We know, weight = mass * gravity
10 = m * 9.8
m = 10/9.8 = 1.02 Kg
Now, Let, the gravity of that planet = g'
g' = m/r² [m,r = mass & radius of that planet ]
g' = M/10 / (1/2R)² [M, R = mass & radius of Earth ]
g' = 4M / 10R²
g' = 2/5 * M/R²
g' = 2/5 * g
g' = 2/5 * 9.8
g' = 3.92
Weight on that planet = planet's gravity * mass
W' = 3.92 * 1.02
W' = 4 N
In short, Your Answer would be 4 Newtons
Hope this helps!
<u>Answer:</u>
Velocity of the dog relative to the road = 26.04 m/s 3.15⁰ north of east.
<u>Explanation:</u>
Let the east point towards positive X-axis and north point towards positive Y-axis.
Speed of truck = 25 m/s north = 25 j m/s
Speed of dog = 1.75 m/s at an angle of 35.0° east of north = (1.75 cos 35 i + 1.75 sin 35 j)m/s
= (1.43 i + 1.00 j) m/s
Velocity of the dog relative to the road = 25 j + 1.43 i + 1.00 j = 1.43 i + 26.00 j
Magnitude of velocity = 26.04 m/s
Angle from positive horizontal axis = 86.85⁰
So Velocity of the dog relative to the road = 26.04 m/s 86.85⁰ east of north = 26.04 m/s 3.15⁰ north of east.
Complete Question
A football coach walks 18 meters westward, then 12 meters
eastward, then 28 meters westward, and finally 14 meters
eastward.
a
From this motion what is the distance covered
b
What is the magnitude and direction of the displacement
Answer:
a

b
Magnitude
Direction
West
Explanation:
From the question we are told that
The first distance covered westward is ![d_w_1 = 18 \ m /tex] The first distance covered eastward is [tex]d_e1 = 12 \ m /tex] The second distance covered westward is [tex]d_w_2 = 28 \ m /tex] The second distance covered eastward is [tex]d_e2 = 14 \ m /tex] Generally the distance covered is mathematically represented as [tex]D = d_w1 + d_w2 + d_e1 + d_e2](https://tex.z-dn.net/?f=d_w_1%20%20%3D%20%2018%20%5C%20%20m%20%2Ftex%5D%3C%2Fp%3E%3Cp%3E%20%20%20%20%20The%20%20first%20distance%20covered%20eastward%20is%20%5Btex%5Dd_e1%20%3D%20%2012%20%5C%20%20m%20%2Ftex%5D%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20The%20second%20distance%20covered%20westward%20is%20%5Btex%5Dd_w_2%20%20%3D%20%2028%20%5C%20%20m%20%2Ftex%5D%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20The%20%20second%20distance%20covered%20eastward%20is%20%5Btex%5Dd_e2%20%3D%20%2014%20%5C%20%20m%20%2Ftex%5D%3C%2Fp%3E%3Cp%3E%20%20%20%3C%2Fp%3E%3Cp%3EGenerally%20the%20distance%20covered%20is%20mathematically%20represented%20as%20%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20%5Btex%5DD%20%3D%20%20d_w1%20%2B%20d_w2%20%2B%20d_e1%20%2B%20d_e2)
=> 
=> 
For the second question eastward is in the direction of the positive x-axis so it would be positive and westward is in the direction of the negative x-axis so it would be negative
The magnitude of the displacement is

=>
=>
The direction is west
<h3>16.</h3>
Your answer is correct.
___
<h3>17.</h3>
The fractional change in resistance is equal to the given temperature coefficient multiplied by the change in temperature.
R = R₀×(1 + α×ΔT)
R = (10.0 Ω)×(1 + 0.004×(65 -20)) = 11.8 Ω