PHYSICS MIDDLE SCHOOL

The power in an electrical circuit is given by the equation P= RR, where /isthe current flowing through the circuit and Ris the resistance of the circuit.
What is the power in a circuit that has a current of 12 amps and a resistance
of 100 ohms?

A. 144 watts
B. 8.3 watts
c. 1200 watts
D. 14,400 watts

Answers

Answer 1

Answer:

Answer:D

Explanation:Electric power=I*I*R

=12*12*100

=14400watts

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An artificial satellite circling the Earth completes each orbit in 129 min(a) Find the altitude of the satellite.

(b) What is the value of g at the location of this satellite?

Answers

(a) $2.09\cdot 10^6 m$ above Earth's surface

The orbital speed of a satellite orbiting the Earth can be found using the equation

$v=\sqrt{(GM)/(r)}$

where

G is the gravitational constant

$M=5.98\cdot 10^(24) kg$ is the Earth's mass

r is the radius of the satellite's orbit

The orbital speed can also be rewritten as the ratio between the circumference of the orbit and the orbital period, T:

$(2\pi r)/(T)$

where

T = 129 min = 7740 s is the period

Combining the two equations,

$(2\pi r)/(T)=\sqrt{(GM)/(r)}$

And solving for r,

$((2\pi)^2 r^2)/(T^2)=(GM)/(r)\nr=\sqrt[3]{(GMT^2)/((2\pi)^2)}=\sqrt[3]{((6.67\cdot 10^(-11))(5.98\cdot 10^(24))(7740)^2)/((2\pi)^2)}=8.46\cdot 10^6 m$

This is, however, the orbital radius: this means we have to subtract the Earth's radius to find the altitude of the satellite, which is

$R=6.37\cdot 10^6 m$

therefore, the altitude of the satellite is

$h=r-R=8.46\cdot 10^6 - 6.37\cdot 10^6 =2.09\cdot 10^6 m$

b) $5.57 m/s^2$

The value of g at the location of the satellite is given by

$g=(GM)/(r^2)$

where:

G is the gravitational constant

$M=5.98\cdot 10^(24) kg$ is the Earth's mass

$r=8.46 \cdot 10^6 m$ is the radius of the satellite's orbit

Substituting into the equation, we find

$g=((6.67\cdot 10^(-11))(5.98\cdot 10^(24)))/((8.46\cdot 10^6)^2)=5.57 m/s^2$

Final answer:

The satellite orbits at an altitude of approximately 800 km. The gravitational constant, 'g', at this location is approximately 8.66 m/s^2.

Explanation:

The orbital period of an artificial satellite can be used to calculate the altitude at which it orbits. For a satellite that completes each orbit in 129 min (or approximately 2.15 hr), we can apply Kepler's third law which states that the square of the period of a satellite is proportional to the cube of its semi-major axis (distance from the center of the Earth to the satellite).

The formula for the altitude is given by: h = [(GMT^2)/(4π^2)]^(1/3) - R, where G is the gravitational constant, M the mass of Earth, T the orbital period, and R the Earth's radius. With the values G=6.67 x 10^-11 N(m/kg)^2, M=5.98 x 10^24 kg, T=2.15 hr = 7740s, and R=6.371 x 10^6 m, we get h approximately equals 800 km.

The value of 'g' at the satellite's location is given by g = GM/(R+h)^2. Substituting the aforementioned values, we get g to be approximately 8.66 m/s^2. This is less than the 9.81 m/s^2 at Earth's surface due to the increased distance from the Earth's center.

Learn more about Artificial Satellite Orbit here:

brainly.com/question/31204877

#SPJ11

Displacement vectors of 4 km south, 2 km north, 5 km south, and 5 km north combine to atotal displacement of
16 km north.
11 km west.
6km south.
2 km south

Answers

Answer:

The displacement = 2 km south.

Explanation:

Given:

Displacement vectors.

(i) Of 4 km south.
(ii) 2 km north.
(iii) 5 km south.
(iv) 5 km north.

Total displacement :Displacement is the shortest distance from the initial point of any vector.

We have to work this in a Cartesian co-ordinate where we have y axis as North and South (negative y-axis).

Lets move 4 km towards south that is 4 units on negative y-axis.

The points will be $(0,-4)$ .

Now move 2 km upward (towards north) that is toward positive y-axis.

So the points are $(0,-2)$ .

From here again 5 km downward (towards south).

Points of the co-ordinates are $(0,-7)$ .

Finally move 5 km upwards (towards north).

Points of the Cartesian are $(0,-2)$ .

So (0,-2) means that the displacement vector is toward south and is 2 km way from the origin that is $(0,0)$ .

Total displacement is of 2 km in south.

A square block of steel with volume 10 cm3 and mass of 75 g is cut precisely in half. The density of the two smaller pieces is now...a. the same as the original density.
b. one-half the original density.
c. two times the original density.
d. one-fourth the original density.

Answers

a. Density only depends on the substance. It doesn't matter whether you have a little chip of it or a supertanker full of it ... the density doesn't change.

Answer:

a. the same as the original density.

Explanation:

The density of a material depends only on the properties of the material, so if a block of steel is cut in a half, it is still made of the same material (steel), so it still has the same density.

We can verify it by doing some calculations. In fact, the initial density of the block of steel is given by the ratio between its mass (75 g) and its volume (10 cm^3):

$d=(m)/(V)=(75 g)/(10 cm^3)=7.5 g/cm^3$

When the block is cut in a half, its mass becomes half:

$m' = (m)/(2)=(75 g)/(2)=37.5 g$

and its volume also becomes half:

$V'=(V)/(2)=(10 cm^3)/(2)=5 cm^3$

So the new density is

$d'=(m')/(V')=(37.5 g)/(5 cm^3)=7.5 g/cm^3$

So, the density has not changed.

How is it possible for a single sedimentary rock to contain rock particles animal shells and minerals that crystallized from water?

Answers

"Sedimentary rocks are types of rock that are formed by the deposition of material at the Earth's surface and within bodies of water. Sedimentation is the collective name for processes that cause mineral and/or organic particles (detritus) to settle and accumulate or minerals to precipitate from a solution."

Because as the process is making the sediments the fossils and other shrub is washed up together then formed into the rock. When the process is making all sediments the sediments have broken up particles which can be fossils and other things.

3. You shoot an arrow straight up into the air with a velocity of 25 m/s.a. What is the velocity of the arrow at its peak?
b. How high did the arrow go?
c. How long is the arrow in the air?

Answers

a) The velocity of the arrow at its peak is zero

b) The maximum height of the arrow is 31.9 m

c) The time of flight is 5.10 s

Explanation:

The motion of the arrow fired straight upward into the air is a uniformly accelerated motion, with constant acceleration $g=9.8 m/s^2$ (acceleration of gravity) towards the ground.

The initial velocity of the arrow when it is fired is upward: since the acceleration is downward, this means that as the arrow moves upward, its velocity decreases in magnitude.

Eventually, at some point, the velocity of the arrow will become zero, and then it will change direction (downward) and will start increasing in magnitude. The moment when the velocity raches zero corresponds to the peak of the trajectory of the arrow: therefore, at the peak the velocity is zero.

Since the motion of the arrow is a uniformly accelerated motion, we can use the following suvat equation:

$v^2-u^2=2as$