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3 years ago

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You are on a boat which is crossing the prime meridian. the altitude of polaris is 50Âº.explain how you know the boat's location

is 50Âº north latitude and 0Âº longitude.

1 answer:

Lerok [7]3 years ago

7 0

The definition for the prime meridian is: The prime meridian with 0 degrees longitude runs through Greenwich. Polaris is the north star <span>at 50 degrees above the horizon, which means 50 degrees latitude.
</span>If you observe Polaris at 50 degrees of altitude, you are at latitude 50 North. You that's why you know that your boat's location is <span>50Âº north latitude and 0Âº longitude.</span>

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A constant force of 12N is applied for 3.0s to a body initially at rest. The final velocity of the body is 6.0ms–1. What is the

sp2606 [1]

From the question,
$u = 0m {s}^{ - 1}$
$v = 6m {s}^{ - 1}$

$t = 3s$
$F=12N$

Using Impulse, the product of the constant force, F and time t equals the product of the mass of the body and change in velocity.

$Ft =m(v-u)$

$12(3.0)=m(6.0- \: 0)$
This implies that

$36.0 = 6m$
$m = \frac{36.0}{6.0}$
$\therefore \: m = 6.0kg$

You can also use the equation of linear motion,
$v = u + at$
$6 = 0 + a(3)$
$6 = 3a$
$a = \frac{6}{3}$

$a = 2 {ms}^{ - 2}$
But
$F=ma$
$12 = m(2)$
$12 = 2m$
$\frac{12}{2} = m$
$\therefore \: m = 6kg$

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3 years ago

In any energy transformation, there is always some energy that gets wasted as non-useful heat.

Nady [450]

It is a completely false statement that in <span>any energy transformation, there is always some energy that gets wasted as non-useful heat. The correct option among the two options that are given in the question is the second option. I hope that this is the answer that has actually come to your desired help.</span>

8 0

3 years ago

Read 2 more answers

Clay Matthews, a linebacker for the Green Bay Packers, can reach a speed of 10.0 m/s. At the start of a play, Matthews runs down

MA_775_DIABLO [31]

Answer:

a) D_ total = 18.54 m, b) v = 6.55 m / s

Explanation:

In this exercise we must find the displacement of the player.

a) Let's start with the initial displacement, d = 8 m at a 45º angle, use trigonometry to find the components

sin 45 = y₁ / d

cos 45 = x₁ / d

y₁ = d sin 45

x₁ = d sin 45

y₁ = 8 sin 45 = 5,657 m

x₁ = 8 cos 45 = 5,657 m

The second offset is d₂ = 12m at 90 of the 50 yard

y₂ = 12 m

x₂ = 0

total displacement

y_total = y₁ + y₂

y_total = 5,657 + 12

y_total = 17,657 m

x_total = x₁ + x₂

x_total = 5,657 + 0

x_total = 5,657 m

D_total = 17.657 i^+ 5.657 j^ m

D_total = Ra (17.657 2 + 5.657 2)

D_ total = 18.54 m

b) the average speed is requested, which is the offset carried out in the time used

v = Δx /Δt

the distance traveled using the pythagorean theorem is

r = √ (d1² + d2²)

r = √ (8² + 12²)

r = 14.42 m

The time used for this shredding is

t = t1 + t2

t = 1 + 1.2

t = 2.2 s

let's calculate the average speed

v = 14.42 / 2.2

v = 6.55 m / s

8 0

3 years ago

An electric motor consumes 8.40 kJ of electrical energy in 1.00 min. Part A If one-third of this energy goes into heat and other

Salsk061 [2.6K]

Answer:

The torque is 0.31 Nm.

Explanation:

Electrical energy, E = 8400 J

time, t = 1 min

Angular speed, w = 2900 rpm = 303.53 rad/s

efficiency = 2/3 of input power

The toque is given by

$P =\tau w\\\\\frac{2}{3}\times \frac{E}{t}=\tau w\\\\\frac{2}{3}\times \frac{8400}{60}=\tau \times 303.53\\\\\tau =0.31 Nm$

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3 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.0 m/s at ground level.

BARSIC [14]

Answer:

There is an interval of 24.28s in which the rocket is above the ground.

Explanation:

$y_{i}=0m$

$v_i=80\frac{m}{s}$

$a=4\frac{m}{s^2}$

$y_{e}=1000m$

$g=9.8\frac{m}{s^2}$

From Kinematics, the position $y$ as a function of time when the engine still works will be:

$y(t)=v_it+\frac{1}{2}at^2$

At what time the altitud will be $y_{e}=1000m$ ?

$v_it+\frac{1}{2}at^2=y_{e}$ ⇒ $\frac{1}{2}at^2+v_it-y_{e}=0$

Using the quadratic formula: $t_1=10s$ .

How much time does it take for the rocket to touch the ground? No the function of position is:

$y(t)=y_{e}+v_et-\frac{1}{2}gt^2$

Where our new initial position is $y_{max}$ , the velocity when the engine breaks is $v_e=v_i+at=120\frac{m}{s}$ and the only acceleration comes from gravity (which points down).

Now, when the rocket tounches the ground:

$y_{e}+v_et-\frac{1}{2}gt^2=0$

Again, using the quadratic ecuation:

$t_2=24.49s$

Now, the total time from the moment it takes off and the moment it tounches the ground will be:

$t_T=t_1+t_2=34.49s$ .

6 0

3 years ago