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2 years ago

9

A 20 kg mass is hanging on one side of a pulley and a 5 kg mass is hanging on the other. Determine the acceleration of each mass

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1 answer:

Ann [662]2 years ago

5 0

Answer:

i believe it's 7

Explanation:

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The radius of curvature is smaller at the top than on the sides so that the downward centripetal acceleration at the top will be

kap26 [50]

Answer:

$v = 14.86 m/s$

Explanation:

As we know that the force equation at the top is given as

$\frac{mv^2}{R} = ma$

now we know that

$a_c = 1.5 g$

so we have

$\frac{v^2}{R} = 1.5 g$

$v = \sqrt{1.5 Rg}$

so we will have

$v = \sqrt{1.5(15)(9.81)}$

$v = 14.86 m/s$

3 0

2 years ago

A child and sled with a combined mass of 50.0 kg slide down a frictionless slope. if the sled starts from rest and has a speed o

Furkat [3]

<span> Using conservation of energy

Potential Energy (Before) = Kinetic Energy (After)

mgh = 0.5mv^2

divide both sides by m

gh = 0.5v^2

h = (0.5V^2)/g

h = (0.5*2.2^2)/9.81

h = 0.25m

</span>

4 0

2 years ago

Read 2 more answers

An open system starts with 52 J of mechanical energy. The energy changes

spayn [35]

Answer:

I think the answer is D,54 joules

5 0

2 years ago

A firecracker breaks up into two pieces , one has a mass of 200 g and files off along the x –axis with a speed of 82.0 m/s and t

Readme [11.4K]

Answer:

A) 21.2 kg.m/s at 39.5 degrees from the x-axis

Explanation:

Mass of the smaller piece = 200g = 200/1000 = 0.2 kg

Mass of the bigger piece = 300g = 300/1000 = 0.3 kg

Velocity of the small piece = 82 m/s

Velocity of the bigger piece = 45 m/s

Final momentum of smaller piece = 0.2 × 82 = 16.4 kg.m/s

Final momentum of bigger piece = 0.3 × 45 = 13.5 kg.m/s

since they acted at 90oc to each other (x and y axis) and also momentum is vector quantity; then we can use Pythagoras theorems

Resultant momentum² = 16.4² + 13.5² = 451.21

Resultant momentum = √451.21 = 21.2 kg.m/s at angle 39.5 degrees to the x-axis ( tan^-1 (13.5 / 16.4)

5 0

2 years ago

During a baseball game, a batter hits a high pop-up. If the ball remains in the air for 6.22 s, how high does it rise? The accel

BigorU [14]

Answer:

47.4 m

Explanation:

When an object is thrown upward, it rises up, it reaches its maximum height, and then it goes down. The time at which it reaches its maximum height is half the total time of flight.

In this case, the time of flight is 6.22 s, so the time the ball takes to reach the maximum height is

$t=\frac{6.22}{2}=3.11 s$

Now we consider only the downward motion of the ball: it is a free fall motion, so we can find the vertical displacement by using the suvat equation

$s=ut+\frac{1}{2}gt^2$

where

s is the vertical displacement

u = 0 is the initial velocity

t = 3.11 s is the time

$g=9.8 m/s^2$ is the acceleration of gravity (taking downward as positive direction)

Solving the formula, we find

$s=\frac{1}{2}(9.8)(3.11)^2=47.4 m$

7 0

3 years ago