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2 years ago

A 1300 kg car traveling with a speed of 3.5 m/s executes a turn with a 8.5 m radius of curvature.

Physics

1 answer:

Y_Kistochka [10]2 years ago

7 0

Answer:

1.4 m/s/s (2.s.f)

Explanation:

The formula for centripetal acceleration is:

$a=\frac{v^{2} }{r}$ , where v is velocity and r is the radius.

In the question we are given the information that the car has a mass of 1300kg, a velocity of 2.5m/s, and a turn radius of 8.5m which are all the values we need. Therefore we can simply substitute in the values to solve the question:

$a=\frac{3.5^{2} }{8.5} \\a=1.4$

Therefore the centripetal acceleration of the car is 1.4m/s/s. (2.s.f)

Hope this helped!

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Two forces F1 and F2 are applied at the same point of an object. F1 has a magnitude of 20 N and is directed at an angle of 120 d

butalik [34]

Answer:

The value is $\vec F_{n} = -5.67 i +14.82 j \ \ N$

Explanation:

From the question we are told that

The magnitude of the first force is $F_1 = 20 \ N$

The angle which it makes with the x-axis is $\theta_1 = 120 ^o$

The magnitude of the second force is $F_2 = 5 \ N$

The angle which it make with x-axis is $\theta_2 = -30 ^o$

Generally the x-component of the first force is

$F__{1x}} = 20 cos (120 )$ '

=> $F__{1x}} = -10 \ N$

Generally the y-component of the first force is

$F__{1y}} = 20 sin (120 )$

=> $F__{1y}} = 17.32 \ N$

Generally the vector representation of the first force is mathematically represented as

$\vec F_1 = [ -10 i + 17.32 j \ ] \ N$

Generally the x-component of the second force is

$F__{2x}} = 20 cos (-30 )$

=> $F__{2x}} = 4.33 \ N$

Generally the y-component of the second force is

$F__{2y}} = 20 sin (-30 )$

=> $F__{2y}} = -2.5 \ N$

Generally the vector representation of the first force is mathematically represented as

$\vec F_2 = [ 4.33 i - 2.5 j \ ] \ N$

Generally the net force acting that point is mathematically represented as

$\vec F_{n} = -10 i + 17.32 j + 4.33 i - 2.5 j$

=> $\vec F_{n} = -5.67 i +14.82 j \ \ N$

6 0

2 years ago

The x and y components of vector S are -30.0 m and +40.0 m, respectively. Find the magnitude of S and the angle between the dire

34kurt

Answer:

The given vector can be represented in unit vector as

$\overrightarrow{w}=-30\widehat{i}+40\widehat{j}$

The magnitude of any vector $\overrightarrow{r}=u\widehat{i}+v\widehat{j}$ is given by

$|w|=\sqrt{u^{2}+v^{2}}$

Applying values we get

$|w|=\sqrt{-30^{2}+40^{2}}\\\\|r|=50$

We know that positive x axis in vertorial form is represented as

$\overrightarrow{r}=\widehat{i}$

taking dot product of both the vector's we get

$\overrightarrow{r}.\overrightarrow{w}=|r||w|cos(\theta )\\\\\therefore cos(\theta )=\frac{(-30\widehat{i}+40\widehat{j}).\widehat{i}}{50}\\\\\therefore \theta =cos^{-1}(\frac{-30}{50})=126.86^{o}$

5 0

2 years ago

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How many electrons will a single atom of sulfur with no charge and no bonds have in its valence shell?

Svet_ta [14]

Answer:

There are 6 electrons in the outermost shell.

Explanation:

Sulphur is a non-mettalic element which is in the period 3 and group .6on the periodic table. It has an atomic number of 16 and a Mass number of 32. Atomic number tells you the number of electrons in an electrically neutral atom. It has the electronic configuration of 1s2 2s2 2p6 3s2 3p4.

The orbitals have a formula 2n^2 where n = 0, 1, 2, 3 etc.

In the shells, n = 1 so there are 2 electrons. For n = 2, 2*(2)^2 = 8 electrons. So, 16 - (8 + 2) = 6 electrons in the 3 shell (outermost shell)

Therefore from the electronic confriguration above, there are 6 electrons in the outermost shell.

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3 years ago

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Newton's third law says that for every action force there is a reaction force which is

olga nikolaevna [1]

A. Equal and opposite For every action, there is an equal and opposite reaction

4 0

3 years ago

PHYSICS! I AM TIMED!