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Murrr4er [49]
3 years ago
15

Suppose a boat moves at 12.0m/s relative to the water. If the boat is in a river with the current directed east of 2.50m/s, what

is the boat's speed relative to the groiund when it is heading (a) east, with the current, and (b) west, against the current?
Physics
1 answer:
inessss [21]3 years ago
8 0

Answer:

(a) 14.5 m/s

(b) 9.5 m/s

Explanation:

Let the speed of the boat in the still water (i.e relative to ground) is v_0.

Given that the speed of the boat relative to water, v_r=12.0 m/s.

Speed of the water current of the river relative to the ground, u_0=2.50 m/s towards east.

(a) When the boat is heading toward the east (in the same direction of the current of the river)

The relative velocity, v_r, of the boat with respect to ground is

v_r=v_0-u_0

\Rightarrow 12=v_0-2.5

\Rightarrow v_0=12+2.5=14.5 m/s

Hence, the boat's speed relative to the ground when it is heading east, with the current, is 14.5 m/s.

(b) When the boat is heading toward the west (in the opposite direction of the current of the river)

The relative velocity, v_r, of the boat with respect to ground is

v_r=v_0-(-u_0)

\Rightarrow 12=v_0+2.5

\Rightarrow v_0=12-2.5=9.5 m/s

Hence, the boat's speed relative to the ground when it is heading west, against the current, is 9.5 m/s.

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Charge flows from low potential to high potential.<br> O True<br> or<br> O False
inn [45]

Answer:

False

Explanation:

Think of the electric potential in terms of potential energy. If you imagine a place with high elevation (A) and another one at sea level (B), a ball will roll from high potential to low potential (A-->B).

Everything in our universe wants to reach a lower state of energy if no external force is acted upon it. Every object tends to slow down (friction), a radioactive element dissipates energy (an unstable element releases energy to get to a stable state), water in the clouds comes down to the ground (rain experiencing difference in potential energy).

Electric potential is exactly the same, you just can't see it! It flows from higher voltage (which is a synonym for electric potential) to lower voltage.

8 0
3 years ago
Calculate the force of gravity on a 1–kilogram box located at a point 1.3 × 107 meters from the center of Earth if the force on
Sati [7]

Since you already gave us the weight of the 2.5-kg box,
we don't even need to know what the distance is, just
as long as it doesn't change.

Look at the formula for the gravitational force:

                           F = G  m₁ m₂ / R² .

If 'G', 'm₁' (mass of the Earth), and 'R' (distance from the Earth's center)
don't change, then the Force is proportional to  m₂ ... mass of the box,
and you can write a simple proportion:

                       (6.1 N) / (2.5 kg)  =  (F) / (1 kg)

Cross-multiply:  (6.1 N) (1 kg)  =  (F) (2.5 kg)

Divide each side by (2.5 kg):  F = (6.1N) x (1 kg) / (2.5 kg)  =  2.44 N .

5 0
3 years ago
7. Two people are pushing a 40.0kg table across the floor. Person 1 pushes with a force of 490N
artcher [175]

Answer:

20.4 m/s^{2}

Explanation:

To start doing this problem, first draw a free body diagram of the table. My teacher always tells us to do this, and I find that it is very helpful. I have attached a free body diagram to this answer- take a look at it.

First, let us see if Net force = MA. To do that, we need to determine whether the object is at equilibrium horizontally. For an object to be at equilibrium, it either needs to be moving at a constant velocity or not moving at all. Also, if an object is at equilibrium, there will not be any acceleration. But we know that there IS acceleration horizontally, so it cannot be in equilibrium. If it is not in equilibrium, we can use the formula ∑F= ma.

Let us determine the net force. Since the object is moving horizontally, we can ignore the weight and normal force, because they are vertical forces. The only horizontal forces we need to worry about are the applied force and force of friction.

Applied force = 1055 N (490 + 565)

Friction force= Unknown

To find the friction force, use the kinetic friction formula, Friction = μkN

μk is the coefficient, which the problem includes- it is 0.613.

N is the normal force, which we have to find.

*To find the normal force, we have to determine if the object is at equilibrium VERTICALLY. Since it has no acceleration vertically (it's not moving up/down), it is at equilibrium. Now, when an object is at equilibrium in one direction, it means that all the forces in that direction are equal. What are our vertical forces? Weight (mg) and Normal force (N). So it means that the Normal force is equal to the Weight.

Weight = mg = (40)(9.8) = 392 N

Normal force = 392 N

Now, plug it back into the formula (μkN): (0.613)(392) = 240.296 N

Friction = 240.296 N

Now that we know the friction, we can find the horizontal net force. Just subtract the friction force, 240.296 from the applied force, 1055 N

Horizontal Net Force: 814.704 N

Now that we know the net force, plug in the numbers for the formula

∑F= ma.

814.704 = (40.0)(a)

*Divide on both sides)

a = 20.3676 m/s^2

Round it to 3 significant figures, to get:

20.4 m/s^{2}

7 0
3 years ago
The equation T^2=A^3 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A
kobusy [5.1K]

Answer:

D. 2^(3/2)

Explanation:

Given that

T² = A³

Let the mean distance between the sun and planet Y be x

Therefore,

T(Y)² = x³

T(Y) = x^(3/2)

Let the mean distance between the sun and planet X be x/2

Therefore,

T(Y)² = (x/2)³

T(Y) = (x/2)^(3/2)

The factor of increase from planet X to planet Y is:

T(Y) / T(X) = x^(3/2) / (x/2)^(3/2)

T(Y) / T(X) = (2)^(3/2)

3 0
3 years ago
Hey guys, i need some help. I'm having a physics test tmmrow and I understand nothing :(. Can anyone plz explain or give me a br
professor190 [17]

We think of sound as something we hear—something that makes noise. But in pure physics terms, sound is just a vibration going through matter.

The way a vibration “goes through” matter is in the form of a sound wave. When you think of sound waves, you probably think of something like this:1

But that’s not how sound waves work. A wave like that is called a transverse wave, where each individual particle moves up and down to create a snake situation.

A sound wave is more like an earthworm situation:2

Like an earthworm, sound moves by compressing and decompressing. This is called a longitudinal wave. A slinky can do both kinds of waves:13

Sound starts with a vibration of some kind creating a longitudinal wave through matter. Check this out:4

That’s what sound looks like—except picture an expanding ripple of spheres doing that. In this animation, the sound wave is being generated by that vibrating grey bar on the left. The bar might be your vocal chords, a guitar string, or a waterfall continually pounding down into the river below. By looking at the red dots, you can see that even though the wave moves in one direction, each individual particle only moves back and forth, mimicking the vibration of the gray bar.

So instead of a curvy snake wave, sound is a pressure wave, which causes each piece of the air to be at either higher-than-normal pressure or lower-than-normal pressure. So when you see a snake-like illustration of a sound wave, it’s referring to the measure of pressure, not the literal path of movement of the particles:5

6 0
3 years ago
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