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3 years ago

A supply bag is dropped from a rescue plane. After the bag falls for 3.2 seconds , what is the velocity of the bag?

Physics

1 answer:

loris [4]3 years ago

8 0

Answer: -31.36 m/s

Explanation:

This is a problem of motion in one direction (specifically vertical motion), and the equation that best fulfills this approach is:

$V_{f}=V_{o}+a.t$ (1)

Where:

$V_{f}$ is the final velocity of the supply bag

$V_{o}=0$ is the initial velocity of the supply bag (we know it is zero because we are told it was "dropped", this means it goes to ground in free fall)

$a=g=-9.8m/s^{2}$ is the acceleration due gravity (the negtive sign indicates the gravity is downwards, in the direction of the center of the Earth)

$t=3.2s$ is the time

Knowing this, let's solve (1):

$V_{f}=0+(-9.8m/s^{2})(3.2s)$ (2)

Finally:

$V_{f}=-31.36m/s$ Note the negative sign is because the direction of the bag is downwards as well.

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Mrs. Lynn's art class is making pencil holders out of soup cans. Each student is first covering a can with felt. What will the c

DochEvi [55]

Answer:

25.13 cm

Explanation:

Mrs. Lynn's art class is making pencil holders out of soup cans

These soup cans are definitely in cylindrical forms.

Each student is first covering a can with felt.

A felt is a type of material that is used for covering other objects. Here a felt is used to cover soup cans.

So if the diameter of the bottom of the can is 8 cm (i.e d = 8cm)

What will the circumference of the piece of felt need to be in order to cover the bottom of the can?

The felt will also take the circumference of a cylinder; and which is given by the expression

C = 2 πr

since diameter (d) = 8 cm

radius (r) = $\frac{d}{2}$

r = $\frac{8}{2}$

r = 4 cm

C = 2 × π × 4 cm

C = 2 × 3.142 × 4 cm

C = 25.13 cm

∴ the circumference of the piece of felt need to be 25.13 cm in order to cover the bottom of the can.

6 0

3 years ago

2. It is to bring down the foot forcibly to the f a heavy step) with or without weight. 2 Ormain lateral position - This is done

irga5000 [103]

Im confused bc im not sure if this is a full question

5 0

3 years ago

The rate at which a force displaces a mass a horizontal distance is measured in?

kodGreya [7K]

Answer:

c. joules

Explanation:

The rate at which a force displaces a mass a horizontal distance is measured in joules. In science, this phenomenon is known as work done.

Work done can be defined as the rate at which a force acting on an object or a body causes it to experience a displacement. The work done is a scalar quantity and is measured in joules (J).

Mathematically, work done is given by the formula;

Work done = force * distance

$W = F * d$

Where,

W is the work done

F represents the force acting on a body.

d represents the distance covered by the body.

For example, a bull pulling a plough through a farm, a girl pushing a shopping cart down the aisle of a supermarket etc.

6 0

3 years ago

Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 25.0 °C for the following reacti

Yakvenalex [24]

The equilibrium constant of the reaction at 25⁰c will be 426827.5.

Theory-

Equilibrium constant :The equilibrium constant comes from the chemical equilibrium law. For the chemical equilibrium state, at a fixed constant temperature, the ratio of the product of the reaction's multiplication to the concentration of its reactants' multiplication, and each is raised to the power to the corresponding coefficients of the elements in the reaction.

The chemical equilibrium is given by for a general chemical reaction.

a. A+ b. B ⇌ c. C+ d. D,.

Kc =[C]c [D]d/[A]a [B]b.

Gibb's free energy :The second law of thermodynamics can be arranged in such a way that it gives a new expression when a chemical reaction happens at a constant temperature and constant pressure.

G=H-TS

Calculations:-

T=25⁰c

G=51.4 x 10³J

$\\\\k=GR+\frac{nRT}{Z} \\$

k= equilibrium constant ,G=Gibbs free energy ,n= no. of moles ,R=Gas constant ,T=temperature ,Z=compressibility

$Ideal.Situation=\left \{ {{Z=1} \atop {n=1}} \right.$

$\\\\\\k=GR+RT$

k=51.4 x 10³ x 8.3 + 8.3 x 25

k=426827.5

To learn equilibrium constant-

brainly.com/question/19669218

#SPJ4

6 0

1 year ago

A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t

Yuliya22 [10]

Answer:

a) $W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

b) $W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

c) $V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

d) $d_1 =0.183m$ or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

$W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

Part b

For this case first we can convert the spring constant to N/m like this:

$2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}$

And the work donde by the spring on this case is given by:

$W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

$W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i$

And if we solve for the initial velocity we got:

$V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

$-1/2mV^2_i = W_g + W_{spring}$

And replacing we got:

$-1/2mV^2_i =mg d_1 -1/2 k d^2_1$

And we can put the terms like this:

$\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0$

If we multiply all the equation by 2 we got:

$k d^2_1 -2 mg d_1 -m V^2_i =0$

Now we can replace the values and we got:

$200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0$

$200 d^2_1 -2.058 d_1 -6.3525=0$

And solving the quadratic equation we got that the solution for $d_1 =0.183m$ or 18.3 cm because the negative solution not make sense.

5 0

3 years ago