How long will it take to travel 200,000 m traveling 100 m/s?

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

3 years ago

A blue line with 5 orange tick marks then one red tick mark then 4 orange tick marks. The number zero is above the red tick mark

Digiron [165]

Answer:

16 cm

Explanation:

Given that,

The object begins from 0 and moves 3cm towards left side followed by 7 cm towards the right and then, 6 cm towards the left side.

Let the x-axis to be the +ve and on the right side and -ve on the left

Thus, displacement would be:

= 0 -3 + 7 -6

= -2 cm

This implies that the object displaces 2cm towards the left.

While the total distance covered by the object equal to,

= 0cm + 3cm + 7cm + 6cm

= 16 cm

Thus, <u>16 cm</u> is the total distance.

3 0

3 years ago

How long does for one rotation of earth

stellarik [79]

Answer: 24 hours

Explanation: One day is one rotation of the Earth, but it takes the earth 365 days to do one revolution or one year

8 0

2 years ago

WIL GIVE BRAINLISEST PLEASE HELP SPEED ACCELERATION GRAPHS LABEL THE GRAPHS PLEASEEeeee

Varvara68 [4.7K]

Answer:

Take the numbers and fill them in graph.

Plot time as X and speed as Y and afterward join then like a line graph.

4 0

2 years ago

The star Rho1 Cancri is 57 light-years from the earth and has a mass 0.85 times that of our sun. A planet has been detected in a

s344n2d4d5 [400]

Answer:

82780.42123 m/s

14.45 days

Explanation:

m = Mass of the planet

M = Mass of the star = $0.85\times 1.989\times 10^{30}\ kg=1.69065\times 10^{30}\ kg$

r = Radius of orbit of planet = $0.11\times 149.6\times 10^{9}\ m=16.456\times 10^{9}\ m$

v = Orbital speed

The kinetic and potential energy balance is given by

$\frac{GMm}{r^2}=\frac{mv^2}{r}\\\Rightarrow v=\sqrt{\dfrac{Gm}{r}}\\\Rightarrow v=\sqrt{\dfrac{6.67\times 10^{-11}\times 1.69065\times 10^{30}}{16.456\times 10^{9}}}\\\Rightarrow v=82780.42123\ m/s$

The orbital speed of the star is 82780.42123 m/s

The orbital period is given by

$t=\frac{2\pi r}{v}\\\Rightarrow t=\dfrac{2\pi \times 16.456\times 10^{9}}{82780.42123}\\\Rightarrow t=1249040.48419\ seconds=\dfrac{1249040.48419}{24\times 60\times 60}=14.45\ days$

The orbital period is 14.45 days

5 0

2 years ago