1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
goblinko [34]
3 years ago
14

A material's ability to allow heat or electricity to flow through it...

Physics
1 answer:
NikAS [45]3 years ago
8 0

Answer:

a

Explanation:

how well a material shines or reflects light

You might be interested in
Please help me......
snow_tiger [21]
A. The statement is true
6 0
3 years ago
Select the correct answer.
Gekata [30.6K]

Answer:

A.

Explanation:

an atom that has most of its mass is electrons.

6 0
3 years ago
A student pushed a box 32.0 m across a smooth, horizontal floor using a constant force of 124 N. If the force was applied for 8.
Brilliant_brown [7]

The power developed is 500 W ( to the nearest Watt)

Power(P) is the rate at which work is done. Work done (W) is the product of the force applied on the object and the displacement (s) made by the point of application of the force.

P = \frac{W}{t}

W= F*s

Therefore,

P=\frac{F*s}{t}

Substitute the given values of force , displacement and time

F = 124 N,s = 32.0 m,t = 8.0 s

P =\frac{W*s}{t} =\frac{124N*22.0s}{8.0s} =496 W

Thus the Power can be rounded off to the nearest value of 500 W

3 0
3 years ago
Read 2 more answers
1.)Two objects, one of m=20,000 kg, and another of 12,500 kg, are placed at a distance of 5 meters apart. What is the force of g
Delvig [45]

1) 6.67\cdot 10^{-4} N

The force of gravitation between the two objects is given by:

F=G\frac{m_1 m_2}{r^2}

where

G=6.67\cdot 10^{-11} kg^{-1} m^{3} s^{-2} is the gravitational constant

m1 = 20,000 kg is the mass of the first object

m2 = 12,500 kg is the mass of the second object

r = 5 m is the distance between the two objects

Substituting the numbers inside the equation, we find

F=(6.67\cdot 10^{-11})\frac{(20,000 kg)(12,500 kg)}{(5 m)^2}=6.67\cdot 10^{-4} N


2)  2.7\cdot 10^{-3} N

From the formula in exercise 1), we see that the force is inversely proportional to the square of the distance:

F \sim \frac{1}{r^2}

this means that if we cut in a half the distance without changing the masses, the magnitude of the forces changes by a factor

F'\sim \frac{1}{(r/2)^2}=4 \frac{1}{r^2}=4F

So, the gravitational force increases by a factor 4. Therefore, the new force will be

F' = 4 F=4(6.67\cdot 10^{-4} N)=2.7\cdot 10^{-3} N


3)  12.5 Nm

The torque is equal to the product between the magnitude of the perpendicular force and the distance between the point of application of the force and the centre of rotation:

\tau=Fd

Where, in this case:

F = 25 N is the perpendicular force

d = 0.5 m is the distance between the force and the center

By using the equation, we find

\tau=(25 N)(0.5 m)=12.5 Nm


4) 0.049 kg m^2/s

The relationship between angular momentum (L), moment of inertia (I) and angular velocity (\omega) is:

L=I\omega

In this problem, we have

I=0.007875 kgm^2

\omega=6.28 rad/s

So, the angular momentum is

L=I\omega=(0.007875 kgm^2)(6.28 rad/s)=0.049 kg m^2/s

6 0
3 years ago
A satellite is in a circular orbit around Mars, which has a mass M = 6.40 × 1023 kg and radius R = 3.40 ×106 m.
Pepsi [2]

Answer:

a) The orbital speed of a satellite with a orbital radius R (in meters) will have an orbital speed of approximately \displaystyle \sqrt\frac{4.27 \times 10^{13}}{R}\; \rm m \cdot s^{-1}.

b) Again, if the orbital radius R is in meters, the orbital period of the satellite would be approximately \displaystyle 9.62 \times 10^{-7}\, R^{3/2}\; \rm s.

c) The orbital radius required would be approximately \rm 2.04 \times 10^7\; m.

d) The escape velocity from the surface of that planet would be approximately \rm 5.01\times 10^3\; m \cdot s^{-1}.

Explanation:

<h3>a)</h3>

Since the orbit of this satellite is circular, it is undergoing a centripetal motion. The planet's gravitational attraction on the satellite would supply this centripetal force.

