<h2>Astronaut travels to different planets - Option 4 </h2>
If an astronaut travels to different planets, none of the planets will the astronaut’s weight be the same as on Earth. On jupiter, weight will be more than the weight on earth. For instance if an astronaut has 100kg on earth then he will have 252 kg on jupiter.
On Mars, weight will be less than the weight on the earth. For instance, if an astronaut has 68 kg on earth then he will has 26 kg on mars. On Mercury, weight of an astronaut will be less than the weight on earth. Example if he has 68 kg on earth then he will have 25.7kg on mercury.
Hence, none of these planets the weight of astronaut will be same as on earth.
Use the right equation. To calculate the normal force of an object at an angle, you need to use the formula: N = m * g * cos (x) For this equation, N refers to the normal force, m refers to the object's mass, g refers to the acceleration of gravity, and x refers to the angle of incline.
Answer:
Explanation:
If the dragster attains the speed equal to that of the car which is moving with constant velocity of v₀ , before the two close in contact with each othe , there will not be collision .
So the dragster starting from rest , must attain the velocity v₀ in the maximum time given that is tmax .
v = u + a t
v₀ = 0 + a tmax
tmax = v₀ / a
The value of tmax is v₀ / a .
Answer:
The answer is "Option C".
Explanation:
It's evident from the figure below that after thirty minutes, not no more hydrogen can be created because all of the reactants have converted into products.
hydrogen gas created in cm cubes per period x = 20 seconds, y = 45 centimeters squared, and so on.
A reaction's terminus (the graph's flat line) indicates that no further products are being created during the reaction.
Answer:
202.8m
Explanation:
Given that A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 120 m/s. He misses his target and the cannonball splashes into the briny deep.
First calculate the total time travelled by using the second equation of motion
h = Ut + 1/2gt^2
Let assume that u = 0
And h = 3.5
Substitute all the parameters into the formula
3.5 = 1/2 × 9.8 × t^2
3.5 = 4.9t^2
t^2 = 3.5/4.9
t^2 = 0.7
t = 0.845s
To know how far the cannonball travel, let's use the equation
S = UT + 1/2at^2
But acceleration a = 0
T = 2t
T = 1.69s
S = 120 × 1.69
S = 202.834 m
Therefore, the distance travelled by the cannon ball is approximately 202.8m.