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4 years ago

a charge of 6.4x10^-7 C experiences an electric force of 1.8x10^-1 N what is the electric field strength

1 answer:

5 0

Field strength = force ÷ charge (E=F/Q)

E= 1.8×10^-1 ÷ 6.4×10^-7... to obtain ur answer

the formula is derived from the definition
Electric field : A region in space where a charged body experiences a force.

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A crumb of bread, of mass 0.056 kg, is pulled upon by ants from rival anthills. They exert the following forces: 0.06 N to the n

Valentin [98]

Answer:

a) 1.855m/s^2, 9.71\° to the east-north

b) 0.103N, 9.46° to the west-south

Explanation:

To find the acceleration of the system you can assume that the forces are applied in a xy plane, where force toward north are directed in the +y direction, and forces to the east in the +x direction. BY taking into account the components of the acceleration for each axis you obtain the following systems of equations:

$0.06N-0.06Nsin(45\°)=ma_y\\\\0.08N-0.02N+0.06cos(45\°)=ma_x$

m: mass of the crumb of bread = 0.056kg

you simplify the equations an replace the values of the mass in order to obtain the acceleration components:

$a_y=\frac{0.017N}{0.056kg}=0.313\frac{m}{s^2}\\\\a_x=\frac{0.102N}{0.056kg}=1.829\frac{m}{s^2}$

$\theta=tan^{-1}(\frac{0.313}{1.829})=9.71\°\\\\a=\sqrt{a_x^2+a_y^2}=\sqrt{(1.829)^2+(0.313)^2}\frac{m}{s^2}=1.855\frac{m}{s^2}$

then, the acceleration of the system has a magnitude of 1.855m/s^2 and a direction of 9.71\° to the east-north

The fifth force must cancel both x an y components of the previous net force, that is:

$0.06N-0.06Nsin(45\°)+F_y=0\\\\F_y=-0.017N\\\\0.08N-0.02N+0.06cos(45\°)+F_x=0\\\\F_x=-0.102N\\\\\phi=tan^{-1}(\frac{-0.017}{-0.102N})=9.46\°$

$F=\sqrt{(0.102)^2+(0.017)^2}N=0.103N$

the, the force needed to reach the equilibrium has a magnitude of 0.103N and a direction of 9.46° to the west-south

4 0

3 years ago

A neutron star has a mass of 2.0 × 1030 kg (about the mass of our sun) and a radius of 5.0 × 103 m (about the height of a good-s

Nitella [24]

Answer:

$v=516526.9m/s$

Explanation:

The force to which the object of mass m is attracted to a star of mass M while being at a distance r is:

$F=\frac{GMm}{r^2}$

Where $G=6.67\times10^{-11}Nm^2/Kg^2$ is the gravitational constant.

Also, Newton's 2nd Law tells us that this object subject by that force will experiment an acceleration given by F=ma.

We have then:

$ma=\frac{GMm}{r^2}$

Which means:

$a=\frac{GM}{r^2}$

The object departs from rest ( $v_0=0m/s$ ) and travels a distance d, under an acceleration a, we can calculate its final velocity with the formula $v^2=v_0^2+2ad$ , which for our case will be:

$v^2=2ad=\frac{2GMd}{r^2}$

$v=\sqrt{\frac{2GMd}{r^2}}$

We assume a constant on the vecinity of the surface because d=0.025m is nothing compared with $r=5\times10^3m$ . With our values then we have:

$v=\sqrt{\frac{2GMd}{r^2}}=\sqrt{\frac{2(6.67\times10^{-11}Nm^2/Kg^2)(2\times10^{30}Kg)(0.025m)}{(5\times10^3m)^2}}=516526.9m/s$

7 0

3 years ago

an object of mass 1 g is hung from a spring and set in oscillatory motion .At t=0 the displacement is 43.75cm and the accelerati

charle [14.2K]

The spring constant is $4.0\cdot 10^{-5} N/m$

Explanation:

For an object in a simple harmonic motion, the acceleration of the object is related to the displacement by

