is the acceleration

s= 20 ×20 +(-1×20×20)/2

<u>s= 200 m</u>


3 0
3 years ago
Read 2 more answers
What is the longest wavelength of light that will emit electrons from a metal whose work function is 3.70 eV
SSSSS [86.1K]

Answer:

h f = Wf + K

where the total energy available is h f, Wf is the work function or the work needed to remove the electron and K is the kinetic energy of the removed electron

If K = zero then hf = Wf

Wf = h f = h c / λ    or

λ = h c / Wf = 6.63E-34 * 3.0E8 / (3.7 * 1.6E-19)

λ = 6.63 * 3 / (3.7 * 1.6) E-7 = 3.36E-7

This would be 3360 angstroms or 336 millimicrons

Visible light = 400-700 millimicrons

6 0
2 years ago
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