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bazaltina [42]
3 years ago
6

What magnitude charge creates a 1.0 n/c electric field at a point 1.0 m away?

Physics
1 answer:
Stolb23 [73]3 years ago
8 0

Answer:

1.1\cdot 10^{-10}C

Explanation:

The electric field produced by a single point charge is given by:

E=k\frac{q}{r^2}

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have

E = 1.0 N/C (magnitude of the electric field)

r = 1.0 m (distance from the charge)

Solving the equation for q, we find the charge:

q=\frac{Er^2}{k}=\frac{(1.0 N/c)(1.0 m)^2}{9\cdot 10^9 Nm^2c^{-2}}=1.1\cdot 10^{-10}C

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One speed skater starts across a frozen lake at an average speed of 8 m/s. Ten seconds later, a second speed skater starts from
Luba_88 [7]

Answer:

40 s

Explanation:

After 10 seconds, the first skater would have a 8m/s * 10s = 80 m head start

Let t be the number of seconds after the second skater starts will the second skater overtake the first skater

The distance traveled by the first skater after t seconds is

s_1 = v_1t = 8t

Similarly the distance traveled by the 2nd skater after t seconds is

s_2 = v_2t = 10t

Since the 2nd skater catches up to the 1st one after 80 m behind, the distance traveled by the 2nd one must be 80m greater than the distance of the 1st skater

s_2 = s_1 + 80

We can substitute s_1 = 8t, s_2 = 10t

10t = 8t + 80

2t = 80

t = 80 / 2 = 40 s

7 0
3 years ago
Water is boiled at sea level in a coffeemaker equipped with an immersion-type electric heating element. The coffee maker contain
Luden [163]

Answer:

P=1362\ W

t'=251.659\ s is time required to heat to boiling point form initial temperature.

Explanation:

Given:

initial temperature of water, T_i=18^{\circ}C

time taken to vapourize half a liter of water, t=18\ min=1080\ s

desity of water, \rho=1\ kg.L^{-1}

So, the givne mass of water, m=1\ kg

enthalpy of vaporization of water, h_{fg}=2256.4\times 10^{-3}\ J.kg^{-1}

specific heat of water, c=4180\ J.kg^{-1}.K^{-1}

Amount of heat required to raise the temperature of given water mass to 100°C:

Q_s=m.c.\Delta T

Q_s=1\times 4180\times (100-18)

Q_s=342760\ J

Now the amount of heat required to vaporize 0.5 kg of water:

Q_v=m'\times h_{fg}

where:

m'=0.5\ kg= mass of water vaporized due to boiling

Q_v=0.5\times 2256.4

Q_v=1.1282\times 10^{6}\ J

Now the power rating of the boiler:

P=\frac{Q_s+Q_v}{t}

P=\frac{342760+1128200}{1080}

P=1362\ W

Now the time required to heat to boiling point form initial temperature:

t'=\frac{Q_s}{P}

t'=\frac{342760}{1362}

t'=251.659\ s

6 0
3 years ago
Two bicycle tires are set rolling with the same initial speed of 3.30 m/s along a long, straight road, and the distance each tra
vredina [299]

Answer:

At low pressure- \mu_{k}=0.02315

At high pressure- \mu_{k}=0.00445

Explanation:

Initial speed, V_{i}=3.3 m/s

Final speed, V_{f}=3.3/2= 1.65 m/s

Net horizontal force due to rolling friction F_{net}=\mu_{k} mg where m is mass, g is acceleration due to gravity, \mu_{k} is coefficient of rolling friction

From kinematic relation, V_{f}^{2}- V_{i}^{2}=2ad

For each tire,

V_{f}^{2}- V_{i}^{2}=2\mu_{k}gd

Making \mu_{k} the subject

\mu_{k}=\frac {V_{f}^{2}- V_{i}^{2}}{2gd}

Under low pressure of 40 Psi, d=18 m

\mu_{k}=\frac {1.65^{2}- 3.3^{2}}{2*9.8*18}=-0.02315

Therefore, \mu_{k}=0.02315

At a pressure of 105 Psi, d=93.7

\mu_{k}=\frac {1.65^{2}- 3.3^{2}}{2*9.8*93.7}=-0.00445

Therefore, \mu_{k}=0.00445

4 0
3 years ago
a stomp rocket takes 1.5 seconds to reach its maximum height what was the initial velocity and what was the maximum height ?
Zinaida [17]
1) Vf = Vo - gt; Vf = 0 => Vo = gt = 9.8m/s^2 * 1.5s = 14.7 m/s

2) d = Vo*t - gt^2 /2 = 14.7m/s*1.5 - 9.8m/s^2 * (1.5s)^2 / 2 = 11.02 m

 
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3 years ago
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In general, which type of compound tends to have the lowest melting point?
Liono4ka [1.6K]

A. Molecular solids tend to have lower melting points than Ionic so it would be Ionic if it weren't for Molecular.

5 0
3 years ago
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