All categories

3 years ago

is this true or false? an object with a mass of 9.0kg has an acceleration of 3m/s2. the force acting on it is 3N

Physics

1 answer:

ludmilkaskok [199]3 years ago

5 0

The answer is "False". The force acting on the object is 27 N.

According to Newton's second law, when a force F acts on am object of mass m, it produces an acceleration a. The force is given by the expression,

$F=ma$

Thus, if the body has a mass of 9.0 kg and if it has an acceleration of 3 m/s², then, on substituting the values in the equation for force,

$F=ma\\ =(9.0kg)(3m/s^2)\\ =27N$

Thus, it can be seen that the force acting on the body is 27 N and not 3 N as is mentioned in the statement. Hence the statement is false.

You might be interested in

An apple with a mass of 0.95 kilograms hangs from a tree branch 3.0 meters above the ground.if it falls to the ground, what is t

NeTakaya

KE=1/2mv^2
KE=1/2x0.95x3=1.425

5 0

4 years ago

How does the total amount of energy change as the spring goes up and down?

NeX [460]

The total amount of energy always stays the same because of the law of conservation of energy, meaning that there is always the same amount of energy, it just gets turned into different forms like potential energy, kinetic energy, sound, thermal, etc.

7 0

3 years ago

Suppose that the dipole moment associated with an iron atom of an iron bar is 2.8 × 10-23 J/T. Assume that all the atoms in the

masya89 [10]

To solve this exercise it is necessary to apply the equations related to the magnetic moment, that is, the amount of force that an image can exert on the electric currents and the torque that a magnetic field exerts on them.

The diple moment associated with an iron bar is given by,

$\mu = \alpha *N$

Where,

$\alpha =$ Dipole momento associated with an Atom

N = Number of atoms

$\alpha$ y previously given in the problem and its value is 2.8*10^{-23}J/T

$L = 5.8cm = 5.8*10^{-2}m$

$A = 1.5cm^2 = 1.5*10^{-4}m^2$

The number of the atoms N, can be calculated as,

$N = \frac{\rho AL}{M_{mass}}*A_n$

Where

$\rho =$ Density

$M_{mass} =$ Molar Mass

A = Area

L = Length

$A_n =$ Avogadro number

$N = \frac{(7.9g/cm^3)(1.5cm)(5.8cm^2)}{55.9g/mol}(6.022*10^{23}atoms/mol)$

$N = 7.4041*10^{23}atoms$

Then applying the equation about the dipole moment associated with an iron bar we have,

$\mu = \alpha *N$

$\mu = (2.8*10^{-23})*(7.4041*10^{23})$

$\mu = 20.72Am^2$

PART B) With the dipole moment we can now calculate the Torque in the system, which is

$\tau = \mu B sin(90)$

$\tau = (20.72)(2.2)$

$\tau = 45.584N.m$

Note: The angle generated is perpendicular, so it takes 90 ° for the calculation made.

3 0

3 years ago

Given three different locations on Earth's surface, where will the weight of a person be greatest?

Feliz [49]

Answer:

Explanation:

In order to answer this question, we simply have to refer to the laws of the equations of gravitational mechanics.

The equation given by Newton tells us that

$F = \frac{Gm_{1} m_{2} }{r^{2} }$

In the case where we compare a specific place where the Force of Gravity is greater or lesser, we focus on the term assigned to the Planet's Radius.

In the case of $G, m_{1} ,m_{2}$ , we understand that they are constant.

We can easily notice that the more the Radius (Height seen from a viewer on the ground), the lower the force will be.

In other words, the smaller the radius in which the measurement is made with respect to the center of the earth, the greater the gravitational force.

In that order of ideas the smallest radio has South Pole, which is about 6356 km from the center of the Earth on the Equator line

4 0

3 years ago

A horizontal sheet of negative charge has a uniform electric field E = 3000N/C. Calculate the electric potential at a point 0.7m

vesna_86 [32]

Answer:

Electric potential, E = 2100 volts

Explanation:

Given that,

Electric field, E = 3000 N/C

We need to find the electric potential at a point 0.7 m above the surface, d = 0.7 m

The electric potential is given by :

$V=E\times d$

$V=3000\ N/C\times 0.7\ m$

V = 2100 volts

So, the electric potential at a point 0.7 m above the surface is 2100 volts. Hence, this is the required solution.

6 0

3 years ago