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2 years ago

How much force would cause a 120 kg object to decelerate from a rate of 16m/s to 13 m/s in 5 seconds

Physics

1 answer:

Yuki888 [10]2 years ago

6 0

$a=\frac{13m/s-16m/s}{5s} \\a=-0.6 m/s^2\\then\\F=m.a\\F=(120kg)(-0.6 m/s^2)\\F=72 N$

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Determine the length of the tungsten filament in a 100 watt lightbulb given: (i) the resistivity of tungsten is 5.6 × 10−8 Ω · m

svetoff [14.1K]

Answer:

0.34148 m

Explanation:

$\rho$ = Resistivity of tungsten = $5.6\times 10^{-8}\ \Omega m$

d = Diameter = 0.0018 inch

r = Radius = $\dfrac{r}{2}=\dfrac{0.0018}{2}=0.0009\ in$

$r=0.0009\times 0.0254=0.00002286\ m$

$\alpha$ = Temperature coefficient of tungsten = $0.0045 /^{\circ}C$

Power is given by

$P=\dfrac{V^2}{R}\\\Rightarrow R=\dfrac{V^2}{P}\\\Rightarrow R=\dfrac{120^2}{100}\\\Rightarrow R=144\ \Omega$

We have the equation

$R_2=R_1[1+\alpha(T_2-T_1)]\\\Rightarrow R_1=\dfrac{R_2}{1+\alpha(T_2-T_1)}\\\Rightarrow R_1=\dfrac{144}{1+0.0045(2550-25)}\\\Rightarrow R_1=11.64812\ \Omega$

Resistance is given by

$R=\rho\dfrac{l}{A}\\\Rightarrow l=\dfrac{RA}{\rho}\\\Rightarrow l=\dfrac{11.64812\times \pi (0.00002286)^2}{5.6\times 10^{-8}}\\\Rightarrow l=0.34148\ m$

The length of the filament is 0.34148 m

7 0

3 years ago

Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.

padilas [110]

Explanation:

We have,

Semimajor axis is $4\times 10^{12}\ m$

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

$T^2=\dfrac{4\pi ^2}{GM}a^3$

G is universal gravitational constant

M is solar mass

Plugging all the values,

$T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s$

Since,

$1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}$

So, the orbital period of a dwarf planet is 138.52 years.

3 0

2 years ago

What needed to be present in marine mud to form fossils fuels?

garri49 [273]

It would be oraganic matter I think.

6 0

3 years ago

Read 2 more answers

Describe how the student should measure the time taken for the toy parachute to

PolarNik [594]

The time must be measured with respect to gravity. As it falls, it has free fall that is the force acting on it will be the gravity.With the distance in account, d = 1/2 gt²

t = √(2d/g)

4 0

2 years ago

A box has a mass of 400 g. What minimum force is needed to lift it?

mart [117]

Answer: 3.92 N.

Explanation:

Your box weighs 400g, or 0.4kg. In order to lift it, you need to overcome the force of gravity. F = ma, and acceleration due to gravity is -9.8 m/s^2. So gravity acts on the box with a force of 0.4 kg * -9.8 m/s^2 = -3.92 N. A force of +3.92 N is required to overcome this.

5 0

3 years ago