A popular physics lab involves a hand generator and an assortment of wires with different values of resistance. In the lab, the leads of the generator are connected across each wire in turn. For each wire, students attempt to turn the generator handle at the same constant rate. Students must push harder on the handle when the leads of the generator are connected. This is because turning the handle at a given constant rate produces , regardless of what is connected to the leads. So, when turning the handle at a constant rate, lab students must push harder in cases where there is

Answers

Answer 1

Answer:

Answer:

Explanation:

Students must push harder on the handle when the leads of the generator are connected across the wire with the lowest resistance.

This is because turning the handle at a given constant rate produces a constant voltage across the leads, regardless of what is connected to the leads.

So, when turning the handle at a constant rate, lab students must push harder in case where there is a greater current through the connected wire.

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Given a double slit apparatus with slit distance 1 mm, what is the theoretical maximum number of bright spots that I would see when I shine light with a wavelength 400 nm on the slits
You perform a double‑slit experiment in order to measure the wavelength of the new laser that you received for your birthday. You set your slit spacing at 1.09 mm and place your screen 8.61 m from the slits. Then, you illuminate the slits with your new toy and find on the screen that the tenth bright fringe is 4.53 cm away from the central bright fringe (counted as the zeroth bright fringe). What is your laser's wavelength ???? expressed in nanometers?
1. Explain the change of state from solid dry ice to carbon dioxide gas.2. The motion of the particles in dry ice and carbon dioxide gas.3. Explain how the original mass of dry ice compares with the mass of carbon dioxide gas.
A baseball player standing on a platform throws a baseball out over a level playing field. The ball is released from a point 3.50 m above the field with an initial speed of 14.3 m/s at an upward angle of 27.0 degrees above the horizontal. What horizontal distance will the ball travel before hitting the ground?

A positive magnification means the image is inverted compared to the object

Answers

False

Explanation:

A positive magnification means the image is erect compared to the object. Magnifications with values greater than one represent images that are smaller than the object. A magnification of 1 (plus or minus) means that the image is the same size as the object. If m has a magnitude greater than 1 the image is larger than the object, and an m with a magnitude less than 1 means the image is smaller than the object. If the magnification is positive, the image is upright compared to the object; if m is negative, the image is inverted compared to the object.

A section of highway has the following flowdensity relationship q = 50k − 0.156k2 [with q in veh/h and k in veh/mi]. What is the capacity of the highway section, the speed at capacity, and the density when the highway is at one-quarter of its capacity?

Answers

(a) The capacity will be "4006.4 veh/h".

(b) The speed at capacity be "25 mph".

(c) The density will be "299 veh/mi".

Given:

$q = 50k - 0.156 k^2$

At max. flow density,

$(dd)/(dk) =0$

$((dq)/(dt) ) = 50-0.321k =0$

(a)

→ $k = ((50)/(0.312) )$

$= 160.3 \ or \ 160 \ veh/mi$

By substituting the value,

→ $q = 50k-0.156k^2$

$= 50* 160.3-0.156* (160.3)^2$

$= 4006.4 \ veh/h$

(b)

The speed will be:

→ $U = (q)/(k)$

$= (4006.4)/(160.3)$

$= 25 \ mph$

(c)

The density be:

→ $1001.6 = 50k-0.156k^2$

$0.156k^2-50k+1001.6 =0$

$k = 21.5 \ veh/mi \ or \ 299 \ veh/mi$

Thus the responses above are correct.

Find out more information about density here:

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Answer:

a) capacity of the highway section = 4006.4 veh/h

b) The speed at capacity = 25 mph

c) The density when the highway is at one-quarter of its capacity = k = 21.5 veh/mi or 299 veh/mi

Explanation:

q = 50k - 0.156k²

with q in veh/h and k in veh/mi

a) capacity of the highway section

To obtain the capacity of the highway section, we first find the k thay corresponds to the maximum q.

q = 50k - 0.156k²

At maximum flow density, (dq/dk) = 0

(dq/dt) = 50 - 0.312k = 0

k = (50/0.312) = 160.3 ≈ 160 veh/mi

q = 50k - 0.156k²

q = 50(160.3) - 0.156(160.3)²

q = 4006.4 veh/h

b) The speed at the capacity

U = (q/k) = (4006.4/160.3) = 25 mph

c) the density when the highway is at one-quarter of its capacity?

Capacity = 4006.4

One-quarter of the capacity = 1001.6 veh/h

1001.6 = 50k - 0.156k²

0.156k² - 50k + 1001.6 = 0

Solving the quadratic equation

k = 21.5 veh/mi or 299 veh/mi

Hope this Helps!!!

A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's initial speed is 1.37 m/s and it stops after 2.8 s. Determine the magnitude of the friction force exerted on the box.A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's initial speed is 1.37 m/s and it stops after 2.8 s. Determine the magnitude of the friction force exerted on the box.

Answers

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

$\Sigma F = -f = m\cdot a$ (Eq. 1)

Where:

$f$ - Kinetic friction force, measured in newtons.

$m$ - Mass of the box, measured in kilograms.

