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3 years ago

Question 9

Physics

1 answer:

egoroff_w [7]3 years ago

6 0

Answer:

Explanation:

F=ma

given solution

v=12m/s a=v/t

s=6 sec =12m/s÷6sec

=2m/s^2 then we get acceleration now we will find the mass. first derive the the formula of mass by crisis cross then you will get this formula which is m=F/a

=36÷2

= 18

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What are the two forces involved in an interaction called

natka813 [3]

Answer:These two forces are called action and reaction forces and are the subject of Newton's third law of motion. Formally stated, Newton's third law is: For every action, there is an equal and opposite reaction. The statement means that in every interaction, there is a pair of forces acting on the two interacting objects.

Explanation:

8 0

3 years ago

A sled of mass 50 kg is pulled along a snow-covered, flat ground. The static friction coefficient is 0.3 and the kinetic frictio

Diano4ka-milaya [45]

Answer:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled.

b) The weight of the sled is 490.35 newtons.

c) A force of 147.105 newtons is needed to start the sled moving.

d) A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

Explanation:

a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled. All forces are listed:

$F$ - External force exerted on the sled, measured in newtons.

$f$ - Friction force, measured in newtons.

$N$ - Normal force from the ground on the mass, measured in newtons.

$W$ - Weight, measured in newtons.

b) The weight of the sled is determined by the following formula:

$W = m\cdot g$ (1)

Where:

$m$ - Mass, measured in kilograms.

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

If we know that $m = 50\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , the weight of the sled is:

$W = (50\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$W = 490.35\,N$

The weight of the sled is 490.35 newtons.

c) The minimum force needed to start the sled moving on the horizontal ground is:

$F_{min,s} = \mu_{s}\cdot W$ (2)

Where:

$\mu_{s}$ - Static coefficient of friction, dimensionless.

$W$ - Weight of the sled, measured in newtons.

If we know that $\mu_{s} = 0.3$ and $W = 490.35\,N$ , then the force needed to start the sled moving is:

$F_{min,s} = 0.3\cdot (490.35\,N)$

$F_{min,s} = 147.105\,N$

A force of 147.105 newtons is needed to start the sled moving.

d) The minimum force needed to keep the sled moving at constant velocity is:

$F_{min,k} = \mu_{k}\cdot W$ (3)

Where $\mu_{k}$ is the kinetic coefficient of friction, dimensionless.

If we know that $\mu_{k} = 0.1$ and $W = 490.35\,N$ , then the force needed to keep the sled moving at a constant velocity is:

$F_{min,k} = 0.1\cdot (490.35\,N)$

$F_{min,k} = 49.035\,N$

A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.

8 0

3 years ago

You want the current amplitude through a inductor with an inductance of 4.70 mH (part of the circuitry for a radio receiver) to

goldenfox [79]

Answer:

f = 1.69*10^5 Hz

Explanation:

In order to calculate the frequency of the sinusoidal voltage, you use the following formula:

$V_L=\omega iL=2\pi f i L$ (1)

V_L: voltage = 12.0V

i: current = 2.40mA = 2.40*10^-3 A

L: inductance = 4.70mH = 4.70*10^-3 H

f: frequency = ?

you solve the equation (1) for f and replace the values of the other parameters:

$f=\frac{V_L}{2\pi iL}=\frac{12.0V}{2\pi (2.4*10^{-3}A)(4.70*10^{-3}H)}=1.69*10^5Hz$

The frequency of the sinusoidal voltage is f

3 0

3 years ago

A positively charged objectwith a mass of 0.114 kg oscillates at the end of a spring, generating ELF (extremely low frequency) r

katen-ka-za [31]

Answer:

k = 167.33 N/m

Explanation:

The radio waves have a fixed relationship between the propagation speed (the speed of light in vacuum), the frequency and the wavelength, as follows:
v = c = λ*f

where c= speed of light in vacuum = 3*10⁸ m/s, λ = wavelength =

4.92*10⁷ m.

Solving for f, we get the frequency of the radio waves:

f = 6.1 Hz

Now, from the Hooke's law, we know that the mass attached at the end of the spring oscillates with an angular frequency defined by a fixed relationship between the spring constant k and the mass m, as follows:

$\omega_{o}^{2} =\frac{k}{m} (1)$

Now, we know that there exists a fixed relationship between the angular frequency and the frequency, as follows:

$\omega = 2*\pi *f (2)$

We also know that f in (2) is the same that we got for the radio waves, so replacing (2) in (1), and rearranging terms, we can solve for k, as follows:
$k = 4*\pi ^{2}*f^{2} *m = 4*\pi ^{2} * (6.1Hz)^{2} * 0.114 kg = 167.33 N/m$

3 0

3 years ago

Which of the following explains why metallic bonding only occurs between

Y_Kistochka [10]

Answer:

D. Metallic atoms have valence shells that are mostly empty, which

means these atoms are more likely to give up electrons and allow

them to move freely.

Explanation:

Metals usually contain very few electrons in their valence shells hence they easily give up these few valence electrons to yield metal cations.

In the metallic bond, metal cations are held together by electrostatic attraction between the metal ions and a sea of mobile electrons.

Since metals give up their electrons easily, it is very easy for them to participate in metallic bonding. They give up their electrons easily because their valence shells are mostly empty, metal valence shells usually contain only a few electrons.

3 0

3 years ago