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3 years ago

Does increasing impulse means that theres less momentum?

1 answer:

4 0

in every collision there will be always an impulse I which represents change in momentum and is always constant. Thus, I = F x t = constant. Hence whenever, there is a change in impact ie force, there will also be equal change in time or vice versa. Thus, increae of time will decrease impact ie., force.

e.g., a person gets less injured on sand floor when compared to cemented floor when jumping from same height due to increase of time and decrease of force.

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Un neumático sin cámara, soporta una presión de 1.5 atm cuando la temperatura ambiente es de 300°K. ¿Qué presión llegará a sopor

arlik [135]

Answer:

El neumático soportará una presión de 1.7 atm.

Explanation:

Podemos encontrar la presión final del neumático usando la ecuación del gas ideal:

$PV = nRT$

En donde:

P: es la presión

V: es el volumen

n: es el número de moles del gas

R: es la constante de gases ideales

T: es la temperatura

Cuando el neumático soporta la presión inicial tenemos:

P₁ = 1.5 atm

T₁ = 300 K

$V_{1} = \frac{nRT_{1}}{P_{1}}$ (1)

La presión cuando T = 67 °C es:

$P_{2} = \frac{nRT_{2}}{V_{2}}$ (2)

Dado que V₁ = V₂ (el volumen del neumático no cambia), al introducir la ecuación (1) en la ecuación (2) podemos encontrar la presión final:

$P_{2} = \frac{nRT_{2}}{V_{2}} = \frac{nRT_{2}}{\frac{nRT_{1}}{P_{1}}} = \frac{P_{1}T_{2}}{T_{1}} = \frac{1.5 atm*(67 + 273)K}{300 K} = 1.7 atm$

Por lo tanto, si en el transcurso de un viaje las ruedas alcanzan una temperatura de 67 ºC, el neumático soportará una presión de 1.7 atm.

Espero que te sea de utilidad!

4 0

3 years ago

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

Distance of the center from each corners $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

Force by each of the charges at the remaining corners:

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

Now, net electric field in the vertical direction:

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

Now, net electric field in the horizontal direction:

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

8 0

3 years ago

Find the direction and magnitude of Ftot, the total force exerted on her by the others, given that the magnitudes F1 and F2 are

Mars2501 [29]

Answer:

θ = 36°

Explanation:

given,

F₁ = 22.8 N

F₂ = 16.6 N

magnitude of force = ?

direction of force = ?

$F = \sqrt{F_1^2 + F_2^2}$

$F = \sqrt{22.8^2 + 16.6^2}$

$F = \sqrt{795.4}$

F = 28.20 N

direction

$\theta = tan^{-1}(\dfrac{F_2}{F_1})$

$\theta = tan^{-1}(\dfrac{16.6}{22.8})$

$\theta = tan^{-1}(0.728)$

θ = 36°

5 0

3 years ago

Can someone pls help me answer this I’ll give brainliest to whoever actually answers it

sergejj [24]

Answer:

Limestone classifacation: sedimentary rock

Sandstone: sedimentary rock

7 0

3 years ago

1. Strzała o masie 20 g tuż po wystrzale ma prędkość 50 m/s. Oblicz pracę wykonaną

kow [346]

Answer:

Explanation:

Question 1

An arrow weighing 20g shortly after firing has a speed of 50m / s. Calculate the work done by the athlete. What is the potential energy of the elasticity of the tensed string?

mass m = 20g = 20/1000 = 0.02kg

speed v = 50m / s

P.E = K.E = ½mv²

P.E = ½ × 0.02 × 50²

P.E = 25 J

work done = P.E = 25J

Qestion 2

A 80 kg athlete stood on a trampoline with a coefficient of elasticity of k = 2 kN / m. As far as the edge of the trampoline lowers.

force of elasticity

F = -kx

x = F / k

in our case F will be the force of pressure or gravity

F = mg

g is gravitational acceleration, and according to Newton's second law, acceleration is force through mass - unit of force N, unit of mass kg. Acceleration either in m / s ^ 2 or N / kg

F = 80kg * 10N / kg = 800 N

x = 800N / -2000N = -0.4

The trampoline will lower, so from the level by 0.4 meters and hence this minus

7 0

2 years ago