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Mila [183]
3 years ago
12

Describe the acceleration of your bicycle as you ride it from your home to the store

Physics
1 answer:
erik [133]3 years ago
8 0
<span>While you're going to the store, your acceleration changes. Some times it increases your overall speed sometimes it reduces it. Constant acceleration does not occur because it would mean that you would constantly accelerate and eventually go past the store. Even reduction of speed is a type of acceleration in physics. When you reach it, we can then calculate how much your velocity was on average and analyze how changing acceleration would've affected it.</span>
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Question in pic asap plz
REY [17]
We can’t see anything
5 0
3 years ago
You work at a garden store for the summer, and you lift a 14 kg bag of fertilizer with a force of 227 N.
nika2105 [10]

Answer:

(a) Acceleration of the bag will be a=16.214m/sec^2  

(B) Weight of the bag will be 137.2 N

Explanation:

We have given mass of the bag m = 14 kg

Force with which bag is lifted = 227 N

(A) According to newtons law we force is equal to F = ma , here m is mass and a is acceleration

So 227=14\times a

a=16.214m/sec^2

(b) Acceleration due to gravity g=9.8m/sec^2

We know that weight is given by W = mg , here m is mass and g is acceleration due to gravity

So weight W=mg=14\times 9.8=137.2N

So weight of the bag will be 137.2 N

8 0
3 years ago
A 3 Kg ball at the end of a string is spun in a circle. If the length of the string is 2 m and the velocity of the ball is 6 m/s
gulaghasi [49]
M is 3kg
r is 2m
v is 6m/s

Fc = mv^2 / r
Fc = (3x6^2)/2 =54

a = v^2/r
a = 6^2 /2 = 18
3 0
2 years ago
6. A .25 kg arrow with a velocity of 12 m/s to the west strikes and pierces the center of a 6.8 kg target. a. What is the final
Alenkasestr [34]

Answer:

(a) the final velocity of the combined mass is 9.43 m/s

(b) the decrease in kinetic energy during the collision is 386.1 J

Explanation:

Given;

mass of arrow, m₁ = 25 kg

initial velocity of arrow, u₁ = 12 m/s

mass of target, m₂ = 6.8 kg

initial velocity of the target, u₂ = 0

Part (a)

From the principle of conservation of linear momentum;

Total momentum before collision = Total momentum after collision

m₁u₁ + m₂u₂ = v(m₁+m₂)

where;

v is the final velocity of the combined mass

25 x 12 + 0 = v(25 + 6.8)

300 = v(31.8)

v = 300/31.8

v = 9.43 m/s

Part(b)

Kinetic Energy, K.E = ¹/₂mv²

Initial kinetic energy =  ¹/₂m₁u₁² + ¹/₂m₂u₂²  = ¹/₂ x 25 x (12)² + 0 = 1800 J

Final kinetic energy = ¹/₂m₁v² + ¹/₂m₂v² = ¹/₂v²(m₁ + m₂)

                                                               = ¹/₂ x (9.43)²(25+6.8)

                                                               = 1413.91 J

Decrease in kinetic energy = Initial K.E - Final K.E

Decrease in kinetic energy = 1800J - 1413.91 J = 386.1 J

                               

4 0
3 years ago
A flare is launched from a boat. The height, , in meters, of the flare above the water is approximately modelled by the function
Tpy6a [65]

Answer:

10 seconds

Explanation:

If the height is modeled by the function h(t)=-15t^{2} +150t, then the seconds it takes to reach the water (when the height equals 0) is modeled by the following.

0=-15t^{2} +150t\\0=-15t*(t-10)\\\\t=0\\t=10

6 0
3 years ago
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