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abruzzese [7]
4 years ago
15

The weight of a bucket is 186 N. The bucket is being raised by two ropes. The free-body diagram shows the forces acting on the b

ucket.
The acceleration of the bucket, to the nearest tenth, is
___ m/s2.

Physics
1 answer:
amm18124 years ago
4 0
Fnet=(115+106)-186= 34 N

mass=Force/g= 186N/9.8m/s^2 = 18.98 kg

a=fnet/mass => 34N/18.98kg = 1.79 m/s^2

so A= 1.8m/s^2
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Electrically inert metal ball A is connected to the ground by a wire. What happens to the charge of this ball if you bring a neg
kaheart [24]

Explanation:

They will repel, meaning that they are made of an electrical conductor.

7 0
3 years ago
A 1200 Kg car rounds a corner of radius r = 45m. If the coefficient of static friction between the ties and the road is us = 0.8
Vaselesa [24]

Answer:

The greatest speed of the car is 19.36m/s

Explanation:

The maximum speed the car will attain without skidding is given by:

F= uN = umg ...eq1

But F = mv^2/r

mv^2/r = umg

Dividing both sides by m, leaves you with:

V= Sqrt(ugr)

Where u = coefficient of static friction

g = acceleration due to gravity

r = raduis

Given:

U = 0.82

r=0.82

g= 9.8m/s

V = Sqrt(0.82 × 9.8 × 45)

V = Sqrt(374.85)

V = 19.36m/s

5 0
4 years ago
A free falling object has the velocity time graph shown. What is the objects displacement between 0.0 and 6.0s
zloy xaker [14]

For free fall motion the displacement can be found by graphically as well as by kinematics equation

Here acceleration of object is constant as it fall due to gravity so we can use

d = v_i * t + \frac{1}{2}at^2

here if body starts with zero initial speed then we can say

d = 0 + \frac{1}{2}*9.8*t^2

here we need to find the displacement from t = 0 to t = 6s

so we can say

d = \frac{1}{2}*9.8*6^2

d = 176.4 m

so the displacement will be 176.4 m

in order to find the displacement from the graph of velocity and time we need to find the area under the graph for given time interval that will also give us same displacement for given period of time.

6 0
4 years ago
Read 2 more answers
calculate the initial velocity of an object displaced 33 meters while accelerating 8 m/s^2 to a final velocity of 68 m/s how do
Grace [21]

Use the equation

{v_f}^2-{v_i}^2=2a\Delta x

The final velocity v_f is 68 m/s, the acceleration a is 8 m/s^2, and the net displacement \Delta x is 33 m, so you can solve for the initial velocity v_i:

\left(68\,\dfrac{\mathrm m}{\mathrm s}\right)^2-{v_i}^2=2\left(8\,\dfrac{\mathrm m}{\mathrm s^2}\right)(33\,\mathrm m)

{v_i}^2=4096\,\dfrac{\mathrm m^2}{\mathrm s^2}

v_i=64\,\dfrac{\mathrm m}{\mathrm s}

8 0
3 years ago
Spud Webb, height 5'7", was one of the shortest basketball players to play in the NBA. But he had an impressive vertical leap; h
nlexa [21]

Answer:

a)   v = 4.64 m / s , b)    t = 0.947 s , c)    t = 0.947 s

Explanation:

We will work on this exercise with vertical launch kinematics, let's start by calculating the height of the jumper in the SI system

         y₀ = 5 ’(0.3048 m / 1’) + 7 ”(2.54 10-2 m / 1”) = 1.70 m

The distance they give is the height of the jump

          y = 1.10 m

Let's use energy conservation

Starting point.  On the floor

          Em₀ = K = ½ m v²

Final point. Maximum height

           Em_{f} = U = m g y

           Em₀ = Em_{f}

           ½ m v² = m g y

           v = √2gy

         

Let's calculate

          v = √(2 9.8 1.10)

          v = 4.64 m / s

b) Air time is the time to go up plus the time to go down, which is the same

For maximum height the speed is zero

          v = v₀ - g t₁

          t₁ = v₀ / g

          t₁ = 4.64 /9.8

          t₁ = 0.4735 s

The total time is

           t = 2 t₁

           t = 2 0.4735

           t = 0.947 s

c) if it takes a distance of 0.40 to reach speed, what is the acceleration, as it stands on the floor its initial speed is zero

          v² = v₀² + 2 a x

         a = v² / 2x

         a = 4.64²/2 0.40

         a = 26.9 m / s²

7 0
4 years ago
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