Answer:
Dx = -0.5
Dy = -0.25
Explanation:
Two vectors are given in rectangular components form as follows:
A = i + 6j
B = 3i - 7j
It is also given that:
A - B - 4D = 0
so, we solve this to find D vector:
(i + 6j) - (3i - 7j) - 4D = 0
- 2i - j = 4D
D = - (2/4)i - (1/4)j
D = - (1/2)i - (1/4)j
<u>D = - 0.5i - 0.25j</u>
Therefore,
<u>Dx = -0.5</u>
<u>Dy = -0.25</u>
Answer:
Option D) 4A
Explanation:
As the cycle of the wave passes by, the amplitude gives the longest journey when the spot travels from the undistributed position. During each cycle the spot travels "Four times" .
Considering one of this cycle, if it begins to travel from it's undistributed position , there would be four movements i.e
* Upward movement through distance A
*Downward movement through distance A
*Downward again through distance A
*Upward through distance A.
Then it would travel back to its undistributed position held
Answer:
5.66 × 10⁻²³ m/s
Explanation:
If i assume i can jump as high as h = 2 m, my initial velocity is gotten from v² = u² + 2gh. Since my final velocity v = 0, u = √2gh = √(2 × 9.8 × 2) = √39.2 m/s = 6.26 m/s.
Since initial momentum = final momentum,
mv₁ + MV₁ = mv₂ + MV₂ where m, M, v₁, V₁, v₂ and V₂ are my mass, mass of earth, my initial velocity, earth's initial velocity, my final velocity and earth's final velocity respectively.
My mass m = 54 kg, M = 5.972 × 10²⁴ kg, v₁ = 6.26 m/s, V₁ = 0, v₂ = 0 and V₂ = ?
So mv₁ + M × 0 = m × 0 + MV₂
mv₁ = MV₂
V₂ = mv₁/M = 54kg × 6.26 m/s/5.972 × 10²⁴ kg = 338.093/5.972 × 10²⁴ = 56.61 × 10⁻²⁴ m/s = 5.661 × 10⁻²³ m/s ≅ 5.66 × 10⁻²³ m/s
Answer:
The object will travel at the speed of 16 m/s.
Explanation:
Given
To determine
How fast is the object traveling?
<u>Important Tip:</u>
The product of the mass and velocity of an object — momentum.
Using the formula
where
Thus, in order to determine the speed of the object, all we need to do is to substitute p = 64 and m = 4 in the formula
switch the equation
divide both sides by 4
simplify
m/s
Therefore, the object will travel at the speed of 16 m/s.
