How many micrograms make 1 kg

The opposite poles of two magnets will do which of the following

madam [21]

To what i see, the answer is....

C.

6 0

2 years ago

ONLINE CALCULATOR .A force of 187 pounds makes an angle of 73 degrees 36 ' with a second force. The resultant of the two forces

saul85 [17]

Answer:

The magnitudes of the second force is $Z = 129.9 N$

The magnitudes of the resultant force is $R = 256.047 N$

Explanation:

From the question we are told that

The force is $F = 187 \ lb$

The angle made with second force $\theta_o = 73 ^o 36' = 73 + \frac{36}{60} = 73.6^o$

The angle between the resultant force and the first force $\theta _1 = 29 ^o 1 ' = 29 + \frac{1}{60} = 29.0167^o$

For us to solve problem we are going to assume that

The magnitude of the second force is Z N

The magnitude of the resultant force is R N

According to Sine rule

$\frac{F}{sin (\theta _o - \theta_1 } = \frac{Z}{\theta _1}$

Substituting values

$\frac{187}{sin(73.3 - 29.01667)} =\frac{Z}{sin (29.01667)}$

$267.82 =\frac{Z}{0.4851}$

$Z = 129.9 N$

According to cosine rule

$R = \sqrt{F ^2 + Z^2 + 2(F) (Z) cos (\theta _o) }$

Substituting values

$R = \sqrt{187^2 + 129.9 ^2 + 2 (187 ) (129.9) cos (73.6)}$

$R = 256.047 N$

3 0

2 years ago

To calculate the change in kinetic energy, you must know the force as a function of _______. The work done by the force causes t

aliya0001 [1]

To calculate the change in kinetic energy, you must know the force as a function of position. The work done by the force causes the kinetic energy change

Explanation:

The work-energy theorem states that the change in kinetic enegy of an object is equal to the work done on the object:

$\Delta E_k = W$

where the work done is the integral of the force over the position of the object:

$W=\int F(x) dx$

As we see from the formula, the magnitude of the force F(x) can be dependent from the position of the object, therefore in order to solve correctly the integral and find the work done on the object, it is required to know the behaviour of the force as a function of the position, x.

6 0

2 years ago

In 2005 astronomers announced the discovery of a large black hole in the galaxy Markarian 766 having clumps of matter orbiting a

IRISSAK [1]

A. $4.64\cdot 10^{11}m$

The orbital speed of the clumps of matter around the black hole is equal to the ratio between the circumference of the orbit and the period of revolution:

$v=\frac{2\pi r}{T}$

where we have:

$v=30,000 km/s = 3\cdot 10^7 m/s$ is the orbital speed

r is the orbital radius

$T=27 h \cdot 3600 =97,200 s$ is the orbital period

Solving for r, we find the distance of the clumps of matter from the centre of the black hole:

$r=\frac{vT}{2\pi}=\frac{(3\cdot 10^7 m/s)(97200 s)}{2\pi}=4.64\cdot 10^{11}m$

B. $6.26\cdot 10^{36}kg, 3.13\cdot 10^6 M_s$

The gravitational force between the black hole and the clumps of matter provides the centripetal force that keeps the matter in circular motion:

$m\frac{v^2}{r}=\frac{GMm}{r^2}$

where

m is the mass of the clumps of matter

G is the gravitational constant

M is the mass of the black hole

Solving the formula for M, we find the mass of the black hole:

$M=\frac{v^2 r}{G}=\frac{(3\cdot 10^7 m/s)^2(4.64\cdot 10^{11} m)}{6.67\cdot 10^{-11}}=6.26\cdot 10^{36}kg$

and considering the value of the solar mass

$M_s = 2\cdot 10^{30}kg$

the mass of the black hole as a multiple of our sun's mass is

$M=\frac{6.26\cdot 10^{36} kg}{2\cdot 10^{30} kg}=3.13\cdot 10^6 M_s$

C. $9.28\cdot 10^9 m$

The radius of the event horizon is equal to the Schwarzschild radius of the black hole, which is given by

$R=\frac{2MG}{c^2}$

where M is the mass of the black hole and c is the speed of light.

Substituting numbers into the formula, we find

$R=\frac{6.26\cdot 10^{36} kg)(6.67\cdot 10^{-11})}{(3\cdot 10^8 m/s)^2}=9.28\cdot 10^9 m$

8 0

2 years ago

Which is a warm countercurrent that periodically flows southward along the coasts of Ecuador and Peru

Maksim231197 [3]

The "El-nino current is awarm current that periodically flows southward along the coast of Ecuador and Peru. El-nino start out from the central and east central of the pacific, which include pacific coast of this south America. this ocean event is associated with fluctuation of inter tropical surface pressure pattern and circulation in the ocean.

8 0

3 years ago

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