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barxatty [35]
1 year ago
15

The y component of a vector R of magnitude k = Bcm shown in the figure below is Ky = +6 cm. What is the direction of this vector

(The angle 8 that makes K with the x axis)?

Physics
1 answer:
WARRIOR [948]1 year ago
6 0

Given

The y-component of vector K is

K_y=6\text{ cm}

The magnitude of vector K is , K=8 cm

To find

The angle

\theta

Explanation

Resolving K along its y-component we have,

\begin{gathered} K_y=Ksin\theta \\ \Rightarrow6=8sin\theta \\ \Rightarrow sin\theta=\frac{3}{4} \\ \Rightarrow\theta=48.59\text{ }^o \end{gathered}

Conclusion

The angle made with the x-axis is

48.59^o

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Two objects, one with a mass of 2 m and one with a mass of 4 m are attracted to each other
Nadusha1986 [10]
It would be 7 M 4 m x 2 m equals 8 M - 1 =7
3 0
2 years ago
A 21.6−g sample of an alloy at 93.00°C is placed into 50.0 g of water at 22.00°C in an insulated coffee-cup calorimeter with a h
IrinaK [193]

Answer : The specific heat capacity of the alloy 1.422J/g^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

C_1 = specific heat of alloy = ?

C_2 = specific heat of water = 4.18J/g^oC

m_1 = mass of alloy = 21.6 g

m_2 = mass of water = 50.0 g

T_f = final temperature of system = 31.10^oC

T_1 = initial temperature of alloy = 93.00^oC

T_2 = initial temperature of water = 22.00^oC

Now put all the given values in the above formula, we get

21.6g\times c_1\times (31.10-93.00)^oC=-50.0g\times 4.18J/g^oC\times (31.10-22.00)^oC

c_1=1.422J/g^oC

Therefore, the specific heat capacity of the alloy 1.422J/g^oC

6 0
2 years ago
True or False:
bekas [8.4K]
I think it false. Sorry if i'm wrong.

5 0
2 years ago
Read 2 more answers
Two charges are placed on the x axis. One of the charges (q1 = +7.7 µC) is at x1 = +3.1 cm and the other (q2 = -19 µC) is at x2
Alinara [238K]

Answer:

a)E=50.53\times 10^{6}\ N/C

The direction will be negative direction.

b)E=268.22\times 10^{6}\ N/C

The direction will be positive direction.

Explanation:

Given that

q1 = +7.7 µC is at x1 = +3.1 cm

q2 = -19 µC is at x2 = +8.9 cm

We know that electric filed due to a charge given as

E=K\dfrac{q}{r^2}

E_1=K\dfrac{q_1}{r_1^2}

E_2=K\dfrac{q_2}{r_2^2}

Now by putting the va;ues

a)

E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.031^2}\ N/C

E_1=72.11\times 10^{6}\ N/C

E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.089^2}\ N/C

E_2=21.58\times 10^{6}\ N/C

The net electric field

E=E_1-E_2

E=50.53\times 10^{6}\ N/C

The direction will be negative direction.

As we know that electric filed line emerge from positive charge and concentrated at negative charge.

b)

Now

distance for charge 1 will become =5.5 - 3.1 = 2.4 cm

distance for charge 2 will become =8.9-5.5 = 3.4 cm

E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.024^2}\ N/C

E_1=120.3\times 10^{6}\ N/C

E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.034^2}\ N/C

E_2=147.92\times 10^{6}\ N/C

The net electric field

E=E_1+E_2

E=268.22\times 10^{6}\ N/C

The direction will be positive direction.

   

7 0
3 years ago
You pull a solid nickel ball with a density of 8.91 g/cm3 and a radius of 1.40 cm upward through a fluid at a constant speed of
Sunny_sXe [5.5K]

Answer:

P = 1.090\,N

Explanation:

The constant speed means that ball is not experimenting acceleration. This elements is modelled by using the following equation of equilibrium:

\Sigma F = P - W + F_{D}

\Sigma F = P - \rho \cdot V \cdot g + c\cdot v = 0

Now, the exerted force is:

P = \rho \cdot V \cdot g - c\cdot v

The volume of a sphere is:

V = \frac{4\cdot \pi}{3}\cdot R^{3}

V = \frac{4\cdot \pi}{3}\cdot (0.014\,m)^{3}

V = 1.149\times 10^{-5}\,m^{3}

Lastly, the force is calculated:

P = (8910\,\frac{kg}{m^{3}} )\cdot (1.149\cdot 10^{-5}\,m^{3})\cdot (9.81\,\frac{m}{s^{2}} )+(0.950\,\frac{kg}{s})\cdot (0.09\,\frac{m}{s} )

P = 1.090\,N

5 0
3 years ago
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