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Paladinen [302]
3 years ago
8

9. A Veggie meatball with v = 5.0 m/s rolls off a 1.0 m high table. How long does it take to hit the floor if no one sneezes? Wh

at is its velocity when it hits the floor?
Physics
1 answer:
Paha777 [63]3 years ago
8 0

By Considering the vertical distance and both vertical and horizontal final velocity, the time t = 0.45 s and Velocity V = 6.7 m/s

Given that a Veggie meatball with v = 5.0 m/s rolls off a 1.0 m high table.

Height h = 1.0 m

As the ball rolls off the table, it will be fallen under gravity. Where

g = 9.8 m/s^{2}

Initial vertical velocity u_{y} = 0

Initial horizontal velocity u_{x} = 5 m/s

Considering the vertical distance, the formula to use to calculate the time will be;

h = ut + 1/2gt^{2}

1 = 0 + 1/2 x 9.8t^{2}

1 = 4.9t^{2}

t^{2} = 1/4.9

t = \sqrt{0.204}

t = 0.45 seconds

It takes 0.45 seconds to hit the floor if no one sneezes.

To calculate its velocity when it hits the floor, we will need to calculate for both vertical and horizontal final velocity and find the resultant velocity of the two.

Vertical component

V_{y} = U_{y} + gt

V_{y} = 0 + 9.8(0.45)

V_{y} = 4.41 m/s

Horizontal component

V_{x} = u_{x} + at

but a = 0

V_{x} = 5 m/s

Final velocity V = \sqrt{5^{2} + 4.41^{2}  }

V = 6.67 m/s

Therefore, it will hit the floor at a velocity of 6.7 m/s

Learn more here: brainly.com/question/5063616

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A stockroom worker pushes a box with mass 11.2 kg on a horizontal surface with a constant speed of 3.30 m/s . The coefIficient o
Artemon [7]

Answer:

Explanation:

Mass =11.2kg

Constant velocity =3.3m/s

μk=0.25

Since the body is moving in constant velocity, then the acceleration is zero(0).

ΣF = Σ(ma)

The normal force acting on the body is upward and the weight is acting downward

Then ΣFy=0

Therefore, N=W

W=mg=11.2×9.8=109.76N

So, N=W=109.76N

Frictional force is given as

Fr=μkN

Fr=0.25×109.76

Fr=27.44N

Frictional force acting against the motion is 27.44N

Then the forward force moving the body forward

ΣF = Σ(ma)

Since a = 0

Then,

ΣF = 0

F-Fr=0

Then F=Fr

So the force moving the body forward is 27.44N

8 0
3 years ago
A 15.0-m uniform ladder weighing 500 N rests against a frictionless wall. The ladder makes a 60.08 angle with the horizontal. (a
grigory [225]

Answer:

a)  fr = 266.92 N,   fy = 1300 N,  b)    μ = 0.36

Explanation:

a) This is a balancing act.

Let's write the rotational equilibrium relations, where the turning point is the bottom of the ladder and the counterclockwise rotations are positive

             -w x - W x₂ + R y = 0         (1)

usemso trigonometry to find distances

            cos 60.08 = x / 7.5

            x = 7.5 cos 60.08

            x = 3.74 m

fireman

           cos 60.08 = x₂ / 4

           x2 = 4 cos 60

           x2 = 2 m

wall support

           sin 60.08 = y / 15

           y = 15 are 60.08

           y = 13 m

we substitute in equation 1

           R y = w x + W x2

            R = (w x + W x2) / y

            R = (500 3.74 +800 2) / 13

            R = 266.92 N

now let's write the expressions for the translational equilibrium

X axis

           R -fr = 0

           R = fr

           fr = 266.92 N

Y Axis  

           Fy - w-W = 0

           fy = 500 + 800

           fy = 1300 N

b) ask the friction coefficient

the firefighter's distance is

          cos 60.08 = x₃ / 9.00

          x₃ = 9 cos 60

          x₃ = 5.28 m

from equation 1

          R = (w x + W x₃) / y

          R = 500 3.74 + 800 5.28) / 13

          R = 468.769 N

we saw that

          fr = R = 468.769

The expression for the friction force is

          fr = μ N

in this case the normal is the ratio to pesos

        N = Fy

       N = 1300 N

        μ N = fr

        μ = fr / N

        μ = 468,769 / 1300

         μ = 0.36

7 0
3 years ago
An airplane flying at a speed of 130 mi/h (58.1 m/s) and at an altitude of 4.9 km drops a food package. Without a parachute, at
telo118 [61]

The food package will strike the ground at 11 degrees below the horizontal.

<h3>Time for the food package to hit the ground</h3>

The time for the food package to hit the ground is calculated as follows;

h = vt + ¹/₂gt²

<em>let the initial velocity be horizontal</em>

4900 = 0(t) + (0.5 x 9.8)t²

4900 = 4.9t²

t² = 4900/4.9

t² = 1,000

t = √1,000

t = 31.62 s

<h3> Final speed of the food package when it hits ground</h3>

vf(y) = vo + gt

vf(y) = 0 + (31.62 x 9.8)

vf(y) = 309.88 m/s

<h3>Angle of projection</h3>

The horizontal component of the speed will be constant, while vertical component will change

tan(\theta ) = \frac{V_y}{V_x} \\\\\theta = tan^{-1} (\frac{V_y}{V_x})\\\\\theta = tan^{-1} (\frac{309.88}{58.1} )\\\\\theta = 79^0

Angle below the horizontal = 90 - 79 = 11⁰

Thus, the food package will strike the ground at 11 degrees below the horizontal.

Learn more about angle of projection here: brainly.com/question/10671136

4 0
2 years ago
Which of these warnings refers to a chemical property of the material?
Wewaii [24]

Answer:

Please show the warning's

Explanation:

6 0
3 years ago
What is the net force of an object with a mass of 90.0 and accelerating at 3.0 m/s
Ksivusya [100]

The net force is 270 N

Explanation:

We can solve this problem by using Newton's second law, which states that the net force on an object is equal to the product between its mass and its acceleration:

F=ma

where

F is the force

m is the mass

a is the acceleration

In this problem, we have

m = 90.0 kg

a=3.0 m/s^2

Substituting, we find the net force on the object:

F=(90.0)(3.0)=270 N

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

3 0
3 years ago
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