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Brut [27]
3 years ago
13

Find the magnitude and direction for 101m,60.0,85.0m

Physics
1 answer:
Crank3 years ago
4 0

Answer:

magnitude = 161.3m, ∅ = 32.9°

Explanation:

Vector addition always works the same. Add two vectors by adding their respective components.

vector A: \left[\begin{array}{c}85.0&0.0\end{array}\right]

vector B:101.0\left[\begin{array}{c} cos60.0&sin 60.0\end{array}\right] =\left[\begin{array}{c}50.5&87.5\end{array}\right]

Adding vector A and B: \left[\begin{array}{c}85.0&0.0\end{array}\right] +\left[\begin{array}{c}50.5&87.5\end{array}\right] = \left[\begin{array}{c}135.5&87.5\end{array}\right]

The magnitude of any vector \left[\begin{array}{c}a&b\end{array}\right] is given by the Pythagorean theorem:

magnitude = \sqrt{a^2+b^2}

In the case of the vector A+B:

magnitude = \sqrt{135.5^2+87.5^2}

The angle ∅ of the vector can by found by using trigonometric functions:

For instance, the angle ∅ for a vector \left[\begin{array}{c}a&b\end{array}\right] is given by the equation:

tan\phi= \frac{b}{a}

The direction ∅ can be found by solving the trigonometric function.

In the example of vector A+B:

tan\phi = \frac{87.5}{135.5}

Solving for ∅:

\phi = tan^{-1} (\frac{87.5}{135.5})=32.9

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Small, slowly moving spherical particles experience a drag force given by Stokes' law: Fd = 6πηrv where r is the radius of the p
Dominik [7]

Answer:

Explanation:

At the time of a body achieving terminal velocity, the drag force becomes equal to the weight of the body less the buoyant force by the surrounding medium which can be represented by the following equation

\frac{4\pi\times r^3(d-\rho)}{3} =6\pi\times n\times r\times v

Where r is radius of the body , d is density of the material of the body σ is density of the medium and n is coefficient of viscosity of the medium and v is terminal velocity.

Simplifying

v = \frac{2\times r^2(d-\rho)}{9\times n}

Assuming the value of density of air as 1.225 kg/m³ and putting other given values in the formula we get

v = [tex]\frac{2\times (1.2\times10^{-5})^2(2182-1.225)}{9\times 1.8\times10^{-5}}[/tex]

v = 387 x 10⁻⁵ m/s

Terminal velocity = 387 x 10⁻⁵ m/s

Time taken to fall a distance of 100 m

= \frac{100}{387\times10^{-5}}

= 2.6 x 10⁴ s.

5 0
2 years ago
What are the two categories of waves​
fredd [130]

answer: transverse and longitudinal

6 0
3 years ago
The circuit has a 3 volt EMF and two ohm resistors. How much power in watts does this circuit draw? A) 4.5 , B) 24, C) 1.13 D) 2
dlinn [17]

Answer:

P = 4.5 watts

Explanation:

Given that,

EMF of the circuit, E = 3 volt

The resistance  of the resistors, R = 2 ohms

We need to find the power of this circuit. The relation between power, emf and resistance is given by the formula as follows :

P=\dfrac{V^2}{R}

Substitute all the values,

P=\dfrac{3^2}{2}\\\\P=4.5\ W

So, the power of this circuit is equal to 4.5 watts.

5 0
3 years ago
If a system has 225 kcal of work done to it, and releases 5.00 × 102 kj of heat into its surroundings, what is the change in int
vovikov84 [41]

We can solve the problem by using the first law of thermodynamics:

\Delta U = Q-W

where

\Delta U is the change in internal energy of the system

Q is the heat absorbed by the system

W is the work done by the system on the surrounding


In this problem, the work done by the system is

W=-225 kcal=-941.4 kJ

with a negative sign because the work is done by the surrounding on the system, while the heat absorbed is

Q=-5 \cdot 10^2 kJ=-500 kJ

with a negative sign as well because it is released by the system.


Therefore, by using the initial equation, we find

\Delta U=Q-W=-500 kJ+941.4 kJ=441.4 kJ

8 0
3 years ago
If a plane has an airspeed of 40 m/s and is experiencing a crosswind of 30 m/s, what is its ground speed in m/s?
Anon25 [30]
<h2>Answer:50ms^{-1}</h2>

Explanation:

Let v_{a} be the airspeed.

Let v_{w} be the cross wind speed.

We know that,ground speed is the vector sum of airspeed and cross wind speed and airspeed is perpendicular to cross wind speed.

If v_{1} and v_{2} are two perpendicular vectors,the resultant vector has the magnitude \sqrt{|v|_{1}^{2}+|v|_{2}^{2}}

Given,

v_{a}=40ms^{-1}\\v_{c}=30ms^{-1}

So,the ground speed is \sqrt{40^{2}+30^{2}}=\sqrt{2500}=50ms^{-1}

6 0
3 years ago
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