What is the center of a electromagnet called?

Which of the following statements are true of an object in orbit around Earth? (Select all that apply.) The gravity force on the

faltersainse [42]

Answer:

All these statement are true

Explanation:

Gravity will be acting like a centripetal force for the circular motion of object around earth, which makes it perpendicular to the velocity vector. In the case of elliptical motion, gravity can still be divided into 2 vectors, one parallel and the other perpendicular to the velocity. At the nearest point in elliptical motion, gravity is directly perpendicular to velocity just like in circular motion. At the farthest point, the potential energy is minimized and has been converted into kinetic energy. Therefore at this point the speed is greatest.

5 0

4 years ago

A fire hose held near the ground shoots water at a speed of 6.8 m/s. At what angle(s) should the nozzle point in order that the

Vesna [10]

So, based on the angle values that have been found, the angle of elevation of the nozzle can be <u>16° or 74°</u>.

<h3>Introduction</h3>

Hi ! This question can be solved using the principle of parabolic motion. Remember ! When the object is moving parabolic, the object has two points, namely the highest point (where the resultant velocity is 0 m/s in a very short time) and the farthest point (has the resultant velocity equal to the initial velocity). At the farthest distance, the object will move with the following equation :

$\boxed{\sf{\bold{x_{max} = \frac{(v_0)^2 \cdot \sin(2 \theta)}{g}}}}$

With the following condition :

$\sf{x_{max}}$ = the farthest distance of the parabolic movement (m)
$\sf{v_0}$ = initial speed (m/s)
$\sf{\theta}$ = elevation angle (°)
g = acceleration due to gravity (m/s²)

<h3>Problem Solving :</h3>

We know that :

$\sf{x_{max}}$ = the farthest distance of the parabolic movement = 2.5 m
$\sf{v_0}$ = initial speed = 6.8 m/s
g = acceleration due to gravity = 9.8 m/s²

<h3>What was asked :</h3>

$\sf{\theta}$ = elevation angle = ... °

Step by Step :

Find the equation value $\sf{\bold{theta}}$ (elevation angle)

$\sf{x_{max} = \frac{(v_0)^2 \cdot \sin(2 \theta)}{g}}$

$\sf{x_{max} \cdot g = (v_0)^2 \cdot \sin(2 \theta)}$

$\sf{\frac{x_{max} \cdot g}{(v_0)^2} = \sin(2 \theta)}$

$\sf{\frac{2.5 \cdot 9.8}{(6.8)^2} = \sin(2 \theta)}$

$\sf{\frac{24.5}{46.24} = \sin(2 \theta)}$

$\sf{\sin(2 \theta) \approx 0.53}$

$\sf{\cancel{\sin}(2 \theta) \approx \cancel{\sin}(32^o)}$

Find the angle value of the equation by using trigonometric equations. Provided that the parabolic motion has an angle of elevation 0° ≤ x ≤ 90°.

First Probability

$\sf{2 \theta = 32^o + k \cdot 360^o}$

$\sf{\theta = 16^o + k \cdot 180^o}$

→ $\sf{k = 0 \rightarrow 16^o + 0 = 16^o}$ (T)

→ $\sf{k = 1 \rightarrow 16^o + 180^o = 196^o}$ (F)

Second Probability

$\sf{2 \theta = (180^o - 32^o) + k \cdot 360^o}$

$\sf{2 \theta = 148^o + k \cdot 360^o}$

$\sf{\theta = 74^o + k \cdot 180^o}$

→ $\sf{k = 0 \rightarrow 74^o + 0 = 74^o}$ (T)

→ $\sf{k = 1 \rightarrow 74^o + 180^o = 254^o}$ (F)

$\boxed{\sf{\therefore \theta \{16^o , 74^o\} }}$

<h3>Conclusion</h3>

So, based on the angle values that have been found, the angle of elevation of the nozzle can be 16° or 74°.

8 0

3 years ago

Only kealani57 can have it no one can have this points if you take the points you will get reported

MatroZZZ [7]

Hi sorry she is a little mean sorry

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7 0

3 years ago

Which of the following is an example of charging by friction?

taurus [48]

Answer: where is the examples?

Explanation:

6 0

3 years ago

Read 2 more answers

lace each example of a collision in correct category. bumper cars colliding man jumping in a cab mud sticking to car hat being s

Marrrta [24]

Answer:

bumper cars colliding- inelastic

man jumping in a cab- perfectly inelastic

mud sticking to car - perfectly inelastic

hat being sat on door being slammed- inelastic

ball bouncing- elastic

Explanation:

In a perfectly inelastic collision, the objects stick together after collision and move with a common velocity. Maximum kinetic energy is lost during such collision.

For an inelastic collision, kinetic energy is partly lost and the colliding objects move apart at different velocities. This is often encountered in real life situations.

For an elastic collision, both momentum and kinetic energy are conserved. The object rebounds with the same relative velocity with which it approached.

3 0

4 years ago