All categories

3 years ago

What are the three points of the fire triangle

Physics

1 answer:

prisoha [69]3 years ago

5 0

1-fuel
2-heat
3-oxygen

You might be interested in

What is a tornado? Details

ohaa [14]

A tornado is a natural force It is a very strong wind that causes thing to damage

6 0

4 years ago

Read 2 more answers

How do I calculate how many meters are in 7.2 light years?

Vikentia [17]

Answer:

68.2 Quadrillion meters

Explanation:

A lightyear is the distance that light travels in one year.

Speed of light is $3*10^8\ m/s$

So light covers 300,000,000 meters in one second.

One year has 31536000 seconds so , light covers

$9.461*10^{15}\ meters\ in\ one\ year$

so 7.2 light years is

$7.2*(9.461*10^{15})\\6.82*10^{16}$

so 7.2 light years is

6.82 x 10^(16) meters or

68.2 Quadrillion meters

5 0

3 years ago

A charge of 7.00 mC is placed at opposite corners corner of a square 0.100 m on a side and a charge of -7.00 mC is placed at oth

andrew-mc [135]

Answer:

4.03\times10^{7}N[/tex], 135°

Explanation:

charge, q = 7 mC = 0.007 C

charge, - q = - 7 mC = - 0.007 C

d = 0.1 m

Let the force on charge placed at C due to charge placed at D is FD.

$F_{D}=\frac{kq^{2}}{DC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FD is along C to D.

Let the force on charge placed at C due to charge placed at B is FB.

$F_{B}=\frac{kq^{2}}{BC^{2}}$

$F_{B}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FB is along C to B.

Let the force on charge placed at C due to charge placed at A is FA.

$F_{A}=\frac{kq^{2}}{AC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1 \times\sqrt{2} \times 0.1 \times\sqrt{2}}=2.205 \times 10^{7}N$

The direction of FA is along A to C.

The net force along +X axis

$F_{x}=F_{A}Cos45-F_{D}$

$F_{x}=2.205\times10^{7}Cos45-4.41\times10^{7}=-2.85\times10^{7}N$

The net force along +Y axis

$F_{y}=F_{B}-F_{A}Sin45$

$F_{y}=4.41\times10^{7}-2.205\times10^{7}Sin45=2.85\times10^{7}N$

The resultant force is given by

$F=\sqrt{F_{x}^{2}+F_{y}^{2}}=\sqrt{(-2.85\times10^{7})^{2}+(2.85\times10^{7})^{2}}$

$F = 4.03\times10^{7}N$

The angle from x axis is Ф

tan Ф = - 1

Ф = -45°

Angle from + X axis is 180° - 45° = 135°

5 0

3 years ago

The rate of change of an object's velocity is called _______.

tamaranim1 [39]

Answer:

The rate of change of an object's velocity is called acceleration.

4 0

3 years ago

Read 2 more answers

Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to o

Juliette [100K]

(a) 0.473

The potential energy of a satellite orbiting around Earth is given by

$U=-\frac{GMm}{R+h}$

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

R is the Earth's radius

h is the altitude of the satellite above the Earth's surface

So the potential energy of satellite A is

$U_A=-\frac{GMm}{R+h_A}$

while potential energy of satellite B is

$U_B=-\frac{GMm}{R+h_B}$

Therefore the ratio of the potential energy of satellite B to that of satellite A is

$\frac{U_B}{U_A}=\frac{R+h_A}{R+h_B}$

and using

hA = 5920 km

hB = 19600 km

R = 6370 km

we find

$\frac{U_B}{U_A}=\frac{6370+5920}{6370+19600}=0.473$

(b) 0.473

The kinetic energy of a satellite orbiting around Earth instead is given by

$K=\frac{GMm}{2(R+h)}$

So the kinetic energy of satellite A is

$K_A=\frac{GMm}{2(R+h_A)}$

while kinetic energy of satellite B is

$K_B=\frac{GMm}{2(R+h_B)}$

Therefore the ratio of the kinetic energy of satellite B to that of satellite A is

$\frac{K_B}{K_A}=\frac{R+h_A}{R+h_B}$

which is identical to before, so it gives

$\frac{K_B}{K_A}=\frac{6370+5920}{6370+19600}=0.473$

(c) Satellite B

The total energy of a satellite in orbit is given by

$E=U+K = -\frac{GMm}{R+h}+\frac{GMm}{2(R+h)}=-\frac{GMm}{2(R+h)}$

We see that the total energy is:

1) negative (because the satellite is on a bound orbit)

2) inversely proportional to the distance of the satellite from the Earth's center, R+h

So the magnitude of the fraction in the equation is larger for the satellite which is closer to the Earth's surface (satellite A), but since the energy is negative, this means that the total energy of this satellite is smaller than that of satellite B. So, satellite B has a greater total energy.

(d) $1.03\cdot 10^7 J$