$a=-\omega^2 x$

where

a is the acceleration

$\omega$ is the angular frequency

x is the displacement

The angular frequency is defined as

$\omega=\sqrt{\frac{k}{m}}$

where

k is the spring constant

m is the mass

Substituting the second equation into the first one, we get

$a=-\frac{k}{m}x$

In this problem we have

m = 1 g = 0.001 kg

And at t=0,

x = 43.75 cm

a = -1.754 cm/s

Therefore, we can re-arrange the equation above to find the spring constant:

$k=-\frac{ma}{x}=-\frac{(0.001)(-1.754)}{43.75}=4.0\cdot 10^{-5} N/m$

#LearnwithBrainly

6 0

3 years ago

Speedy Sue, que conduce a 30.0 m/s, entra a un túnel de un carril. En seguida observa una camioneta lenta 155 m adelante que se

Romashka [77]

Answer:

Si ocurre una colisión. 5.201 segundos después de que Speedy Sue ingrese al túnel y a una distancia de 128.995 metros.

Explanation:

Supongamos que el vehículo de Speedy Sue decelera a razón constante, mientras que la camioneta se desplaza a velocidad constante. Se requiere conocer si ambos vehículos colisionarán, lo cual implica conocer si existe algún instante tal que ambos tengan la misma posición. Consideremos además que la posición de referencia se encuentra en la posición inicial de Sppedy Sue. Entonces, las ecuaciones cinemáticas son:

Speedy Sue

$x_{S} = x_{S,o}+v_{S,o}\cdot t+\frac{1}{2}\cdot a_{S}\cdot t^{2}$

Camioneta lenta

$x_{C} = x_{C,o} +v_{C}\cdot t$

Donde:

$x_{S,o}$ , $x_{C,o}$ - Posiciones iniciales de Speedy Sue y la camioneta lenta, medidas en metros.

$v_{S,o}$ - Velocidad inicial de Speedy Sue, medida en metros por segundo.

$v_{S}, v_{C}$ - Velocidades actuales de Speedy Sue y la camioneta lenta, medidas en metros por segundo.

$a_{S}$ - Deceleración de Speedy Sue, medida en metros por segundo cuadrado.

$t$ - Tiempo, medido en segundos.

Si conocemos que $x_{S} = x_{C}$ , $x_{S,o} = 0\,m$ , $x_{C,o} = 155\,m$ , $v_{S,o} = 30\,\frac{m}{s}$ , $v_{C} =-5\,\frac{m}{s}$ y $a_{S} = -2\,\frac{m}{s^{2}}$ , encontramos la siguiente función cuadrática:

$155\,m + \left(-5\,\frac{m}{s} \right)\cdot t = 0\,m+\left(30\,\frac{m}{s} \right)\cdot t +\frac{1}{2}\cdot (-2\,\frac{m}{s^{2}} )\cdot t^{2}$

$-t^{2}+35\cdot t-155 = 0$ (Ec. 1)

Las raíces de esta función son:

$t_{1}\approx 29.798\,s$ , $t_{2} \approx 5.201\,s$

La colisión ocurriría en la raíz positiva más pequeña, es decir:

$t \approx 5.201\,s$

Ahora, la posición en que ocurriría la colisión se determina a partir de la ecuación de desplazamiento de la camioneta lenta, es decir: ( $v_{C,o} = -5\,\frac{m}{s}$ , $x_{C,o} = 155\,m$ , $t \approx 5.201\,s$ )

$x_{C} = 155\,m +\left(-5\,\frac{m}{s}\right)\cdot (5.201\,s)$

$x_{C} = 128.995\,m$

En síntesis, si ocurre una colisión. 5.201 segundos después de que Speedy Sue ingrese al túnel y a una distancia de 128.995 metros.

0 0

3 years ago

Describe two physical environmental factors in your environment at the moment

Anna11 [10]

Answer:

The factors in the physical environment that are important to health include harmful substances, such as air pollution or proximity to toxic sites the focus of classic environmental epidemiology access to various health-related resources e.g. healthy or unhealthy foods, recreational resources, medical care.

7 0

3 years ago