A loop with a break in it that prevents current form flowing is called a(n) blank circuit

The velocity of a projectile at launch has a horizontal component vh and a vertical component vv. When the projectile is at the

Usimov [2.4K]

Answer:

- horizontal component of the velocity: $v_h$ (because it is constant)

- vertical component of the velocity: 0

- vertical component of the acceleration: $-g=-9.8 m/s^2$ (downward)

Explanation:

The motion of a projectile consists of two independent motions:

- Along the horizontal direction, there are no forces acting on the projectile (if we neglect air resistance), therefore the horizontal acceleration is zero and the horizontal component of the velocity, vh, is constant

- Along the vertical direction, there is only one force acting on the projectile: the force of gravity, downward, which produces a constant downward acceleration of $g=9.8 m/s^2$ . As a consequence, the vertical component of the velocity changes according to

$v_v(t) = v_v-gt$

where vv is the initial vertical velocity and t the time. According to this equation, the vertical component of the velocity decreases first, then becomes zero at the point of maximum height, then becomes negative (= changes direction and points downward)

So, in summary, at the highest point of the trajectory we have:

- horizontal component of the velocity: $v_h$ (because it is constant)

- vertical component of the velocity: 0

- vertical component of the acceleration: $-g=-9.8 m/s^2$ (downward)

3 0

3 years ago

Two geological field teams are working in a remote area. A global positioning system (GPS) tracker at their base camp shows the

Masja [62]

Answer:

distance of 2nd team from 1st team will be: 58.2

Direction of 2nd team from 1st team will be: 14.90 deg North of east

Explanation:

ASSUME Vector is R and makes angle A with +x-axis,

therefore component of vector R is

$R_x = Rcos A$

$R_y = Rsin A$

From above relation

Assuming base camp as the origin, location of 1st team is

$R_1 = 37 km$ away at 21 deg North of west (North of west is in 2nd quadrant, So x is -ve and y is positive)

$R_{1x} = -R_1*cos A_1 = -37*cos 21 deg = -34.54 km$

$R_{1y} = R_1*sin A_1 = 37*sin 21 deg = 13.25 km$

location of 2nd team is at

$R_2 = 32 km$ , at 38 deg East of North = 32 km, at 58 deg North of east (North of east is in 1st quadrant, So x and y both are +ve)

$R_{2x} = R_2*cos A_2 = 32*cos 58 deg = 16.95 km$

$R_{2y} = R_2*sin A_2 = 32*sin 58 deg = 27.13 km$

Now position of 2nd team with respect to 1st team will be given by:

$R_3 = R_2 - R_1$

$R_3 = (R_{2x} - R_{1x}) i + (R_{2y} - R_{1y}) j$

Using above values:

$R_3 = (16.95 - (-34.54)) i + (27.13 - 13.42) j$

$R_3 = 51.49 i + 13.71 j$

distance of 2nd team from 1st team will be:

$\left | R_3 \right | = \sqrt (51.49^2 +13.71^2)$

$\left | R_3 \right | = 53.28 km = 58.2 km$

Direction of 2nd team from 1st team will be:

$Direction = tan^{-1} \frac{R_{3y}}{R_{3x}} = tan^{-1}[ \frac{13.71}{51.49}]$

Direction = 14.90 deg North of east

6 0

3 years ago

In a closed system, what happens to the total energy of the system as energy conversion takes place?

scoundrel [369]

Answer:

In a closed system, the total energy is conserved or remains the same as energy transformations take place.

Explanation:

The law of conservation of energy states that energy cannot be created or destroyed but can be transformed from one form to another.

This law of conservation of energy applies only to a closed system. A closed system is a system which does not exchange energy with its surroundings. All forms of energy conversions occurring within a closed system does not result in an increase or decrease of the total energy of the system, rather, energy remains constant. For example, the universe is a closed system in that all forms of energy conversions occurs within it and energy is not exchanged with an external environment. However, the earth is not a closed system as some of the energy it receives from the sun can be radiated out into space. Since it's an open system, its total energy can change.

7 0

2 years ago

During the motion of the slinky in a transverse wave, what do the particles of the slinky coil do?

Mazyrski [523]

Answer:

C.) The slinky particles move up and down

Explanation:

Transverse Wave-

A wave that has a disturbance perpendicular to the wave motion

Hello! This is the correct answer! Have a blessed day! :)

If you are in K12, please review the lesson! :) It will give you some very helpful definitions! I hope this helped!

5 0

2 years ago

Problem: The frequency of an FM radio station is 89.3 MHz. Calculate its period. Part B: From the Library, select the general eq

vekshin1

Answer:

Time period, $T=1.11\times 10^{-8}\ s$

Explanation:

We have,

The frequency of an FM radio station is 89.3 MHz.

It is required to find the period of the wave.

The reciprocal of frequency is called time period of a wave. It can be given by :

$T=\dfrac{1}{f}\\\\T=\dfrac{1}{89.3\times 10^{6}\ Hz}\\\\T=1.11\times 10^{-8}\ s$

So, the period of the wave is $1.11\times 10^{-8}\ s$ .

5 0

2 years ago