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3 years ago

5

What are nasa's four great observatories and in what parts of the electromagnetic spectrum do they observe?

2 answers:

Hoochie [10]3 years ago

6 0

1) Hubble Space Telescope- Visible and near-ultraviolet
2) Compton Gamma Ray Observatory- Gamma Rays
3) Chandra X-ray Observatory- X-rays
4) Spitzer Space Telescope- infrared

LenaWriter [7]3 years ago

3 0

Answer:

HUbble space obsevatory

Explanation:

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Select the major problems associated with the production of nuclear energy.

liberstina [14]

Hazardous solid wastes

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3 years ago

Read 2 more answers

A positive test charge of 8.5 × 10 negative 7 Columbus experiences a force of 4.1 × 10 negative 1 N calculate the electric field

Sophie [7]

Explanation:

direction of electric field is same as that of force experienced by the test charge

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3 years ago

The block in the figure below has a mass of 5.1 kg and it rests on an incline of angle . You pull on the rope with a force F = 3

viktelen [127]

42.9°

Explanation:

Let's assume that the x-axis is aligned with the incline and the positive direction is up the incline. We can then apply Newton's 2nd law as follows:

$x:\;\;\;\;F - mg\sin{\theta} = 0\;\;\;\;$

$\Rightarrow mg\sin{\theta} = F$

Note that the net force is zero because the block is moving with a constant speed when the angle of the incline is set at $\theta.$ Solving for the angle, we get

$\sin{\theta} = \dfrac{F}{mg}$

or

$\theta = \sin^{-1}\left(\dfrac{F}{mg}\right)$

$\;\;\;= \sin^{-1}\left[\dfrac{34\:\text{N}}{(5.1\:\text{kg})(9.8\:\text{m/s}^2)}\right]$

$\;\;\;=42.9°$

6 0

2 years ago

Question Part Points Submissions Used A car is stopped for a traffic signal. When the light turns green, the car accelerates, in

olya-2409 [2.1K]

(a) 328.6 kg m/s

The linear impulse experienced by the passenger in the car is equal to the change in momentum of the passenger:

$I=\Delta p = m\Delta v$

where

m = 62.0 kg is the mass of the passenger

$\Delta v$ is the change in velocity of the car (and the passenger), which is

$\Delta v = 5.30 m/s - 0 = 5.30 m/s$

So, the linear impulse experienced by the passenger is

$I=(62.0 kg)(5.30 m/s)=328.6 kg m/s$

(b) 404.7 N

The linear impulse experienced by the passenger is also equal to the product between the average force and the time interval:

$I=F \Delta t$

where in this case

$I=328.6 kg m/s$ is the linear impulse

$\Delta t = 0.812 s$ is the time during which the force is applied

Solving the equation for F, we find the magnitude of the average force experienced by the passenger:

$F=\frac{I}{\Delta t}=\frac{328.6 kg m/s}{0.812 s}=404.7 N$

7 0

3 years ago

A boat is traveling at an initial velocity of 2.7 meters per second in the positive direction. It accelerates at a rate of 0.15

cupoosta [38]

Answer:

$\boxed {\boxed {\sf 4.5 \ m/s \ in \ the \ positive \ direction}}$

Explanation:

We are asked to find the final velocity of the boat.

We are given the initial velocity, acceleration, and time. Therefore, we will use the following kinematic equation.

$v_f= v_i + at$

The initial velocity is 2.7 meters per second. The acceleration is 0.15 meters per second squared. The time is 12 seconds.

$v_i$ = 2.7 m/s
a= 0.15 m/s²
t= 12 s

Substitute the values into the formula.

$v_f = 2.7 \ m/s + (0.15 \ m/s^2)(12 \ s)$

Multiply the numbers in parentheses.

$v_f= 2.7 \ m/s + (0.15 \ m/s/s * 12 \ s)$

$v_f = 2.7 \ m/s + (0.15 \ m/s *12)$

$\v_f=2.7 \ m/s + (1.8 \ m/s)$ $v_f=2.7 \ m/s + (1.8 \ m/s)$

Add.

$v_f=4.5 \ m/s$

The final velocity of the boat is <u>4.5 meters per second in the positive direction.</u>

5 0

3 years ago