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3 years ago

Hello!

Physics

1 answer:

Hunter-Best [27]3 years ago

5 0

Answer:

Explanation:

Force = something being pulled or pushed

Motion = what the object is doing, most likely moving

i think thats right

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An object having a mass of 11.0 g and a charge of 8.00 ✕ 10-5 C is placed in an electric field E with Ex = 5.70 ✕ 103 N/C, Ey =

dezoksy [38]

Answer : $F_{x} = 45.6\times10^{-2}\ N$ , $F_{y} = 3040\times10^{-5} N$ and $F_{z} = 0\ N$

Explanation :

Given that,

Charge of the object q = $8.00\times 10^{-5}\ C$

Electric field in x-direction $E_{x} = 5.70\times10^{3}\ N/C$

Electric field in y- direction $E_{y} = 380\ N/C$

Electric field in z - direction $E_{z} = 0$

Now, using formula

$F = q E$

Now, the force on the object in x- direction

$F_{x} = 8.00\times10^{-5} C \times5.70\times10^{3} N/C$

$F_{x} = 45.6\times10^{-2}\ N$

The force on the object in y- direction

$F_{y} = 8.00\times10^{-5}\ C\times 380\ N/C$

$F_{y} = 3040\times10^{-5} N$

The force on the object in z- direction

$F_{z} = 8.00\times10^{-5}\ C\times 0$

$F_{z} = 0\ N$

Hence, this is the required solution.

8 0

4 years ago

Một nguồn O phát sóng cơ dao động theo phương trình u = 4cos(4pi t - pi/4). Tốc độ truyền sóng là 1m/s.

brilliants [131]

Answer:

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jdjdu3jthh

hhhujusbrnog

hhjfjtinrny

ykrjrhrnirjtjjtt

tkrjthr74uu3jt

hri4urjjrjtjjtjtjy

Explanation:

uueuhhhwuejroskanficndui39wn

jebfufkr

5 0

3 years ago

Hi im a little stuck on this question that came from my textbook in class it got a little confusing for me because i really dont

koban [17]

Consult the attached free body diagram.

By Newton's second law, the net force on the crate acting parallel to the surface is

∑ F[para] = (370 N) cos(-20°) - f = 0

(this is the x-component of the resultant force)

where

• (370 N) cos(-20°) = magnitude of the horizontal component of the pushing force

• f = magnitude of kinetic friction

The crate is moving at a constant speed and thus not accelerating, so the crate is in equilibrium.

Solve for f :

f = (370 N) cos(-20°) ≈ 347.686 N

The net force acting perpendicular to the surface is

∑ F[perp] = n - 1480 N - (370 N) sin(-20°) = 0

(this is the y-component of the resultant force)

where

• n = magnitude of normal force

• 1480 N = weight of the crate

• (370 N) sin(-20°) = magnitude of the vertical component of push

The crate doesn't move up or down, so it's also in equilibrium in this direction.

Solve for n :

n = 1480 N + (370 N) sin(-20°) ≈ 1606.55 N ≈ 1610 N

Then the coefficient of kinetic friction is µ such that

f = µn ⇒ µ = f/n ≈ 0.216

7 0

2 years ago

A charge +q is located at the origin, while an identical charge is located on the x axis at x = +0.57 m. A third charge of +2q i

Jlenok [28]

Answer:

The third charge placed is 0.80 m.

Explanation:

Given that,

Distance = 0.57 m

First charge = q

Third charge = 2q

We need to calculate the electrostatic force on charge q₁ due to q₂

Using formula of electrostatic force

$F_{21}=\dfrac{kq_{1}q_{2}}{r_{1}^2}$

When placed another charge q₃ at certain distance from origin, then the net force on charge q₁ due to both charges is

$F_{net}=F_{21}+F_{31}$

The net electrostatic force on the charge at the origin doubles.

$2F_{21}=F_{21}+F_{31}$

$F_{31}=F_{21}$

$\dfrac{kq_{3}q_{1}}{r_{2}^2}=\dfrac{kq_{2}q_{1}}{r_{1}^2}$

$\dfrac{q_{3}}{r_{2}^2}=\dfrac{q_{2}}{r_{1}^2}$

$r_{2}^2=\dfrac{q_{3}}{q_{2}\timesr_{1}^2}$

$r_{2}=\sqrt{\dfrac{q_{3}}{q_{2}}}r_{1}$

Put the value into the formula

$r_{2}=\sqrt{\dfrac{2q}{q}}\times0.57$

$r_{2}=\sqrt{2}\times0.57$

$r_{2}=0.80\ m$

Hence, The third charge placed is 0.80 m from origin in x-axis.

4 0

3 years ago

A body is thrown vertically upwards.Its velocity keeps on decreasing. What happens to its kinetic energy when it reaches the max

Nikolay [14]

The kinetic energy will rise once the body comes back down. As it goes up, the potential energy increases while the kinetic energy decreases. Once the body is at its maximum height, the potential energy is at it’s highest. When it starts falling, it will gain kinetic energy and lose potential energy.

3 0

4 years ago

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