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3 years ago

What is the relationship between the normal force and weight

Physics

1 answer:

GalinKa [24]3 years ago

4 0

Answer: they have the same magnitude.

Explanation:

normal force = mg

weight = mg

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The Doppler effect is an effect produced by a moving source of sound or electromagnetic waves due to the relative motion of a so

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3 years ago

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How did the magnet’s density measurement using the Archimedes’ Principle compare to the density measurement using the calculated

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The two methods will yield different results as one is subject to experimental errors that us the Archimedes method of measurement, the the density measurement method will be more accurate

Explanation:

This is because the density method using the calculated volume will huve room for less errors that's occur in practical method i.e Archimedes method due to human error

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3 years ago

PLEASE HELP ME 45 POINTS

sergij07 [2.7K]

Answer:

a) We kindly invite you to see the explanation and the image attached below.

b) The acceleration of the masses is 4.203 meters per square second.

c) The tension force in the cord is 28.02 newtons.

d) The system will take approximately 0.845 seconds to cover a distance of 1.5 meters.

e) The final speed of the system is 3.551 meters per second.

Explanation:

a) At first we assume that pulley and cord are both ideal, that is, masses are negligible and include the free body diagrams of each mass and the pulley in the image attached below.

b) Both masses are connected to each other by the same cord, the direction of acceleration will be dominated by the mass of greater mass (mass A) and both masses have the same magnitude of acceleration. By the 2nd Newton's Law, we create the following equation of equilibrium:

Mass A

$\Sigma F = T - m_{A}\cdot g = -m_{A}\cdot a$ (1)

Mass B

$\Sigma F = T - m_{B}\cdot g = m_{B}\cdot a$ (2)

Where:

$T$ - Tension force in the cord, measured in newtons.

$m_{A}$ , $m_{B}$ - Masses of blocks A and B, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$a$ - Net acceleration of the each block, measured in meters per square second.

By subtracting (2) by (1), we get an expression for the acceleration of each mass:

$m_{B}\cdot a +m_{A}\cdot a = T-m_{B}\cdot g -T + m_{A}\cdot g$

$(m_{B}+m_{A})\cdot a = (m_{A}-m_{B})\cdot g$

$a = \frac{m_{A}-m_{B}}{m_{B}+m_{A}} \cdot g$

If we know that $m_{A} = 5\,kg$ , $m_{B} = 2\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , then the acceleration of the masses is:

$a = \left(\frac{5\,kg-2\,kg}{5\,kg+2\,kg}\right) \cdot\left(9.807\,\frac{m}{s^{2}} \right)$

$a = 4.203\,\frac{m}{s^{2}}$

The acceleration of the masses is 4.203 meters per square second.

c) From (2) we get the following expression for the tension force in the cord:

$T = m_{B}\cdot (a+g)$

If we know that $m_{B} = 2\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $a = 4.203\,\frac{m}{s^{2}}$ , then the tension force in the cord:

$T = (2\,kg)\cdot \left(4.203\,\frac{m}{s^{2}}+9.807\,\frac{m}{s^{2}} \right)$

$T = 28.02\,N$

The tension force in the cord is 28.02 newtons.

d) Given that system starts from rest and net acceleration is constant, we determine the time taken by the block to cover a distance of 1.5 meters through the following kinematic formula:

$\Delta y = \frac{1}{2}\cdot a\cdot t^{2}$ (3)

Where:

$a$ - Net acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

$\Delta y$ - Covered distance, measured in meters.

If we know that $a = 4.203\,\frac{m}{s^{2}}$ and $\Delta y = 1.5\,m$ , then the time taken by the system is:

$t = \sqrt{\frac{2\cdot \Delta y}{a} }$

$t = \sqrt{\frac{2\cdot (1.5\,m)}{4.203\,\frac{m}{s^{2}} } }$

$t \approx 0.845\,s$

The system will take approximately 0.845 seconds to cover a distance of 1.5 meters.

e) The final speed of the system is calculated by the following formula:

$v = a\cdot t$ (4)

Where $v$ is the final speed of the system, measured in meters per second.

If we know that $a = 4.203\,\frac{m}{s^{2}}$ and $t \approx 0.845\,s$ , then the final speed of the system is:

$v = \left(4.203\,\frac{m}{s^{2}} \right)\cdot (0.845\,s)$

$v = 3.551\,\frac{m}{s}$

The final speed of the system is 3.551 meters per second.

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3 years ago

At what point on the hill will the car have zero gravitational potential energy?

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At the bottom of the hill

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3 years ago

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A proton is initially at rest. After some time, a uniform electric field is turned on and the proton accelerates. The magnitude

marusya05 [52]

Answer:

a) 8.83*10⁵ m/s b) 2.80*10⁶ m/s

Explanation:

a) Assuming no other forces acting on the proton, the acceleration on it is produced by the electric field.

By definition, the force due to the electric field is as follows:

F = q*E = e*E (1)

where e is the elementary charge, the charge carried by only one proton, and is e = 1.6*10⁻¹⁹ C.

According to Newton's 2nd law, this force is at the same time, the product of the mass of the proton, times the acceleration a:

F = mp*a (2)

From (1) and(2), being left sides equal, right sides must be equal too:

$F = e*E = mp*a$

Solving for a:

$a = \frac{e*E}{mp} =\frac{1.6e-19C*1.36e5N/C}{1.67e-27kg} =1.3e13 m/s2$

⇒ a = 1.3*10¹³ m/s²

As we have the value of a (which is constant due to the field is uniform), the displacement x, and we know that the initial velocity is 0, in order to get the value of the speed, we can use the following kinematic equation:

$vf^{2} -vo^{2} = 2*a*x$

Replacing by v₀ = 0, a= 1.3*10¹³ m/s² and x = 0.03 m, we can find vf as follows:

$vf =\sqrt{2*(1.3e13 m/s2)*0.03m} = 8.83e5 m/s$

⇒ vf = 8.83*10⁵ m/s

b) We can just repeat the equation from above, replacing x=0.03 m by x=0.3 m, as follows:

$vf =\sqrt{2*(1.3e13 m/s2)*0.3m} = 2.80e6 m/s$

⇒ vf = 2.80*10⁶ m/s

4 0

3 years ago