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3 years ago

Scott travels north 3 Km and then goes west 3 Km before coming straight

Physics

1 answer:

NikAS [45]3 years ago

8 0

Answer:

you have to times them all ti get the answer

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Multiple choice 1. which of earth's spheres contains mountains, valleys, and other landscapes? (1 point

ycow [4]

Here are the answers:
1. Geosphere (though the term lithosphere is mostly used)
2. Both ice and wind (glaciers, and really strong winds)
3. Water
4. Its inertia (the Earth is constantly "falling" towards the Sun due to its gravitational pull, but its inertia helps the Earth from maintaining its orbit.)
5. The rotating Earth
6. one year
7. The equator
8. It depends on how much of the sunlit side of the Moon faces the Earth
9. When an object in space comes between the Sun and a third object
10. D<span>ifferences in how much the Moon and the Sun pull on different parts of Earth
11. b. False
12. a. True

Hope my answers have come to your help.</span>

4 0

3 years ago

marysya [2.9K]

Answer:

D) shrivel up, since the atmosphere exerts more force on the can as it cools.

Explanation:

As the water in the can is boiled the can gets heated up and contains hot vapour and gases which are rare in density and are in their expanded state. In this state when the can is sealed tightly such that no air leaks in or out of the can. When the temperature of the can drops, the gases shrink in volume and the pressure inside the can become less than the pressure of the atmosphere which leads to shriveling of the can.

5 0

3 years ago

Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150

Sever21 [200]

Answer: $2.89\approx 2.9^{\circ}C/s$

Explanation:

Given

$Power\left ( P\right )=150 MW$

mass of core $\left ( m\right )=1.60\times 10^5 kg$

Average specific heat $\left ( C\right )=0.3349 KJ/kg^{\circ}C$

And rate of increase of temperature = $\frac{\mathrm{d}T}{\mathrm{d} t}$

Now

P= $mc\frac{\mathrm{d}T}{\mathrm{d} t}$

$150\times 10^6=1.60\times 10^5\times 0.3349\times \frac{\mathrm{d}T}{\mathrm{d} t}$

Thus $\frac{\mathrm{d}T}{\mathrm{d} t}=[tex]\frac{1.60\times 10^5\times 0.3349}{150\times 10^6}$

$\frac{\mathrm{d}T}{\mathrm{d} t}=2.89\approx 2.9^{\circ}C/s$

6 0

4 years ago

A spring gun is made by compressing a spring in a tube and then latching the spring at the compressed position. A 4.97-g pellet

dimaraw [331]

Answer:

$v = 2.8898 \frac{m}{s}$

Explanation:

This is a problem easily solve using energy conservation. As there are no non-conservative forces, we know that the energy is conserved.

When the spring is compressed downward, the spring has elastic potential energy. When the spring is relaxed, there is no elastic potential energy, but the pellet will have gained gravitational potential energy and kinetic energy. Lets see what are the terms for each of this.

<h3>Elastic potential energy</h3>

We know that a spring following Hooke's Law has a elastic potential energy:

$E_{ep} = \frac{1}{2} k (\Delta x)^2$

where $\Delta x$ is the displacement from the relaxed length and k is the spring's constant.

To obtain the spring's constant, we know that Hooke's law states that the force made by the spring is :

$\vec{F} = - k \Delta \vec{x}$

as we need 9.12 N to compress 4.60 cm, this means:

$k = \frac{9.12 \ N}{4.6 \ 10^{-2} \ m}$

$k = 198.26 \ \frac{ N}{m}$

So, the elastic energy of the compressed spring is:

$E_{ep} = \frac{1}{2} 198.26 \ \frac{ N}{m} (4.6 \ 10^{-2} \ m)^2$

$E_{ep} = 0.209759 \ Joules$

And when the spring is relaxed, the elastic potential energy will be zero.

<h3>Gravitational potential energy</h3>

To see how much gravitational potential energy will the pellet win, we can use

$\Delta E_{gp} = m g \Delta h$

where m is the mass of the pellet, g is the acceleration due to gravity and \Delta h is the difference in height.

Taking all this together, the gravitational potential energy when the spring is relaxed will be:

$\Delta E_{gp} = 4.97 \ 10^{-3} kg \ 9.8 \frac{m}{s^2} 4.6 \ 10^{-2} m$

$\Delta E_{gp} = 0.00224 \ Joules$

<h3>Kinetic Energy</h3>

We know that the kinetic energy for a mass m moving at speed v is:

$E_k = \frac{1}{2} m v^2$

so, for the pellet will be

$E_k = \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2$

<h3>All together</h3>

By conservation of energy, we know:

$E_{ep} = \Delta E_{gp} + E_k$

$0.209759 \ Joules = 0.00224 \ Joules + \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2$

$\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2 = 0.209759 \ Joules - 0.00224 \ Joules$

$\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2 = 0.207519 \ Joules$

$v = \sqrt{ \frac{ 0.207519 \ Joules}{ \frac{1}{2} \ 4.97 \ 10^{-3} kg } }$

$v = 2.8898 \frac{m}{s}$

7 0

4 years ago

The figure in Figure 1 shows two single-slit diffraction patterns. The distance between the slit and the viewing screen is the s

V125BC [204]

Answer:

"The wavelengths are the same for both. The width of slit 1 is larger than the width of slit 2."

Explanation:

The full question has not been provided, so I just copied this into the web and found this answer and explanation on quizlet:

"The wavelengths are the same for both. The width of slit 1 is larger than the width of slit 2.

D sin θ = m λ

if the wavelengths are the same, then if the angle is smaller, the slit width must be larger. The top photo shows a pattern that is more closely spaced. That means the angle is smaller. The slit width must be larger."

This answer/explanation should be correct, as we are looking at bright fringes and the formula being used corresponds to the parameters of the question.

Hope this helps!

8 0

2 years ago