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Natalija [7]
4 years ago
7

what is the force if a block of wood with mass 20 kg slides along a frictionless surface at 2 m/s squared

Physics
1 answer:
CaHeK987 [17]4 years ago
5 0

Answer:

40 N

Explanation:

F=ma where F is the applied force, m is the mass of object and a is the acceleration.

Since there is no friction, substituting 20 Kg for m and 2 m/s squared for a then we obtain

F=20*2=40 N

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according to Newton's third law what is the equal and opposite force to the downward force of gravity pulling on a man standing
Fudgin [204]

When you're talking about gravity, it's easy to identify the equal
opposite forces.

Gravity ALWAYS produces an equal pair of opposite forces. 
They both act between the centers of the two objects, one in
each direction.

Consider the equal pair of opposite gravitational forces between
you and the Earth.  One force acts on you, and draws you toward
the center of the Earth.  We call that force "your weight". 
The other one acts on the Earth, and draws it toward the center
of you.  Hardly anybody ever talks about that one, but the two
forces are equal ... your weight on Earth is equal to the Earth's
weight on you !

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3 years ago
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8 0
3 years ago
1.The distance moved by objects in a given interval of time can help us to decide which one is faster or slower. Do you agree wi
shutvik [7]

Answer:

The rate of change of distance is defined as speed.

Explanation:

The speed is defined as the rate of change of distance.

Speed = distance/ time

When we know the distance and the time, we get the value of speed. So, e know that who is moving fast or slow.

hen a graph is pltted beteen the distance and time, the slope of the graph gives the value of speed. So, by checking the slopes, hoseslope ismore, the speed is more and thusit is moving faster.

So, i agree with the statement.  

6 0
3 years ago
A copper wire has a diameter of 2.097 mm. what magnitude current flows when the drift velocity is 1.54 mm/s? take the density of
frozen [14]

By using drift velocity of the electron, the current flow is 7.20 ampere.

We need to know about drift velocity of electrons to solve this problem. The drift velocity can be determined as

v = I / (n . A . q)

where v is drift velocity, I is current, n is atom number density, A is surface area and q is the charge.

From the question above, we know that

d = 2.097 mm

r = (0.002097 / 2) m

v = 1.54 mm/s = 0.00154 m/s

ρ = 8.92 x 10³ kg/m³

q = e = 1.6 x 10¯¹⁹C

Find the atom density

n = Na x ρ / Mr

where Na is Avogadro's number (6.022 x 10²³), Mr is the atomic weight of copper (63.5 g/mol = 0.635 kg/mol).

n = 6.022 x 10²³ x 8.92 x 10³ / 0.635

n = 8.46 x 10²⁷ /m³

Find the current flows

v = I / (n . A . q)

0.00154 = I / (8.46 x 10²⁷ . πr² . 1.6 x 10¯¹⁹)

0.00154 = I / (8.46 x 10²⁷ . π(0.002097 / 2)² . 1.6 x 10¯¹⁹)

I = 7.20 ampere

For more on drift velocity at: brainly.com/question/25700682

#SPJ4

6 0
2 years ago
Carbon is allowed to diffuse through a steel plate 9.7-mm thick. The concentrations of carbon at the two faces are 0.664 and 0.3
cupoosta [38]

Answer:

844°C

Explanation:

The problem can be easily solve by using Fick's law and the Diffusivity or diffusion coefficient.

We know that Fick's law is given by,

J = - D \frac{\Delta c}{\Delta x}

Where \frac{\Delta c}{\Delta x} is the concentration of gradient

D is the diffusivity coefficient

and J is the flux of atoms.

In the other hand we have, that

D= D_0 e^{\frac{E_d}{RT}}

Where D_0 is the proportionality constant,

R is the gas constant, T the temperature and E_d is the activation energy.

Replacing the value of diffusivity coefficient in Fick's law we have,

J = -D_0 ^{\frac{E_d}{RT}}\frac{\Delta c}{\Delta x}

Rearrange the equation to get the value of temperature,

T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}

We have all the values in our equation.

\Delta c = 0.664-0.339 = 0.325 C. cm^{-1}

\Delta x = 9.7*10^{-3}m

E_d = 82000J

D_0 = 6.5*10^{-7}m^2/s

J = 3.2*10^{-9}m^2/s

R= 8.31Jmol^{-1}K

Substituting,

T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}

T=-\frac{-82000}{(8.31)ln(\frac{3.2*10^{-9}(9.7*10^{-3})}{6.5*10^{-7} (0.325)})}

T=1118.07K=844\°C

4 0
4 years ago
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