What observations characterize solar maximum?

Help me please, help

Alisiya [41]

Answer:

455,000 Pa

Explanation:

PV = nRT

If n is constant:

PV / T = PV / T

(101,325 Pa) (718 mL) / (273 K) = P (175 mL) / (26 + 273) K

P = 455,000 Pa

6 0

4 years ago

Read 2 more answers

What is the center of a tornado called A.the eye B. The center C. The middle road D. The vortex

Pachacha [2.7K]

A. The eye is correct

3 0

4 years ago

Read 2 more answers

A projectile is fired upward with an initial speed vo on an airless world. A short time later, it comes back down and has a fina

Zinaida [17]

Answer:

$W_{grav} < 0$

Explanation:

When a projectile is fired upwards with some initial speed then the it reaches the top of the projectile and then falls back to the ground.

According to the question we need to find the work done by the gravity which is acting downwards for the projectile when it is at a position just about to hit the ground in course of falling down.

As we know that work is given as:

$W=F.s\cos\theta$

here:

$F=$ force of gravity on the object (which is acting downwards)

$s=$ displacement of the object

Here the work done by the gravity at an instant just before the projectile hits the earth will be negative as the displacement is in the direction opposite to the force of gravity.

7 0

4 years ago

A 290 gg bird flying along at 6.2 m/sm/s sees a 9.0 gg insect heading straight toward it with a speed of 34 m/sm/s (as measured

Murrr4er [49]

Answer:

The bird's speed immediately after swallowing is 4.98 m/s.

Explanation:

Given that,

Mass of bird = 290 g

Speed = 6.2 m/s

Mass of sees = 9.0 g

Speed = 34 m/s

We need to calculate the bird's speed immediately after swallowing

Using conservation of momentum

$m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v_{3}$

Put the value into the formula

$0.290\times6.2+0.009\times(-34)=(0.290+0.009)\times v_{3}$

$v_{3}=\dfrac{0.290\times6.2-0.009\times34}{(0.290+0.009)}$

$v_{3}=4.98\ m/s$

Hence, The bird's speed immediately after swallowing is 4.98 m/s.

6 0

3 years ago

A boat crossing a 153.0 m wide river is directed so that it will cross the river as quickly as possible. The boat has a speed of

Lynna [10]

We have the relation

$\vec v_{B \mid E} = \vec v_{B \mid R} + \vec v_{R \mid E}$

where $v_{A \mid B}$ denotes the velocity of a body A relative to another body B; here I use B for boat, E for Earth, and R for river.

We're given speeds

$v_{B \mid R} = 5.10 \dfrac{\rm m}{\rm s}$

$v_{R \mid E} = 3.70 \dfrac{\rm m}{\rm s}$

Let's assume the river flows South-to-North, so that

$\vec v_{R \mid E} = v_{R \mid E} \, \vec\jmath$

and let $-90^\circ < \theta < 90^\circ$ be the angle made by the boat relative to East (i.e. -90° corresponds to due South, 0° to due East, and +90° to due North), so that

$\vec v_{B \mid R} = v_{B \mid R} \left(\cos(\theta) \,\vec\imath + \sin(\theta) \, \vec\jmath\right)$

Then the velocity of the boat relative to the Earth is

$\vec v_{B\mid E} = v_{B \mid R} \cos(\theta) \, \vec\imath + \left(v_{B \mid R} \sin(\theta) + v_{R \mid E}\right) \,\vec\jmath$

The crossing is 153.0 m wide, so that for some time $t$ we have

$153.0\,\mathrm m = v_{B\mid R} \cos(\theta) t \implies t = \dfrac{153.0\,\rm m}{\left(5.10\frac{\rm m}{\rm s}\right) \cos(\theta)} = 30.0 \sec(\theta) \, \mathrm s$

which is minimized when $\theta=0^\circ$ so the crossing takes the minimum 30.0 s when the boat is pointing due East.

It follows that

$\vec v_{B \mid E} = v_{B \mid R} \,\vec\imath + \vec v_{R \mid E} \,\vec\jmath \\\\ \implies v_{B \mid E} = \sqrt{\left(5.10\dfrac{\rm m}{\rm s}\right)^2 + \left(3.70\dfrac{\rm m}{\rm s}\right)^2} \approx 6.30 \dfrac{\rm m}{\rm s}$

The boat's position $\vec x$ at time $t$ is

$\vec x = \vec v_{B\mid E} t$

so that after 30.0 s, the boat's final position on the other side of the river is

$\vec x(30.0\,\mathrm s) = (153\,\mathrm m) \,\vec\imath + (111\,\mathrm m)\,\vec\jmath$

and the boat would have traveled a total distance of

$\|\vec x(30.0\,\mathrm s)\| = \sqrt{(153\,\mathrm m)^2 + (111\,\mathrm m)^2} \approx \boxed{189\,\mathrm m}$

3 0

2 years ago