Answer:
3.1 m/s²
Explanation:
Given:
Mass of the balloon (m) = 11.4 g = 0.0114 kg ( 1 kg = 1000 g)
Force acting on the balloon (F) = 0.035 N
Acceleration with which the balloon must be hit (a) = ?
Now, we know that, from Newton's second law, net force acting on an object is equal to the product of its mass and acceleration.
Therefore, framing in equation form, we have:
Rewriting in terms of acceleration 'a', we get:
Now, substitute the given values and solve for 'a'. This gives,
Therefore, the acceleration of the water balloon to reach the target must be equal to 3.1 m/s².
Answer:
51.2 mi/h
Explanation:
Total distance, d = 100 miles
First 60 miles with speed 55 mi/h
Next 40 miles with speed 75 mi/h
Time taken for first 60 miles, t1 = 60 / 55 = 1.09 h
Time taken for 40 miles, t2 = 40 / 75 = 0.533 h
Time spent to get stuck, t3 = 20 min = 0.33 h
Total time, t = t1 + t2 + t3 = 1.09 + 0.533 + 0.33 = 1.953 h
The average speed is defined as the ratio of total distance traveled to the total time taken.
Average speed =
Thus, the average speed of the journey is 51.2 mi/h.
Answer:
15.1°
Explanation:
The horizontal velocity of the hockey puck is constant during the motion, since there are no forces acting along this direction:
Instead, the vertical velocity changes, due to the presence of the acceleration due to gravity:
(1)
where
is the initial vertical velocity
g = 9.8 m/s^2 is the gravitational acceleration
t is the time
Since the hockey puck falls from a height of h=2.00 m, the time it needs to reach the ground is given by
Substituting t into (1) we find the final vertical velocity
where the negative sign means that the velocity is downward.
Now that we have both components of the velocity, we can calculate the angle with respect to the horizontal:
Answer:
T₂ = 5646 K
Explanation:
Let's start by finding the power received by the first disc, for this we use Stefan's law
P = σ. A e T⁴
Where next is the Stefam-Bolztmann constant with value 5,670 10-8 W / m² K⁴, A is the area of the disk, T the absolute temperature and e the emissivity that for a black body is 1
The intensity is defined as the amount of radiation that arrives per unit area. For this we assume that the radiation expands uniformly in all directions, the intensity is
I = P / A
Writing this expression for both discs
I₁ A₁ = I₂ A₂
I₂ = I₁ A₁ / A₂
The area of a sphere is
A = 4π r²
I₂ = I₁ (r₁ / r₂)²
r₂ = r₁ ± 5
I₁ = I₂ ( (r₁ ± 5)/r₁)²
.
Let's write the Stefan equation
P / A = σ e T⁴
I = σ e T⁴
This is the intensity that affects the disk, substitute in the intensity equation
σ e₁ T₁⁴ = σ e₂ T₂⁴ (r₂ / r₁)²
The first disc indicates that it is a black body whereby e₁ = 1, the second disc, as it is painted white, the emissivity is less than 1, the emissivity values of the white paint change between 0.90 and 0.95, for this calculation let's use 0.90 matt white
e₁ T₁⁴ = T₂⁴ (r1 + 5)²/r₁²
T₁ = T₂ {(e₂/e₁)}^{1/4} √(1 ± 1/ r₁)
If we assume that r₁ is large, which is possible since the disks are in deep space, we can expand the last term
(1 ±x) n = 1 ± n x
Where x = 5 / r₁ << 1
We replace
T₁ = T₂ {(e₂/e₁)}^{1/4} (1 ± ½ 5/r1)
T₁ = T₂ {(e₂)}^{1/4} (1 ± 5/2 1/r1)
If the discs are far from the star, they indicate that they are in deep space, the distance r₁ from being grade by which we can approximate; this is a very strong approach
T₁ = T₂ {(e₂)}^{1/4} ¼
T<u>₁</u> = T₂ 0.9
5500 = T₂ 0.974
T₂ = 5646 K