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3 years ago

Which of the following statements are true? Select all that apply.

Physics

1 answer:

beks73 [17]3 years ago

5 0

Answer:

The following options are true based on the properties of electric field;

a) Electric field lines near positive point charges radiate outward.

b) The electric force acting on a point charge is proportional to the magnitude of the point charge.

d) In a uniform electric field, the field lines are straight, parallel, and uniformly spaced.

Explanation:

From option b) From coulomb's law F = Kq1q2r/r2

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Please Help, I need to turn this in today

mart [117]

I think it is C , idk why, but I just learn about this and I think it is.... sorry I didn't really help.....

3 0

3 years ago

Read 2 more answers

Please help me with this question

Oksanka [162]

Car at rest:

velocity= 0m/s

Acceleration:

0.2m/s²

Since total time:

3 min = 180s

Formula of acceleration:

acceleration = [final velocity - initial velocity] ÷ [total time]

Velocity at end:

0.2m/s² = [final velocity - 0m/s] ÷ [180s]

0.2m/s² × 180s = [final velocity]

[final velocity] = 36m/s

Distance travelled:

Velocity = displacement(distance) ÷ time

36m/s = displacement(distance) ÷ 180s

displacement(distance) = 36m/s × 180s

displacement(distance) = 6480m

Hey I'm sorry but i do not understand why the answer on your worksheet for distance travelled is 3240m... its half of what my answer is...

6 0

2 years ago

Spitting cobras can defend themselves by squeezing muscles around their venom glands to squirt venom at an attacker. Suppose a s

Alenkasestr [34]

Answer: 1.289 m

Explanation:

The path the cobra's venom follows since it is spitted until it hits the ground, is described by a parabola. Hence, the equations for parabolic motion (which has two components) can be applied to solve this problem:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$x$ is the horizontal distance traveled by the venom

$V_{o}=3.10 m/s$ is the venom's initial speed

$\theta=47\°$ is the angle

$t$ is the time since the venom is spitted until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=0.44 m$ is the initial height of the venom

$y=0$ is the final height of the venom (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Let's begin with (2) to find the time it takes the complete path:

$0=0.44 m+3.10 m/s sin\theta(47\°)+\frac{-9.8m/s^{2} t^{2}}{2}$ (3)

Rewritting (3):

$-4.9 m/s^{2} t^{2} + 2.267 m/s t + 0.44 m=0$ (4)

This is a quadratic equation (also called equation of the second degree) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ (5)

Where:

$a=-4.9 m/s^{2$

$b=2.267 m/s$

$c=0.44 m$

Substituting the known values:

$t=\frac{-2.267 \pm \sqrt{2.267^{2}-4(-4.9)(0.44)}}{2(-4.9)}$ (6)

Solving (6) we find the positive result is:

$t=0.609 s$ (7)

Substituting (7) in (1):

$x=(3.10 m/s)cos(47\°)(0.609 s)$ (8)

We finally find the horizontal distance traveled by the venom:

$x=1.289 m$

7 0

3 years ago

A uniformly charged sphere has a potential on its surface of 450 V. At a radial distance of 7.2 m from this surface, the potenti

Yakvenalex [24]

Answer:

The radius of the sphere is 3.6 m.

Explanation:

Given that,

Potential of first sphere = 450 V

Radial distance = 7.2 m

If the potential of sphere =150 V

We need to calculate the radius

Using formula for potential

For 450 V

$V=\dfrac{kQ}{r}$

$450=\dfrac{kQ}{r}$ ....(I)

For 150 V

$150=\dfrac{kQ}{r+7.2}$ ....(II)

Divided equation (I) by equation (II)

$\dfrac{450}{150}=\dfrac{\dfrac{kQ}{r}}{\dfrac{kQ}{r+7.2}}$

$3=\dfrac{(r+7.2)}{r}$

$3r=r+7.2$

$r=\dfrac{7.2}{2}$

$r=3.6\ m$

Hence, The radius of the sphere is 3.6 m.

3 0

3 years ago

(I think it's D but idontknow)

sukhopar [10]

During an exothermic reaction; light and heat are released into the environment.

An exothermic reaction is one in which heat is released to the environment. This heat can be physically observed sometimes like in an a combustion reaction.

In an exothermic reaction, the enthalpy of the reactants is greater than the enthalpy of the products.

This heat lost is sometimes felt as the hotness of the vessel in which the reaction has taken place.

In conclusion, light and heat are released into the environment in an exothermic reaction.

Learn more: brainly.com/question/4345448

3 0

2 years ago