Answer:
-4×-2y=14 (1)
-10×+7y=-25 (2)
multiplying eq 1 by 7 and eq 2 by 2 and add eq. 1 and 2
-28×-14y=98
-20×+14y=-50
___________
-28×=48
×=48/-28
×=-12/7
now
-4×-2y=14
-4*-12/7-2y=14
48/7-2y=14
-2y=14-48/7
-2y=(98-48)/7
-2y=50/7
y=-50/14
y=-25/7
Answer:
<em>a. The rock takes 2.02 seconds to hit the ground</em>
<em>b. The rock lands at 20,2 m from the base of the cliff</em>
Explanation:
Horizontal motion occurs when an object is thrown horizontally with an initial speed v from a height h above the ground. When it happens, the object moves through a curved path determined by gravity until it hits the ground.
The time taken by the object to hit the ground is calculated by:
The range is defined as the maximum horizontal distance traveled by the object and it can be calculated as follows:
The man is standing on the edge of the h=20 m cliff and throws a rock with a horizontal speed of v=10 m/s.
a,
The time taken by the rock to reach the ground is:
t = 2.02 s
The rock takes 2.02 seconds to hit the ground
b.
The range is calculated now:
d = 20.2 m
The rock lands at 20,2 m from the base of the cliff
Answer: Charge = -2.4x10^-9 Coulombs
Explanation:
The charge of one electron is e = -1.6x10^-19 C
Then, the charge of 1.5 x 10^10 electrons is equal to 1.5 x 10^10 times the charge of one electron:
Here i will use the relation (a^b)*(a^c) = a^(b + c)
Charge = ( 1.5 x 10^10)*( -1.6x10^-19 C) = -2.4x10^(10 - 19) C
Charge = -2.4x10^-9 C
False, the Amazon River is in South America. It's in the countries Brazil, Colombia, and Peru.
Answer:
The distance of the goggle from the edge is 5.30 m
Explanation:
Given:
The depth of pool (d) = 3.2 m
let 'i' be the angle of incidence
thus,
i =
i = 67.75°
Now, Using snell's law, we have,
n₁ × sin(i) = n₂ × 2 × sin(r)
where,
r is the angle of refraction
n₁ is the refractive index of medium 1 = 1 for air
n₂ is the refractive index of medium 1 = 1.33 for water
now,
1 × sin 67.75° = 1.33 × sin(r)
or
r = 44.09°
Now,
the distance of googles = 2.2 + d×tan(r) = 2.2 + (3.2 × tan(44.09°) = 5.30 m
Hence, <u>the distance of the goggle from the edge is 5.30 m</u>