heres free wallpapers and pnts! :)) (phone and PC and chromebook wallpapers) (credits to those who made them !)

The great lakes in north american hold about what percent of earth's fresh surface water?

fenix001 [56]

The answer is 20 percent

8 0

3 years ago

A suspended object A is attracted to a neutral wall. It's also attracted to a positively-charged object B. Which of the followin

olga2289 [7]

There are two particular cases, the first is when Object A is attracted to the neutral wall. This would indicate that the object is not neutral, as there is an attraction.

At the same time we know that Object A is attracted to an object B. And therefore, the load of A must be opposite to that of B. Remember that opposite charges attract each other. If the charge of object B is positive, then the charge of object A will be negative.

Option B is correct: It has a negative charge.

7 0

3 years ago

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0

3 years ago

What is the largest possible magnitude of the acceleration of the electron due to the magnetic field?

Alina [70]

Thank you for posting your question here at brainly. But your question seems incomplete. I will assume you based the situation below:

<span>An electrons moves at 2.0x10^6 m/s through a region in which there is a magnetic field of unspecified direction and magnitude 7.4x10^-2 T.

The </span> largest possible magnitude of the acceleration of the electron due to the magnetic field is <span>= 2.6 × 10 ¹⁶ m / s ²</span>

6 0

3 years ago

Se golpea una pelota de golf de manera que su velocidad inicial forma un ángulo de 45° con la horizontal. La pelota alcanza el s

nordsb [41]

Answer:

42m/s

6.06s

Explanation:

To find the initial velocity and time in which the ball is fling over the ground you use the following formulas:

$x_{max}=\frac{v_o^2sin(2\theta)}{g}\\\\x_{max}=vt_{max}$

θ: angle = 45°

vo: initial velocity

g: gravitational constant = 9.8m/s^2

x_max: max distance = 180 m

t_max: max time

by replacing the values of the parameters and do vo the subject of the first formula you obtain:

$v_o=\sqrt{\frac{gx_{max}}{sin(2\theta)}}\\\\v_o=\sqrt{\frac{(9.8m/s^2)(180m)}{sin(2(45\°))}}=42\frac{m}{s}$

with this value of vo you calculate the max time:

$t_{max}=\frac{x_{max}}{v}=\frac{x_{max}}{v_ocos(45\°)}\\\\t_{max}=\frac{180m}{(42m/s)cos(45\°)}=6.06s$

hence, the initial velocity of the ball is 42m/s and the time in which the ball is in the air is 6.06s

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TRANSLATION:

Para encontrar la velocidad inicial y el tiempo en el que la pelota está volando sobre el suelo, use las siguientes fórmulas:

θ: ángulo = 45 °

vo: velocidad inicial

g: constante gravitacional = 9.8m / s ^ 2

x_max: distancia máxima = 180 m

t_max: tiempo máximo

reemplazando los valores de los parámetros y haciendo el tema de la primera fórmula que obtiene:

con este valor de vo usted calcula el tiempo máximo:

por lo tanto, la velocidad inicial de la pelota es de 42 m / sy el tiempo en que la pelota está en el aire es de 6.06 s

4 0

3 years ago