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Alex_Xolod [135]
3 years ago
14

What does this observation tell you about the relationship between the external force applied and the object’s acceleration (the

rate of change in velocity)?
Physics
1 answer:
Dvinal [7]3 years ago
3 0

Explanation:

Newton’s second law of motion is closely related to Newton’s first law of motion. It mathematically states the cause and effect relationship between force and changes in motion. Newton’s second law of motion is more quantitative and is used extensively to calculate what happens in situations involving a force. Before we can write down Newton’s second law as a simple equation giving the exact relationship of force, mass, and acceleration, we need to sharpen some ideas that have already been mentioned.

First, what do we mean by a change in motion? The answer is that a change in motion is equivalent to a change in velocity. A change in velocity means, by definition, that there is an acceleration. Newton’s first law says that a net external force causes a change in motion; thus, we see that a net external force causes acceleration.

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ASAP please!<br> need physics help on test
givi [52]

Answer:

I think its the second one sorry if im wrong

Explanation:

6 0
3 years ago
A student walks 3 north and 4 m west. The magnitude of the resultant displacement for the student is
evablogger [386]

Answer:

5m

Explanation:

Using Pythagoras theorem,

a^2+ b^2=c^2

3^2+4^2=c^2

25=c^2

√(25)=c

5m=c

6 0
3 years ago
Three boxes in contact rest side-by-side on a smooth, horizontal floor. Their masses are 5.0-kg, 3.0-kg, and 2.0-kg, with the 3.
Ivahew [28]

Answer:

(a)Look at the attached graphic

(b)

(b)-1 Equation 1  : m1= 5kg

       50-F1= 5 *a

(b)-2 Equation 2 : m2= 3kg

        F1-F2= 3 *a

(b)-3 Equation 3 : m3= 2kg

         F2 = 2*a  

(c) F1 =25 N

(d) F2 =10 N

Explanation:

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

(a) Draw the free-body diagrams for each of the boxes

Look at the attached graphic

(b) Write Newton’s equation for each mass along the horizontal direction.

Data: m1=  5.0-kg ,m2= 3.0-kg , ,m3= 2.0-kg

<em>Look</em> <em>m1 free-body diagram:</em>

∑Fx = m1*a

50-F1= 5 *a Equation 1

<em>Look</em> <em>m2 free-body diagram:</em>

∑Fx = m2*a

F1-F2= 3 *a Equation 2

<em>Look</em> <em>m3 free-body diagram:</em>

∑Fx = m3*a

F2 = 2*a     Equation 3

(c) What magnitude force does the 3.0-kg box exert on the 5.0- kg box?

<em>Look</em> <em>Free body diagram of the mass set</em>

∑Fx = m*a   m= m1+m2+m3= 5+3+2 = 10 kg

50 = 10*a

a= 50/10 = 5 m/s²

We replace a = 5 m/s² in the equation 1:

50-F1= 5 *5

50-25= F1

F1 = 25 N

<em> (d) </em><em>What magnitude force does the 3.0-kg box exert on the 2.0kg box?</em>

We replace a= 5 m/s² in the equation 3

F2 = 2*5 = 10 N

4 0
3 years ago
Scientific work is currently underway to determine whether weak oscillating magnetic fields can affect human health. For example
iragen [17]

Given Information:

Magnetic field = B = 1×10⁻³ T

Frequency = f = 72.5 Hz

Diameter of cell = d = 7.60 µm = 7.60×10⁻⁶ m

Required Information:

Maximum Emf = ?

Answer:

Maximum Emf = 20.66×10⁻¹² volts

Explanation:

The maximum emf generated around the perimeter of a cell in a field is given by

Emf = BAωcos(ωt)

Where A is the area, B is the magnetic field and ω is frequency in rad/sec

For maximum emf cos(ωt) = 1

Emf = BAω

Area is given by

A = πr²

A = π(d/2)²

A = π(7.60×10⁻⁶/2)²

A = 45.36×10⁻¹² m²

We know that,

ω = 2πf

ω = 2π(72.5)

ω = 455.53 rad/sec

Finally, the emf is,

Emf = BAω

Emf = 1×10⁻³*45.36×10⁻¹²*455.53

Emf = 20.66×10⁻¹² volts

Therefore, the maximum emf generated around the perimeter of the cell is 20.66×10⁻¹² volts

8 0
3 years ago
Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.
Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

4 0
3 years ago
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