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3 years ago

Which type of elements are likely to have no electical charge at all

Physics

2 answers:

boyakko [2]3 years ago

6 0

Answer:

Noble gas element

Explanation:

because Noble gas element can neither loose nor gain electrons

vaieri [72.5K]3 years ago

3 0

Answer: Neutrons

Explanation:A subatomic particle that has no charge and is located inside the nucleus is called neutrons

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Using a 683 nm wavelength laser, you form the diffraction pattern of a 1.1 mm wide slit on a screen. You measure on the screen t

n200080 [17]

Answer:

10.2 m

Explanation:

The position of the dark fringes (destructive interference) formed on a distant screen in the interference pattern produced by diffraction from a single slit are given by the formula:

$y=\frac{\lambda (m+\frac{1}{2})D}{d}$

where

y is the position of the m-th minimum

m is the order of the minimum

D is the distance of the screen from the slit

d is the width of the slit

$\lambda$ is the wavelength of the light used

In this problem we have:

$\lambda=683 nm = 683\cdot 10^{-9} m$ is the wavelength of the light

$d=1.1 mm = 0.0011 m$ is the width of the slit

m = 13 is the order of the minimum

$y=8.57 cm = 0.0857 m$ is the distance of the 13th dark fringe from the central maximum

Solving for D, we find the distance of the screen from the slit:

$D=\frac{yd}{\lambda(m+\frac{1}{2})}=\frac{(0.0857)(0.0011)}{(683\cdot 10^{-9})(13+\frac{1}{2})}=10.2 m$

6 0

3 years ago

You have a battery marked " 6.00 V 6.00 V ." When you draw a current of 0.383 A 0.383 A from it, the potential difference betwee

Archy [21]

Answer:

V = 4.81 V

Explanation:

As the potential difference between the battery terminals, is less than the rated value of the battery, this means that there is some loss in the internal resistance of the battery.
We can calculate this loss, applying Ohm's law to the internal resistance, as follows:

$V_{rint} = I* r_{int}$

The value of the potential difference between the terminals of the battery, is just the voltage of the battery, minus the loss in the internal resistance, as follows:

$V = V_{b} - V_{rint} = 5.03 V = 6.0 V - 0.383 A* r_{int}$

We can solve for rint, as follows:

$r_{int} = \frac{V_{b}-V}{I} =\frac{6.0V-5.03V}{0.383A} = 2.53 \Omega$

When the circuit draws from battery a current I of 0.469A, we can find the potential difference between the terminals of the battery, as follows:

$V = V_{b} - V_{rint} = 6.0 V - 0.469 A* 2.53 \Omega= 6.0 V - 1.19 V = 4.81 V$

As the current draw is larger, the loss in the internal resistance will be larger too, so the potential difference between the terminals of the battery will be lower.

5 0

2 years ago

An insulated thermos contains 106.0 cm3 of hot coffee at a temperature of 80.0 °C. You put in 11.0 g of ice cube at its melting

Andrei [34K]

Answer:

the final temperature is T f = 64.977 ° C≈ 65°C

Explanation:

Since the thermus is insulated, the heat absorbed by the ice is the heat released by the coffee. Thus:

Q coffee + Q ice = Q surroundings =0 (insulated)

We also know that the ice at its melting point , that is 0 °C ( assuming that the thermus is at atmospheric pressure= 1 atm , and has an insignificant amount of impurities ).

The heat released by coffee is sensible heat : Q = m * c * (T final - T initial)

The heat absorbed by ice is latent heat and sensible heat : Q = m * L + m * c * (T final - T initial)

therefore

m co * c co * (T fco - T ico) + m ice * L + m ice * c wat * (T fwa - T iwa) = 0

assuming specific heat capacity of coffee is approximately the one of water c co = c wa = 4.186 J/g°C and the density of coffee is the same as water

d co = dw = 1 gr/cm³

therefore m co = d co * V co = 1 gr / cm³ * 106 cm³ = 106 gr

m co * c wat * (T f - T ico) + m ice * L + m ice * c wat * (T f - T iwa) = 0

m co * c wat * T f+ m ice * c wat * T f = m ice * c wat * T iwa + m co * c wat * Tico -m ice * L

T f = (m ice * c wat * T iwa + m co * c wat * Tico -m ice * L ) /( m co * c wat * + m ice * c wat )

replacing values

T f = (11 g * 4.186 J/g°C * 0°C + 106 g * 4.186 J/g°C*80°C - 11 g * 334 J/gr) / ( 11 g * 4.186 J/g°C + 106 g * 4.186 J/g°C* ) = 64,977 ° C

T f = 64.977 ° C

7 0

3 years ago

The center of gravity of a loaded truck depends on how the truck is packed. If it is 4.0 m high and 2.4 m wide, and its CG is 2.

Vitek1552 [10]

The slope of the road can be given as the ratio of the change in vertical

distance per unit change in horizontal distance.

The maximum steepness of the slope where the truck can be parked without tipping over is approximately 54.55 %.

Reasons:

Width of the truck = 2.4 meters

Height of the truck = 4.0 meters

Height of the center of gravity = 2.2 meters

Required:

The allowable steepness of the slope the truck can be parked without tipping over.

Solution:

Let, C represent the Center of Gravity, CG

At the tipping point, the angle of elevation of the slope = θ

Where;

$tan\left(\theta \right) = \dfrac{\overline{AM}}{\overline{CM}}$

The steepness of the slope is therefore;

$\mathrm{The \ steepness \ of \ the \ slope}= \dfrac{\overline{AM}}{\overline{CM}} \times 100$

Where;

$\overline{AM}$ = Half the width of the truck = $\dfrac{2.4 \, m}{2}$ = 1.2 m

$\overline{CM}$ = The elevation of the center of gravity above the ground = 2.2 m

$\mathrm{The \ steepness \ of \ the \ slope}= \dfrac{1.2}{2.2} \times 100 \approx 54.55\%$

$tan\left(\theta \right) = \mathbf{\dfrac{2.2}{1.2}} = \dfrac{11}{6}$

$Elevation \ of \ the \ road \ \theta = arctan\left( \dfrac{6}{11} \right) \approx 28.6 ^{\circ}$

The maximum steepness of the slope where the truck can be parked is 54.55 %.

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3 0

2 years ago

In figures (a)-(d), a block moves to the right in the positive x direction through the displacement Δx while under the influence

Elena L [17]

5 0

2 years ago