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2 years ago

AYUDA!!!!

Physics

1 answer:

xeze [42]2 years ago

3 0

Answer:

4. B) 69N

5. B) 7.2kg

Explanation:

Dados los siguientes parámetros;

Masa = 7kg

Peso = 71N

La aceleración debida a la gravedad es igual a 9.8m/s².

a. Para determinar el peso de la siguiente masa;

Matemáticamente, el peso viene dada por la fórmula;

Peso = masa * aceleración debida a la gravedad

Peso = 7 * 9.8

Peso = 68.6 ≈ 69 Newton.

b. ¿Cuál es la masa sobre la tierra de un objeto que pesa.

Peso = masa * aceleración debida a la gravedad

71 = masa * 9.8

Masa = 71/9.8

Masa = 7.2 Kilograms.

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The phosphorus cycle is important to ecosystems. Choose all of these statements that are true concerning the phosphorus cycle.

Lesechka [4]

Answer:

The largest reservoir of phosphorous is sedimentary rock.

Major sources of phosphorous to aquatic ecosystems are fertilizer runoff, sewage leaks, and industrial wastes.

Eccess phosphorous can lead to eutrophication

Explanation:

Phosphorus come from different sources such as aquatic ecosystems and fertilizers used for plants. When these substances containing phosphorus and those from industrial wastes find their way into water bodies, they tend to cause eutrophication, which is the natural enrichment of water bodies.

Also, it is known that a very small portion of phosphoric acid contribute to acid rain in the atmosphere.

7 0

3 years ago

Read 2 more answers

During a rockslide, a 670 kg rock slides from rest down a hillside that is 740 m along the slope and 240 m high. The coefficient

ElenaW [278]

a) $1.58\cdot 10^6 J$

b) $1.15\cdot 10^6 J$

c) $0.43\cdot 10^6 J$

d) 35.8 m/s

Explanation:

The gravitational potential energy of an object is the energy possessed by the object due to its location with respect to the ground.

It is given by:

$U=mgh$

where

m is the mass of the object

g is the acceleration due to gravity

h is the height of the object, relative to a reference level

Here, the reference level is taken at the bottom of the hill (where the potential energy is zero).

So, we have:

m = 670 kg is the mass of the rock

$g=9.8 m/s^2$

h = 240 m is the initial height of the rock

So, the potential energy of the rock just before the slide is

$U=(670)(9.8)(240)=1.58\cdot 10^6 J$

The energy transferred to thermal energy during the slide is equal to the work done by friction, which is:

$W=F_f d$

where

$F_f$ is the force of friction

d = 740 m is the displacement of the rock along the ramp

The force of friction is given by:

$F_f=-\mu mg cos \theta$

where

$\mu=0.25$ is the coefficient of friction

m = 670 kg is the mass of the rock

$\theta$ is the angle of the ramp

Since we know the lenght of the ramp (d = 740 m) and the height (h = 240 m), we can find the angle:

$\theta=sin^{-1}(\frac{h}{d})=sin^{-1}(\frac{240}{740})=18.9^{\circ}$

Therefore, the work done by friction is:

$W=-\mu m g cos \theta d =-(0.25)(670)(9.8)(cos 18.9^{\circ})(740)=-1.15\cdot 10^6 J$

So, the energy transferred to thermal energy is $1.15\cdot 10^6 J$ .

According to the law of conservation of energy, the kinetic energy of the rock as it reaches the bottom of the hill will be equal to the initial potential energy (at the top) minus the energy transformed into thermal energy.

Therefore, we have:

$K_f = U_i -E_t$

where here we have:

$U_i=1.58\cdot 10^6 J$ is the potential energy of the rock at the top of the hill

$E_t=1.15\cdot 10^6 J$ is the energy converted into thermal energy

Substituting, we find

$K_f=1.58\cdot 10^6-1.15\cdot 10^6=0.43\cdot 10^6 J$

So, this is the kinetic energy of the rock at the bottom of the hill.

The kinetic energy of the rock at the bottom of the hill can be rewritten as

$K_f=\frac{1}{2}mv^2$

where

m is the mass of the rock

v is its final speed

In this problem, we have:

$K_f=0.43\cdot 10^6 J$ is the final kinetic energy of the hill

m = 670 kg is the mass of the rock

Therefore, the final speed of the rock is:

$v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(0.43\cdot 10^6)}{670}}=35.8 m/s$

7 0

3 years ago

A flat circular loop of wire of radius 0.50 m that is carrying a 2.0-A current is in a uniform magnetic field of 0.30 T. What is

Luba_88 [7]

Answer:

The magnitude of the magnetic torque on the loop when the plane of its area is perpendicular to the magnetic field is 0.4713 J

Explanation:

Given;

radius of the circular loop of wire = 0.5 m

current in circular loop of wire = 2 A

strength of magnetic field in the wire = 0.3 T

τ = μ x Bsinθ

where;

τ is the magnitude of the magnetic torque

μ is the dipole moment of the magnetic field

θ is the inclination angle, for a plane area perpendicular to the magnetic field, θ = 90

μ = IA

where;

I is current in circular loop of wire

A is area of the circular loop = πr² = π(0.5)² = 0.7855 m²

μ = 2 x 0.7885 = 1.571 A.m²

τ = μ x Bsinθ = 1.571 x 0.3 sin(90)

τ = 0.4713 J

Therefore, the magnitude of the magnetic torque on the loop when the plane of its area is perpendicular to the magnetic field is 0.4713 J

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3 years ago

25 PTS

yarga [219]

The answer would be C

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3 years ago

If you kicked your mom <br><br> would she be mad?

vfiekz [6]

Answer:

Yes !

Explanation:

8 0

2 years ago