$a$ - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight ( $W = m\cdot g$ ) and uniform accelerated motion ( $v = v_(o)+a\cdot t$ ), we expand the previous expression:

$-f = \left((W)/(g) \right)\cdot \left((v-v_(o))/(t)\right)$

And the magnitude of the friction force exerted on the box is calculated by this formula:

$f = -\left((W)/(g) \right)\cdot \left((v-v_(o))/(t)\right)$ (Eq. 1b)

Where:

$W$ - Weight, measured in newtons.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_(o)$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$t$ - Time, measured in seconds.

If we know that $W = 52.4\,N$ , $g = 9.807\,(m)/(s^(2))$ , $v_(o) = 1.37\,(m)/(s)$ , $v = 0\,(m)/(s)$ and $t = 2.8\,s$ , the magnitud of the kinetic friction force exerted on the box is:

$f = -\left((52.4\,N)/(9.807\,(m)/(s^(2)) ) \right)\cdot \left((0\,(m)/(s)-1.37\,(m)/(s) )/(2.8\,s) \right)$

$f = 2.614\,N$

The magnitude of the friction force exerted on the box is 2.614 newtons.

Final answer:

The magnitude of the friction force acting on the box is determined by calculating the box's acceleration, establishing its mass based on its weight information, and applying these values in Newton's second law. The calculated value is 2.62 N.

Explanation:

To determine the magnitude of the friction force, we first have to compute the acceleration of the box. Acceleration (a) can be found using the formula 'final velocity - initial velocity / time'. Since the final velocity is 0 (the box stops), and the initial velocity is 1.37 m/s, and the time is 2.8 s, we get: a = (0 - 1.37) / 2.8 = -0.49 m/s^2. The negative sign indicates deceleration.

Next, we use Newton's second law, which states that the net force acting on an object is equal to its mass times its acceleration. The net force in this case is the frictional force because there is no other force acting on the box in the horizontal direction. However, we do not know the mass of the box, but we do know its weight, and weight = mass x gravitational acceleration (g). So mass = weight/g = 52.4N / 9.8m/s^2 = 5.35 kg.

Lastly, we substitute the mass and deceleration into Newton's second law to find the frictional force (f): f = mass x deceleration = 5.35kg x -0.49m/s^2 = -2.62 N. Again, the negative sign indicates that the force acts opposite to the direction of motion. Thus, the frictional force magnitude is 2.62 N.

Learn more about Friction force here:

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25% part (c) assume that d is the distance the cheetah is away from the gazelle when it reaches full speed. Derive an expression in terms of the variables d, vcmax and vg for the time, tc, it takes the cheetah to catch the gazelle.

Answers

maximum speed of cheetah is

$v_1 = v_(max)$

speed of gazelle is given as

$v_2 = v_(g)$

Now the relative speed of Cheetah with respect to Gazelle

$v_(12) = v_1 - v_2$

$v_(12) = v_(max) - v_g$

now the relative distance between Cheetah and Gazelle is given initially as "d"

now the time taken by Cheetah to catch the Gazelle is given as

$d = v_(12)* t$

so by rearranging the terms we can say

$t = (d)/(v_(12))$

$t = (d)/(v_(max) - v_g)$

so above is the relation between all given variable

What minimum distance would you have to hit a baseball from the center of the earth so that it would eventually reach the moon? Assume you can hit the ball directly along the line that connects the centers of the earth and moon. The distance between the centers of the earth and moon is ???? = 3.82 × 108 m.

Answers

Answer:

d = 3.44 x 10⁸ m

Explanation:

The minimum distance required will be the distance from the centre of the earth to a point where gravitational intensity due to both earth and moon becomes equal . Once this point is reached , moon will attract the baseball on its own .

Let this distance be d from the centre of the earth

So GM / d² = G m / ( 3.82 x 10⁸ - d )²

M is mass of the earth , m is mass of the moon

M / m = ( d / 3.82 x 10⁸ - d )²

5.972 x 10²⁴ / 7.34 x 10²² = ( d / 3.82 x 10⁸ - d )²

81.36 = ( d / 3.82 x 10⁸ - d )²

9.02 = d / 3.82 x 10⁸ - d

34.45 x 10⁸ - 9.02 d = d

34.45 x 10⁸ = 10.02 d

d = 3.44 x 10⁸ m

18. Wind speed on Earth is reduced by which?A. temperature
B. friction
C. weather
D. convergence

Answers

Answer:

Temperaturereduces the wind speed on earth.

Explanation:

Wind speed is an atmospheric quantity that causes the wind to flow from high to low level due to a change in temperature.
Wind speed is controlled by the pressure gradient factor and it increases the wind speed with increase in pressure gradient.
Wind speed decreases with an increase in temperature value as it absorbs the heat developed in the system and wind gets slowed down.
Wind speed is measured in velocity and anemometer is used to measure the rate of wind speed.

Answer:

your answer would be A. temperature

Explanation:

Wind speed on Earth is reduced by temperature, it depends on the temperature because let's say there is a nice sunny day, then obviously it will reduce the wind speed

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A bullet with a mass of 20 g and a speed of 960 m/s strikes a block of wood of mass 4.5 kg resting on a horizontal surface. The bullet gets embedded in the block. The speed of the block immediately after the collision is:________. A) cannot be found because we don't know whether the surface is frictionless. B) is 0.21 km/s. C) is 65 m/s. D) is 9.3 m/s. E) None of these is correct