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3 years ago

Newton’s ______ law states that an object in motion stays in motion unless an unbalanced force acts on that object.

Physics

1 answer:

Varvara68 [4.7K]3 years ago

3 0

First law of motion states that

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a 5.00 kg plane is speeding up on the ground with an applied force of 706 N. If the net force is 450 N, what would be the force

loris [4]

Answer:

The force due to air resistance is 256 N.

Explanation:

Given;

mass of the plane, m = 5 kg

applied force on the plane, Fa = 706 N

the net force on the plane, ∑F= 450 N

Let the force due to air resistance = Fr

The net force on the plane is given as;

Net force = applied force - force due to air resistance

∑F = Fa - Fr

Fr = Fa - ∑F

Fr = 706 - 450

Fr = 256 N.

Therefore, the force due to air resistance is 256 N.

5 0

3 years ago

A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. Af

vekshin1

Answer:

$T=51.64^\circ F$

$t=180.10s$

Explanation:

The Newton's law in this case is:

$T(t)=T_m+Ce^{kt}$

Here, $T_m$ is the air temperture, C and k are constants.

We have

$70^\circ F$ in $t=0$

So:

$T(0)=70^\circ F\\T(0)=10^\circ F+Ce^{k(0)}\\70^\circ F=10^\circ F+C\\C=70^\circ F-10^\circ F=60^\circ F$

And we have $60^\circ F$ in $t=30 s$ , So:

$T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061$

Now, we have:

$T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)$

Applying (1) for $t=1 min=60s$ :

$T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F$

Applying (1) for $T=30^\circ F$ :

$30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s$

8 0

3 years ago

a cone is constructed by cutting a sector from a circular sheet of metal with radius . the cut sheet is then folded up and welde

Svet_ta [14]

The expression for the radius and height of the cone can be obtained from

the property of a function at the maximum point.

$The \ radius, \ of \ the \ base \ of \ the \ cone \ is \ \sqrt{ \dfrac{3}{4}} \times radius \ of \ circular \ sheet \ metal$

The height of the cone is half the length of the radius of the circular sheet metal.

Reasons:

The part used to form the cone = A sector of a circle

The length of the arc of the sector = The perimeter of the circle formed by the base of the cone.

$Volume \ of \ a \ cone = \dfrac{1}{3} \cdot \pi \cdot r^3 \cdot h$

$Volume \ of \ a \ cone, \, V = \dfrac{1}{3} \cdot \pi \cdot r^3 \cdot \sqrt{(s^2- r^2)}$

θ/360·2·π·s = 2·π·r

Where;

s = The radius of he circular sheet metal

h = s² - r²

$\dfrac{dV}{dr} = \dfrac{d}{dr} \left(\dfrac{1}{3} \cdot \pi \cdot r^3 \cdot \sqrt{(s^2- r^2)}\right) = \dfrac{\pi \cdot (3 \cdot r^2 \cdot s^2 - 4 \cdot r^4)}{\sqrt{(s^2- r^2)}} = 0$