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3 years ago

Question 1 of 10

Physics

1 answer:

Novay_Z [31]3 years ago

6 0

Answer:

C. 5.6 × 10^11 N/C

Explanation:

The electric field $E$ at a distance $R$ from a charge $Q$ is given by

$E = k\dfrac{Q}{R^2}$

where $k = 9*10^9Nm/C$ is the coulomb's constant.

Now, in our case

$R = 0.0075m$

$Q = 0.0035C$ ;

therefore,

$E = (9*10^9)\dfrac{0.0035C}{(0.0075m)^2}$

$\boxed{E = 5.6*10^{11}N/C.}$

which is choice C from the options given<em> (at least it resembles it).</em>

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You and your friend live in different states and plant identical hydrangea bulbs in your garden. When they bloom, you send each

Whitepunk [10]

Answer:

sample. rewrite in your own words if you'd like.

Explanation:

Since the hydrangeas are changing color depending on whether the soil is acidic or basic, they must have an indicator. This means that any hydrangea planted in acidic soil will have blue flowers and any hydrangea planted in basic soil will have pink. One common indicator present in many plants is anthocyanin, so maybe hydrangeas contain that chemical.

8 0

3 years ago

Read 2 more answers

To calibrate your calorimeter cup, you first put 45 mL of cold water in the cup, and measure its temperature to be 24.7 °C. You

drek231 [11]

Answer : The heat change of the cold water in Joules is, $1.6\times 10^3J$

Explanation :

First we have to calculate the mass of cold water.

As we know that the density of water is 1 g/mL. The volume of cold water is 45 mL.

$Density=\frac{Mass}{Volume}$

$Mass=Density\times Volume=1g/mL\times 45mL=45g$

Now we have to calculate the heat change of cold water.

Formula used :

$Q=m\times c\times (T_2-T_1)$

where,

Q = heat change of cold water = ?

m = mass of cold water = 45 g

c = specific heat of water = $4.184J/g^oC$

$T_1$ = initial temperature of cold water = $24.7^oC$

$T_2$ = final temperature = $33.4^oC$

Now put all the given value in the above formula, we get:

$Q=45g\times 4.184J/g^oC\times (33.4-24.7)^oC$

$Q=1638.036J=1.6\times 10^3J$

Therefore, the heat change of cold water is $1.6\times 10^3J$

4 0

3 years ago

A ball rolls forward in the grass slowing down as it rolls?

olga nikolaevna [1]

Yes it does, uh huh. It slows down as it rolls. That's a fact.

In order for the ball to roll forward, it has to push grass out of the way. That takes energy. To bend each blade of grass out of its way, the ball has to use a tiny bit of the kinetic energy that it has, so it gradually runs out of kinetic energy. When its kinetic energy is all gone, it stops moving.

3 0

3 years ago

A small sphere with mass m is attached to a massless rod of length L that is pivoted at the top, forming a simple pendulum. The

USPshnik [31]

Answer:

a) see attached, a = g sin θ

c) v = √(2gL (1-cos θ))

Explanation:

In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by

Wₓ = m a

W sin θ = m a

a = g sin θ

b) The diagram is the same, the only thing that changes is the angle that is less

θ' = 9/2 θ

c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.

The easiest way to find linear speed is to use conservation of energy

Highest point

Em₀ = mg h = mg L (1-cos tea)

Lowest point

Emf = K = ½ m v²

Em₀ = Emf

g L (1-cos θ) = v² / 2

v = √(2gL (1-cos θ))

4 0

3 years ago

Can an object have increasing speed while its acceleration is decreasing?

alexdok [17]

The best option is C. This is due to friction.

6 0

3 years ago