Scientists are experimenting with a kind of gun that may eventually be used to fire payloads directly into orbit. In one test, t
1 answer:
Answer:
Explanation:
Project mass m=3.8 kg
Initial speed vi= 0m/s
Final speed vf= 9.3×10³ m/s
Force F=9.3×10⁵N
To find
Time t
Solution
From Newtons second law we know that
∑F=ma
Where m is mass
a is acceleration
We can write this equation as
∑F=m(Δv/Δt)
Rearrange this equation to find time t
So
Substitute the given values
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Answer:
The net force = 0
Explanation:
The given information includes;
The mass of the crate = 250 kg
The way the helicopter lifts the crate = Uniformly (constant rate (speed), no acceleration)
In order to pull the crate upwards, the helicopter has to provide a force equivalent to the weight of the crate keeping the helicopter on the ground.
The weight of the crate = The mass of the crate × The acceleration due gravity acting on the crate
The weight of the crate, ↓ = 250 kg × 9.81 m/s² = 2,452.5 N
The force the helicopter should provide to just lift the crate, ↑ = The weight of the crate = 2,452.5 N
The net force, =
↑ -
↓ = 2,452.5 N - 2,452.5 N = 0
The net force = 0.
Answer:
a) R = 2.5 mi b) To return to your case you must walk in the opposite direction or θ = 98º
This is 8º north west
Explanation:
This is a distance exercise with vectors the best way to work these is to decompose the vectors and perform the sum on each axis separately
To use the Cartesian system all angles must be measured from the positive side of the x-axis or the signs of the components must be assigned manually depending on the quadrant where they are.
First vector A = 2 to 20º north west
Measured from the positive x axis is θ = 180 -20 = 160º
We use trigonometry to find the components
Cos 20 = Aₓ / A
sin 20 = / A
Aₓ = A cos 160 = 2 cos 160
= A sin160 = 2 sin160
Aₓ = -1,879 mi
= 0.684 mi
Second vector B = 4 mi 10º west of the south
Angle θ = 270 - 10 = 260º
cos 2600 = Bₓ / B
sin 260 = / B
Bₓ = B cos 260
= B sin 260
Bₓ = 4 cos 260
= 4 sin 260
Bₓ = -0.6946mi
= - 3,939 mi
Third vector C = 3 mi to 15 north east
cos 15 = Cₓ / C
sin15 = / C
Cₓ = C cos 15
= C sin15
Cₓ = 3 cos 15
= 3 sin 15
Cₓ = 2,898 mi
= 0.7765 mi
Now we can find the final position of the person
X = Aₓ + Bₓ + Cₓ
X = -1.879 -0.6949 + 2.898
X = 0.3241 mi
Y = +
+
Y = 0.684 - 3.939 +0.7765
Y = -2.4785 mi
a) We use Pythagoras' theorem
R = √ (x2 + y2)
R = √ (0.3241 2 + (-2.4785) 2)
R = 2.4996 mi
R = 2.5 mi
b) let's use trigonometry
Tan θ = y / x
Tanθ = -2.4785 / 0.3241
θ = tan⁻¹ (-7,647)
θ = -82
Measured from the positive side of the x axis is Te = 360 - 82 = 278º
(90-82) south east
To return to your case you must walk in the opposite direction or Te = 98º
This is 8º north west