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2 years ago

Cuestionario:

Physics

1 answer:

Sati [7]2 years ago

3 0

Explanation:

1.El objetivo principal de realizar la llamada entrada en calor o calentamiento, es preparar el cuerpo para la actividad física o deportiva. Numerosas lesiones y ciertos problemas cardíacos como algunas arritmias, están asociados a la ejercitación violenta sin mediar un adecuado calentamiento.

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What ideas do you have a bout why different places have lower or higher rates of skin cancer?

AnnZ [28]

Answer: Some places like Florida and California are both very hot and dry places, the sun damages your skin more in sunny places then in colder more breezy places.

Explanation:

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2 years ago

A 403.0-kg copper bar is melted in a smelter. The initial temperature of the copper is 320.0 K. How much heat must the smelter p

aleksklad [387]

The answer is 2.49 x 10^5 KJ. This was obtained (1) use the formula for specific heat to achieve Q or heat then (2) get the energy to melt the copper lastly (3) Subtract both work and the total energy required to completely melt the copper bar is achieved.

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3 years ago

Read 2 more answers

An astronaut on Pluto attaches a small brass ball to a 1.00-m length of string and makes a simple pendulum. She times 10 complet

pickupchik [31]

Answer:

The acceleration due to gravity at Pluto is 0.0597 m/s^2.

Explanation:

Length, L = 1 m

10 oscillations in 257 seconds

Time period, T = 257/10 = 25.7 s

Let the acceleration due to gravity is g.

Use the formula of time period of simple pendulum

$T = 2\pi\sqrt{\frac{L}{g}}\\\\25.7 = 2 \times 31.4\sqrt{\frac{1}{g}}\\\\g = 0.0597 m/s^2$

7 0

3 years ago

How do I calculate the tension in the horizontal string?

matrenka [14]

ANSWER

T₂ = 10.19N

EXPLANATION

Given:

• The mass of the ball, m = 1.8kg

First, we draw the forces acting on the ball, adding the vertical and horizontal components of each one,

In this position, the ball is at rest, so, by Newton's second law of motion, for each direction we have,

$\begin{gathered} T_{1y}-F_g=0_{}_{}_{} \\ T_2-T_{1x}=0 \end{gathered}$

The components of the tension of the first string can be found considering that they form a right triangle, where the vector of the tension is the hypotenuse,

$\begin{gathered} T_{1y}=T_1\cdot\cos 30\degree \\ T_{1x}=T_1\cdot\sin 30\degree \end{gathered}$

We have to find the tension in the horizontal string, T₂, but first, we have to find the tension 1 using the first equation,

$T_1\cos 30\degree-m\cdot g=0$

Solve for T₁,

$T_1=\frac{m\cdot g}{\cos30\degree}=\frac{1.8kg\cdot9.8m/s^2}{\cos 30\degree}\approx20.37N$

Now, we use the second equation to find the tension in the horizontal string,

$T_2-T_1\sin 30\degree=0$

Solve for T₂,

$T_2=T_1\sin 30\degree=20.37N\cdot\sin 30\degree\approx10.19N$

Hence, the tension in the horizontal string is 10.19N, rounded to the nearest hundredth.

8 0

11 months ago

A horizontal force of 750 N is needed to overcome the force of static friction between a level floor and a 250-kg crate. What is

Aleksandr [31]

Answer:

The acceleration of the crate is 1.8 m/s² so the answer is a.

Explanation:

The very first thing you must do when solving this problem is to draw a free body diagram. (The body diagram is attached to this answer)

So once we got the free body diagram, we can analyze it and build our sum of forces in the x and y directions. Notice that according to the diagram, there are 4 forces to this problem, Normal (N), Weight (W), kinetic friction (fk) and the 750N force.

As one may see in the free body diagram, two of the forces are vertical forces: N and W, so we can use them to build a sum of forces:

Starting with the sum of forces in the y-direction, we get:

Σ $F_{y}$ =0

We set the sum equal to zero because there is no movement in the y-direction, so the system is in vertical equilibrium.

so the sum will be:

$N-W=0$

when solving for N we get that:

N=W

where W is found by multiplying the mass of the crate by the acceleration of gravity:

N=250kg*9.8m/s²

N=2450N

Once we found the normal force, we can use it to find the kinetic friction which is given by the following formula:

$f_{k}=N$ μ

where μ is the kinetic friction coefficient.

So we get that the kinetic friction is:

$f_{k}=2450N*0.12$

$f_{k}=294$

With this information we can go ahead and find the sum of horizontal forces:

Σ $F_{x}$ =ma

In this case the sum is equal to mass times acceleration because the crate is moving horizontally due to the action of a force, so it will have an acceleration.

so the sum of forces look like this:

750N- $f_{k}$ =ma

750N-294N=(250kg)a

when solving for a we get:

$a=\frac{759N-294N}{250kg}\\ \\a=1.8m/s^{2}$

so the crate's acceleration is 1.82m/s².

5 0

3 years ago