how to draw a labelled diagram of earth's magnetic field ,show how the solar wind distorts the magnetic field

When an object (like a ball) falls, some of its___ energy changes to ___ energy, due to the law of conservation of energy

Alinara [238K]

Answer:

potential, kinetic

Explanation:

pls give brainliest :p

8 0

3 years ago

The venn diagram shown below compares the nuclear reactions in the sun and nuclear power plants.

BARSIC [14]

Answer:

Formation of new elements

Explanation:

3 0

3 years ago

Say you want to make a sling by swinging a mass M of 2.3 kg in a horizontal circle of radius 0.034 m, using a string of length 0

ycow [4]

Answer:

T = 764.41 N

Explanation:

In this case the tension of the string is determined by the centripetal force. The formula to calculate the centripetal force is given by:

$F_c=m\frac{v^2}{r}$ (1)

m: mass object = 2.3 kg

r: radius of the circular orbit = 0.034 m

v: tangential speed of the object

However, it is necessary to calculate the velocity v first. To find v you use the formula for the kinetic energy:

$K=\frac{1}{2}mv^2$

You have the value of the kinetic energy (13.0 J), then, you replace the values of K and m, and solve for v^2:

$v^2=\frac{2K}{m}=\frac{2(13.0J)}{2.3kg}=11.3\frac{m^2}{s^2}$

you replace this value of v in the equation (1). Also, you replace the values of r and m:

$F_c=(2.3kg)(\frac{11.3m^2/s^2}{0.034})=764.41N$

hence, the tension in the string must be T = Fc = 764.41 N

5 0

3 years ago

When visiting some countries, you may see a person balancing a load on the head. Explain why the center of mass of the load need

Vinvika [58]

Explanation:

If the center of the load is directly above the vertebrae, there is no torque in the system. This is a good thing so that the vertebrae are not put out of alignment over time. (Of course, this still doesn't prevent compression of the vertebrae over time, which is a possibility.)

3 0

4 years ago

The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the

kicyunya [14]

Answer:

Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

The stopping sight distance is given as

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

Here

SSD is the stopping sight distance which is to be calculated.
u is the speed which is given as 60 mi/hr
t is the perception-reaction time given as 2.5 sec.
a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
G is the grade of the road, which is this case is 0 as the road is level

Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft$

So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

Here

SSD is the stopping sight distance which is to be calculated.
u is the speed which is given as 60 mi/hr
t is the perception-reaction time given as 2.5 sec.
a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft$

<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>

For downgrade of 4%, Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft$

<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>

As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.

3 0

3 years ago

how to draw a labelled diagram of earth's magnetic field ,show how the solar wind distorts the magnetic field ​

how to draw a labelled diagram of earth's magnetic field ,show how the solar wind distorts the magnetic field