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sergiy2304 [10]
3 years ago
12

A vector has an x-component of length 10 and a y-component of length 3. What is the angle of the vector? (Hint: Use the inverse

tangent.)
Physics
1 answer:
Sonja [21]3 years ago
5 0
The vector, the x-component and the y-component form a rectangle triangle where the vector is the hypothenuse and the x and y components are the two sides.
Calling \alpha the angle between the vector and the horizontal direction (x), the two sides are related to \alpha by
\tan \alpha  =  \frac{v_y}{v_x}
where vy and vx are the two components on the y- and x-axis. Using vx=10 and vy=3 we find
\tan \alpha  =  \frac{3}{10} =0.3
And so the angle is
\alpha = \arctan (0.3)=16.7^{\circ}
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As the mass of object increased it is density increased
Elan Coil [88]

Answer:

True.

Explanation:

The density of an object is given by its mass divided by its volume. It can be given as follows :

d=\dfrac{m}{V}

It can be seen that the density of an object is directly proportional to its mass. It means if the mass of an object increase, its density will also increase. Hence, the given statement is true.

3 0
2 years ago
In which situation is no work considered to be done by a force?
andre [41]
If the angle is either 0 or 180, that means that there is either negative or positive work, so A and D are not correct.
If the angle is 45, then there is still some work involved.
The only option where there is no work done by a force is B. when the angle is between the force and displacement is 90. 
8 0
3 years ago
At the outer edge of a rotating space habitat, 130 m from the center, the rotational acceleration is g. What is the rotational a
enyata [817]

Answer:

Explanation:

Given:

R1 = 130 m

R2 = 65 m

w^2R = g

Assume, g = 9.81 m/s^2

w^2 = 9.81/130

w = 0.275 rad/s

At R2 = 65 m

g = w^2R

= (0.275^2) × 65

= 4.905 m/s^2

In conclusion,

g × R = k

g1/R1 = g2/R2

g2 = (g1 × 65)/130

= g1 ×1/2

= g1/2

6 0
3 years ago
Solve the inequality 2(n+3) – 4&lt;6. Then graph<br> the solution.
Aloiza [94]
The solution is 22 2(n+3)-4&6
6 0
3 years ago
A concave mirror has a focal length of 13.5 cm. This mirror forms an image located 37.5 cm in front of the mirror. Find the magn
77julia77 [94]

Explanation:

It is given that,

Focal length of the concave mirror, f = -13.5 cm

Image distance, v = -37.5 cm (in front of mirror)

Let u is the object distance. It can be calculated using the mirror's formula as :

\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}

\dfrac{1}{u}=\dfrac{1}{f}-\dfrac{1}{v}

\dfrac{1}{u}=\dfrac{1}{(-13.5)}-\dfrac{1}{(-37.5)}

u = -21.09 cm

The magnification of the mirror is given by :

m=\dfrac{-v}{u}

m=\dfrac{-(-37.5)}{(-21.09)}

m = -1.77

So, the magnification produced by the mirror is (-1.77). Hence, this is the required solution.

7 0
3 years ago
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