Refer to the diagram shown below.

Assume that air resistance is ignored.

Note:
The distance, h, of a falling object with initial vertical velocity of zero at time t is
h = (1/2)gt²
where
g = 9.8 m/s²

The initial vertical velocity of the supplies is 0 m/s.
It the time taken for the supplies to reach the ground is t, then
(50 m) = (1/2)*(9.8 m/s²)*(t s)²

Hence obtain
t² = 50/4.9 = 10.2041
t = 3.1944 s

The horizontal distance traveled at a speed of 100 m/s is
d = (100 m/s)*(3.1944 s) = 319.44 m

Answer: 319.4 m (nearest tenth)

3 0

3 years ago

Is magnitude a<br> vector<br> scalars<br> both<br> neither

Doss [256]

Answer:

scalars

Explanation:

7 0

2 years ago

Eli's gross income as a painter is $30,000 per year, which places him in the 15% approximate tax bracket. How much does he pay i

Umnica [9.8K]

Answer: $4,500

Explanation: 1% of $30,000 would be $300. If you multiply that by 15, you get $4,500, so 15% of 30,000 is 4,500.

3 0

2 years ago

A test charge of -3x10^-7 C is located 7 cm to the right of a charge of -9x10^-6 C and 20 cm to the left of a charge of +10x10^-

musickatia [10]

Answer:

5.634 N rightwards

Explanation:

qo = - 3 x 10^-7 C

q1 = - 9 x 10^-6 C

q2 = 10 x 10^-6 C

r1 = 7 cm = 0.07 m

r2 = 20 cm = 0.2 m

The force on test charge due to q1 is F1 which is acting towards right

According to the Coulomb's law

$F_{1}=\frac{Kq_{1}q_{0}}{r_{1}^{2}}$

F1 = (9 x 10^9 x 9 x 10^-6 x 3 x 10^-7) / (0.07 x 0.07)

F1 = 4.959 N rightwards

The force on test charge due to q2 is F1 which is acting towards right

According to the Coulomb's law

$F_{2}=\frac{Kq_{2}q_{0}}{r_{2}^{2}}$

F2 = (9 x 10^9 x 10 x 10^-6 x 3 x 10^-7) / (0.2 x 0.2)

F2 = 0.675 N rightwards

Net force on the test charge

F = F1 + F2 = 4.959 + 0.675 = 5.634 N rightwards

3 0

3 years ago