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babymother [125]
3 years ago
6

A 100kg bag of sand has a weight on 100 N. When dropped it's acceleration is what?​

Physics
1 answer:
Vesnalui [34]3 years ago
7 0

Answer:

9.8m/s

Explanation:

acceleration due to gravity is independent of mass

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Transfer payments are used to redistribute money to various segments of society. please select the best answer from the choices
mote1985 [20]

Answer:

True

Explanation:

You pay back what you owe for what youve purchased

7 0
2 years ago
Your average speed on the first half of a car trip is 69.0 km/h. How fast do you have to drive on the second half of the trip to
vova2212 [387]

Answer:

13 km/h

Explanation:

Average speed = distance/time

Let the total distance and total time taken for the whole trip be d km and t hours respectively

Average speed for the whole trip = 82 km/h

d = 82t

The distance covered in the first half = d1/2

Time taken = t/2

Average speed = 69 km/h

69 = d1/2 ÷ t/2

d1 = 69t

The distance covered in the second half = d2/2

Time taken = t/2

Let the average sly for the see half be A

A = d2/2 ÷ t/2

d2 = At

d = d1 + d2

82t = 69t + At

At = 82t - 69t

At = 13t

A = 13t/t = 13 km/h

3 0
4 years ago
Constants A capacitor is connected across an ac source that has voltage amplitude 59 0 V and frequency 77 0 Hz Part C What is th
Vladimir79 [104]

Answer:

C = 1.77 \times 10^{-4} F

Explanation:

As we know that in AC circuit we have

V = i x_c

here we have

V = 59 V

i = 5.05 A

so we will have

x_c = \frac{59}{5.05}

x_c = 11.68 ohm

also we know that

x_c = \frac{1}{\omega C}

here we will have

11.68 = \frac{1}{(2\pi 77)C}

C = 1.77 \times 10^{-4} F

7 0
3 years ago
What happens to jetstream’s as they get closer to the equator
MAVERICK [17]

Answer:They stop because jet streams follow boundaries between hot and cold air.

Explanation:

6 0
4 years ago
Read 2 more answers
Object A of mass 0.70 kg travels horizontally on a frictionless surface at 20 m/s. It collides with object B, which is initially
AnnZ [28]

Answer:

25.71 kgm/s

Explanation:

Let K₁ and K₂ be the initial and final kinetic energies of object A and v₁ and v₂ its initial and final speeds.

Given that K₂ = 0.7K₁

1/2mv₂² = 0.7(1/2mv₁²)

v₂ = √0.7v₁ = √0.7 × 20 m/s = ±16.73 m/s

Since A rebounds, its velocity = -16.73 m/s and its momentum change, p₂ = mΔv = m(v₂ - v₁) = 0.7 kg (-16.73 - 20) m/s = 0.7( -36.73) = -25.71 kgm/s.

Th magnitude of object A's momentum change is thus 25.71 kgm/s

6 0
3 years ago
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