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3 years ago

A 2300 kg sailboat is moving west at 5.5 m/s when an eastward wind

1 answer:

5 0

The boat is initially at equilibrium since it seems to start off at a constant speed of 5.5 m/s. If the wind applies a force of 950 N, then it is applying an acceleration <em>a</em> of

950 N = (2300 kg) <em>a</em>

<em>a</em> = (950 N) / (2300 kg)

<em>a</em> ≈ 0.413 m/s²

Take east to be positive and west to be negative, so that the boat has an initial velocity of -5.5 m/s. Then after 11.5 s, the boat will attain a velocity of

<em>v</em> = -5.5 m/s + <em>a</em> (11.5 s)

<em>v</em> = -0.75 m/s

which means the wind slows the boat down to a velocity of 0.75 m/s westward.

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If a child needs interventional service, he or she might have _____.

weqwewe [10]

Answer

Correct Answer is B,D, F and G

Explanation:

If a kid has any type of delays in his physical or mental growth then he need intervention service.

For example if a kid is not able to walk after 18 month, if a kid is not able to speak properly after 3 years, if a kid of 5 years age has lack of response to typical efforts to engage him in an interaction then these kids need intervention service.

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3 years ago

What is the current of the ammeter?

DaniilM [7]

Answer:

.6 A

Explanation:

V = IR

V/R = I

12/20 = I = .6 A

3 0

2 years ago

Read 2 more answers

You throw a ball with a speed of 25.0 m/s at an angle of 40.0â—¦ above the horizontal directly toward a wall. The wall is 22.0 m

Alina [70]

The ball takes about 1.15 seconds to reach the wall

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

This problem is about Projectile Motion

<u>Given:</u>

initial speed = u = 25 m/s

angle of speed = θ = 40.0°

horizontal distance = x = 22.0 m

<u>Unknown:</u>

time taken by the ball = t = ?

<u>Solution:</u>

<em>We will use this following formula to find the time taken by the ball to reach the wall:</em>

$x = u_x t$

$x = u \cos \theta ~t$

$22 = 25 \cos 40^o ~t$

$t = 22 \div ( 25 \cos 40^o )$

$t \approx 1.15 \texttt { seconds}$

$\texttt{ }$

<h2>Conclusion :</h2>

The ball takes about 1.15 seconds to reach the wall

$\texttt{ }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Projectile , Motion , Horizontal , Vertical , Release , Point , Ball , Wall

7 0

3 years ago

Read 2 more answers

a person's lung pressure as recorded by a mercury manometer is 90 mm Hg.Express this pressure in SI units

Sidana [21]

Answer:

11970 pa

Explanation:

A millimetre of mercury is a manometric unit of pressure, formerly defined as the extra pressure generated by a column of mercury one millimetre high, and currently defined as exactly 133.322387415 pascals so if 1 mmHg is 133 pa 90mmHg is 11970 aprox

4 0

3 years ago

A rectangular coil with 50 turns of conducting wire and a total resistance of 10.0 Ω initially lies in the yz-plane at time t =

castortr0y [4]

Answer:

a) 43.20V

b) 2.71W/s

c) 40.25s

d) 7.77Nm

Explanation:

(a) The emf of a rotating coil with N turns is given by:

$emf=NBA\omega sin(\omega t)$

N: turns

B: magnitude of the magnetic field

A: area

w: angular velocity

the emf max is given by:

$emf_{max}=NBA\omega=(50)(1.80T)(0.200m*0.100m)(24.0rad/s)\\\\emf_{max}=43.20V$

(b) the maximum rate of change of the magnetic flux is given by:

$\frac{d\Phi_B}{dt}=\frac{d(A\cdot B)}{dt}=\frac{d}{dt}(ABcos\omega t)=AB\omega sin(\omega t)\\\\\frac{d\Phi_B}{dt}_{max}=(\pi(0.200*0.100))(1.80T)(24.0rad/s)=2.71\frac{W}{s}$

(c) $emf(t=0.050s)=(50)(1.80T)(0.200m*0.100m)(24rad/s)sin(24.0rad/s(0.050s))\\\\emf(t=0.050s)=40.26V$

(d) The torque is given by:

$\tau=NABIsin\theta\\\\NAB\omega=emf_{max}\\\\\tau=\frac{emf_{max}}{\omega}\frac{emf_{max}}{R}\\\\\tau=\frac{(43.20V)^2}{(24.0rad/s)(10.0\Omega)}=7.77Nm$

3 0

3 years ago