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Andre45 [30]
2 years ago
7

What is the length of the X component of the vector plotted below

Physics
1 answer:
dlinn [17]2 years ago
3 0

Answer:

5

Explanation:

If you straighten out the line, it touches the 5, which makes the length 5

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How are the planets sizes related to their surface gravity
dexar [7]

Answer:

The surface gravity is inversely proportional to the square of the radius of the planet

Explanation:

The gravity at the surface of a planet is given by:

g=\frac{GM}{R^2}

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

We see from the formula that the surface gravity is inversely proportional to the square of the radius of the planet, R.

At the Earth's surface, the value of the surface gravity is approximately 9.81 m/s^2.

6 0
2 years ago
Which element is found in all the acids
lana66690 [7]
Hydrogen .When acids touches all metal hydrogen gas is emitted .Strong acids is one that can produce a high concentration of hydrogen ions.Hope this helped!
3 0
3 years ago
Read 2 more answers
A 320 g air track cart traveling at 1.25 m/s collides with a stationary 270 g cart. What is the speed of the 270 g cart after th
Nutka1998 [239]

Answer:

The speed of the 270g cart after the collision is 0.68m/s

Explanation:

Mass of air track cart (m1) = 320g

Initial velocity (u1) = 1.25m/s

Mass of stationary cart (m2) = 270g

Velocity after collision (V) = m1u1/(m1+m2) = 320×1.25/(320+270) = 400/590 = 0.68m/s

7 0
3 years ago
What is the altitude of the Sun at noon on December 22, as seen from a place on the Tropic of Cancer?
scZoUnD [109]

Answer:

108.217 °

Explanation:

Day of year = 356 = d (Considering year of 365 days)

Latitude of Tropic of Cancer = 23.5 °N

Declination angle

δ = 23.45×sin[(360/365)(d+284)]

⇒δ = 23.45×sin[(360/365)(356+284)]

⇒δ = 5.2832 °

Altitude angle at solar noon

90+Latitude-Declination angle

= 90+23.5-5.2832

= 108.217 °

∴ Altitude angle of the Sun as seen from the tropic of cancer on December 22 is 108.217 °

4 0
3 years ago
At an accident scene on a level road, investigators measure a car's skid mark to be 84 m long. It was a rainy day and the coeffi
Flura [38]

The given data is incomplete. The complete question is as follows.

At an accident scene on a level road, investigators measure a car's skid mark to be 84 m long. It was a rainy day and the coefficient of friction was estimated to be 0.36.  Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes. (why does the car's mass not matter?)

Explanation:

Let us assume that v is the final velocity and u is the initial velocity of the car. Let s be the skid marks and \mu be the friction coefficient and m be the mass of car.

Hence, the given data is as follows.

                v = 0,     s = 84 m,     \mu = 0.36

According to Newton's law of second motion the expression for acceleration is as follows.

                      F = ma

                 -\mu N = ma

                 -\mu mg = ma

                      a = -\mu g

Also,    

               v^{2} = u^{2} + 2as

              (0)^{2} = u^{2} + 2(-\mu g)s

                  u^{2} = 2(\mu g)s

                            = \sqrt{2(0.36)(9.81 m/s^{2})(84 m)}

                            = 24.36 m/s

Thus, we can conclude that the speed of the car when the driver slammed on (and locked) the brakes is 24.36 m/s.

4 0
3 years ago
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