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1 year ago

7

What must the charge (sign and magnitude) of a 3.45 g particle be for it to remain stationary when placed in a downward-directed

electric field of magnitude 590 N/C

1 answer:

Pani-rosa [81]1 year ago

8 0

charge must be equal to 5.74 ×10⁻⁵

In the question it is said that the particle remains stationary which means the the net force on the particle is zero. So, the counterbalancing forces must be equal which means weight is equal to upward electric force.

→ Fnet =0

→ mg = qE

substituting the values we get :

0.00345 × 9.81 = q × 590

→ q = 5.74 ×10⁻⁵

Hence the charge must be equal to 5.74 ×10⁻⁵.

Learn more about charges here:

brainly.com/question/26092261

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2 years ago

A tank of volume 0.25 ft is designed to contain 50 standard cubic feet of air. The temperature is 80° F.Calculate the pressure i

slavikrds [6]

Answer:

The inside Pressure of the tank is $4499.12 lb/ft^{2}$

Solution:

As per the question:

Volume of tank, $V = 0.25 ft^{3}$

The capacity of tank, $V' = 50ft^{3}$

Temperature, T' = $80^{\circ}F$ = 299.8 K

Temperature, T = $59^{\circ}F$ = 288.2 K

Now, from the eqn:

PV = nRT (1)

Volume of the gas in the container is constant.

V = V'

Similarly,

P'V' = n'RT' (2)

Also,

The amount of gas is double of the first case in the cylinder then:

n' = 2n

$\]frac{n'}{n} = 2$

where

n and n' are the no. of moles

Now, from eqn (1) and (2):

$\frac{PV}{P'V'} = \frac{nRT}{n'RT'}$

$P' = 2P\frac{T'}{T} = 2\times 2116\times \frac{299.8}{288.2} = 4499.12 lb/ft^{2}$

7 0

3 years ago

Susan, driving north at 53 mphmph , and Shawn, driving east at 63 mphmph , are approaching an intersection. Part A What is Shawn

mafiozo [28]

Answer:

Shawn's speed relative to Susan's speed = 10 mph

Resultant velocity = 82.32 mph

Explanation:

The given data :-

i) Susan driving in north and speed of Susan is ( v₁ ) = 53 mph.

ii) Shawn driving in east and speed of Shawn is ( v₂ ) = 63 mph.

iii) The speed of both Susan and Shawn is relative to earth.

iv) The angle between Susan in north and Shawn in east is 90°.

We have to find Shawn's speed relative to Susan's speed.

v₂₁ = v₂ - v₁ = 63 - 53 = 10 mph

Resultant velocity,

$v = \sqrt{v_{2} ^{2}+ v_{1} ^{2} } =\sqrt{63^{2} +53^{2} }$

v = 82.32 mph

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3 years ago

A block of mass 10.0 kg is pulled to the right along a rough horizontal surface with a constant horizontal force of 20.0 N. The

Yakvenalex [24]

Answer:

the magnitude of acceleration will be 1.50m/s^2

Explanation:

To calculate your acceleration, you can use your formula that states that the net force on an object is equal to the mass of the object multiplied by the acceleration of the object. Fnet=ma

if you draw out this situation and label the forces you will have your vector towards the right with a magnitude of 20.0N and then your friction vector will be pointing to the left (in other words, in the negative direction) (opposing the direction of movement) with a magnitude of 5.00N, with the 10.0 kg box in the middle.

The net force will be calculated using F1+F2=Fnet where your F1=20.0N and F2= -5.00N (since it is towards the negative direction).

you will find that Fnet=15.0N

With that, plug in the values you know to calculate the acceleration of the block:

Fnet=ma

(15.0N)=(10.0kg)a from her you can divide both sides by 10 to isolate a:

1.50=a (and now make sure to label the units of your answer)

a=1.50m/s^2 (which is the typical unit for acceleration)

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2 years ago

An ideal spring hangs from the ceiling. A 1.25-kg mass is hung from the spring. After all vibrations have died away, the spring

ch4aika [34]

The kinetic energy of the mass at the instant it passes back through its equilibrium position is about 1.20 J

$\texttt{ }$

<h3>Further explanation</h3>

Let's recall Elastic Potential Energy formula as follows:

$\boxed{E_p = \frac{1}{2}k x^2}$

where:

<em>Ep = elastic potential energy ( J )</em>

<em>k = spring constant ( N/m )</em>

<em>x = spring extension ( compression ) ( m )</em>

Let us now tackle the problem!

$\texttt{ }$

<u>Given:</u>

mass of object = m = 1.25 kg

initial extension = x = 0.0275 m

final extension = x' = 0.0735 - 0.0275 = 0.0460 m

<u>Asked:</u>

kinetic energy = Ek = ?

<u>Solution:</u>

<em>Firstly , we will calculate the spring constant by using </em><em>Hooke's Law</em><em> as follows:</em>

$F = k x$

$mg = k x$

$k = mg \div x$

$k = 1.25(9.8) \div 0.0275$

$k = 445 \frac{5}{11} \texttt{ N/m}$

$\texttt{ }$

<em>Next , we will use </em><em>Conservation of Energy</em><em> formula to solve this problem:</em>

$Ep_1 + Ek_1 = Ep_2 + Ek_2$

$\frac{1}{2}k (x')^2 + mgh + 0 = \frac{1}{2}k x^2 + Ek$

$Ek = \frac{1}{2}k (x')^2 + mgh - \frac{1}{2}k x^2$

$Ek = \frac{1}{2}k ( (x')^2 - x^2 ) + mgh$

$Ek = \frac{1}{2}(445 \frac{5}{11}) ( 0.0460^2 - 0.0275^2 ) + 1.25(9.8)(0.0735)$

$\boxed {Ek \approx 1.20 \texttt{ J}}$

$\texttt{ }$

<h3>Learn more</h3>

Kinetic Energy : brainly.com/question/692781

Acceleration : brainly.com/question/2283922

The Speed of Car : brainly.com/question/568302
Young Modulus : brainly.com/question/9202964
Simple Harmonic Motion : brainly.com/question/12069840

$\texttt{ }$

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Elasticity

8 0

2 years ago

Read 2 more answers