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2 years ago

Decreasing a telescope's eyepiece focal length will _____.

Physics

2 answers:

krek1111 [17]2 years ago

8 0

Answer:

increase magnification

Explanation:

Novay_Z [31]2 years ago

7 0

I believe the correct answer from the choices listed above is the first option. Decreasing a telescope's eyepiece focal length will increase magnification. The magnification of the telescope image is (focal length of the objective) divided by (focal length of the eyepiece). Hope this answers the question.

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HELP ME PLEASE, I'M BEING TIMED FAST

Tom [10]

Answer:

A) Energy is tranferred from Joey to the water. The temperature of the water increases.

Explanation:

At first Joey jumps and gains a height above the water level of the pool, this way has an energy potential initial, as Joey falls into the water his speed is increased that is to say its energy potential is transformed into kinetic energy, and at the moment of impact with the water, this energy kinetic is transformed into heat which is transferred to the water. Therefore the temperature increment.

Note: This is one of the reasons why space agencies are studying spatial asteroids that are directed toward the earth, as these come with great kinetic energy, and great potential energy, if these are of a considerable size can cause catastrophic damage, even if they fall into the ocean, due to the large amount of energy which can cause the instantaneous evaporation of large amounts of water and collateral damage in other areas.

8 0

2 years ago

Would a measured force of (46.5 0.8 N  ) be in agreement with a theoretically calculated force of (48.4 0.6 N  ) ? Show your w

OverLord2011 [107]

Answer:

A measured force of (46.5 0.8 N  ) would not be in agreement with a theoretically calculated force of (48.4 0.6 N  )

Explanation:

From the question we are told that

Measured force is $F_M = [46.5 \pm 0.8 \ N ]$

Calculated force is $F_c = [48.4 \pm 0.6 \ N ]$

Generally the measured force in interval form is

$46.5 - 0.8 < F_M < 46.5 + 0.8$

=> $45.7 < F_M < 47.3$

Generally the calculated force in interval form is

$48.4 - 0.6 < F_c < 48.4 + 0.6$

=> $47.8 < F_M$

Generally looking both interval we see that they do not intersect at any point Hence

A measured force of (46.5 0.8 N  ) would not be in agreement with a theoretically calculated force of (48.4 0.6 N  )

8 0

2 years ago

A golf ball comes to rest 1.90 seconds after hitting a net. The force of the net slowed the ball down at a rate of 56.3 m/s^2. W

Nadusha1986 [10]

Answer:

The velocity of the ball when it first hit the net is Vi= 106.97 m/s

Explanation:

Vf= 0

Vi= ?

a= 56.3 m/s²

t= 1.9 s

Vf= Vi - a*t

Vi= a*t

Vi= 106.97 m/s

8 0

2 years ago

While flying at an altitude of 5.75 km, you look out the window at various objects on the ground. If your ability to distinguish

Bingel [31]

Answer:

the smallest separation between two objects is 0.8067 m

Explanation:

Given the data in the question;

Altitude h = 5.75 km = 5750 m

Diameter D = 4.0 mm = 0.004 m

λ = 460 nm = 4.6 × 10⁻⁷ m

Now, Using Rayleigh criterion for Airy disks resolution.

we know that, Minimum angular separation for resolving two points is;

θ = 1.22λ / D

so we substitute

θ = (1.22 × 4.6 × 10⁻⁷) / 0.004

θ = 5.612 × 10⁻⁷ / 0.004

θ = 1.403 × 10⁻⁴ rad

so minimum separation $d_{min$ = θh

so we substitute

$d_{min$ = (1.403 × 10⁻⁴) × 5750 m

$d_{min$ = 0.8067 m

Therefore, the smallest separation between two objects is 0.8067 m

6 0

2 years ago

3. A lady walks 15.0 meters east and then 20.0 meters north.

NemiM [27]

Answer:

a. 25m

b. 25m

c. 35m

Explanation:

Given the distances walked as 15m east and 20m north:

Using Pythagoras Theorem, we calculate displacement as:

$a^2+b^2=c2, a=15,b=-20\\\\15^2+20^2=c^2\\\\c=\sqrt{15^2+20^2}=\sqrt{625}\\\\c=25m$

Hence, the resultant displacement is 25m

b. The distance from her start point is obtained by drawing a straight line from her final point to her initial point.

Given the distances walked as 15m east and 20m north:

Using Pythagoras Theorem, we calculate displacement as:

$a^2+b^2=c2, a=15,b=-20\\\\15^2+20^2=c^2\\\\c=\sqrt{15^2+20^2}=\sqrt{625}\\\\c=25m$

Hence, she is 25m away from her starting point.

c. The distance traveled is calculated by summing the total distance walked:

$d=d_1+d_2\\\\=15+20\\\\=35m$

Hence, the total distance traveled is 35m

3 0

2 years ago