The magnitude of gravity between two point or spherical mass is equal to:

\displaystyle \frac{G \cdot M \cdot m}{r^{2}},

where

  • G is the constant of universal gravitation.
  • M is the mass of the first mass. (In this case, let M be the mass of the planet.)
  • m is the mass of the second mass. (In this case, let m be the mass of the satellite.)  
  • r is the distance between the center of mass of these two objects.

On the other hand, the net force on an object in a centripetal motion should be:

\displaystyle \frac{m \cdot v^{2}}{r},

where

  • m is the mass of the object (in this case, that's the mass of the satellite.)
  • v is the orbital speed of the satellite.
  • r is the radius of the circular orbit.

Assume that gravitational force is the only force on the satellite. The net force should be equal to the planet's gravitational attraction on the satellite. Equate the two expressions and solve for v:

\displaystyle \frac{G \cdot M \cdot m}{r^{2}} = \frac{m \cdot v^{2}}{r}.

\displaystyle v^2 = \frac{G \cdot M}{r}.

\displaystyle v = \sqrt{\frac{G \cdot M}{r}}.

Take G \approx 6.67 \times \rm 10^{-11} \; m^3 \cdot kg^{-1} \cdot s^{-2},  Simplify the expression v:

\begin{aligned} v &= \sqrt{\frac{G \cdot M}{r}} \cr &= \sqrt{\frac{6.67 \times \rm 10^{-11} \times 6.40 \times 10^{23}}{r}} \cr &\approx \sqrt{\frac{4.27 \times 10^{13}}{r}} \; \rm m \cdot s^{-1} \end{aligned}.

<h3>b)</h3>

Since the orbit is a circle of radius R, the distance traveled in one period would be equal to the circumference of that circle, 2 \pi R.

Divide distance with speed to find the time required.

\begin{aligned} t &= \frac{s}{v} \cr &= 2 \pi R}\left/\sqrt{\frac{G \cdot M}{R}} \; \rm m \cdot s^{-1}\right. \cr &= \frac{2\pi R^{3/2}}{\sqrt{G \cdot M}} \cr &\approx  9.62 \times 10^{-7}\, R^{3/2}\; \rm s\end{aligned}.

<h3>c)</h3>

Convert 24.6\; \rm \text{hours} to seconds:

24.6 \times 3600 = 88560\; \rm s

Solve the equation for R:

9.62 \times 10^{-7}\, R^{3/2}= 88560.

R \approx 2.04 \times 10^7\; \rm m.

<h3>d)</h3>

If an object is at its escape speed, its kinetic energy (KE) plus its gravitational potential energy (GPE) should be equal to zero.

\displaystyle \text{GPE} = -\frac{G \cdot M \cdot m}{r} (Note the minus sign in front of the fraction. GPE should always be negative or zero.)

\displaystyle \text{KE} = \frac{1}{2} \, m \cdot v^{2}.

Solve for v. The value of m shouldn't matter, for it would be eliminated from both sides of the equation.

\displaystyle -\frac{G \cdot M \cdot m}{r} + \frac{1}{2} \, m \cdot v^{2}= 0.

\displaystyle v = \sqrt{\frac{2\, G \cdot M}{R}} \approx 5.01\times 10^{3}\; \rm m\cdot s^{-1}.

5 0
3 years ago
Other questions:
  • Until a train is a safe distance from the station, it must travel at 5 m/s. Once the train is on open track, it can speed up
    13·1 answer
  • HELP PLZ TIMED TEST
    15·1 answer
  • Two groups wanted to determine which size of parachute, when released, fell to the ground the slowest. Each group's design and r
    8·1 answer
  • At one point in space the Electric potential is measured to be 119 V at a distance of 1 meters away it is measured to be 43 V. F
    9·1 answer
  • 50 points and brainleist
    7·2 answers
  • 100 POINTS.
    15·1 answer
  • What is the electric potential 15 cm above the center of a uniform charge density disk of total charge 10 nC and radius 20 cm?
    10·2 answers
  • A woman holds a frozen smoothie in her hand on a warm
    15·2 answers
  • 5) A moving 2 kg object has an initial velocity of 10 m/s. It comes to a stop on a rough
    14·2 answers
  • A wave travels one complete cycle in20sec and has wavelength of 1000mm.what is the speed​
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!