A student is moving a magnet through a coil of wire. Which of these indicates the strength of the electric current in the coil?

Answer:

C, slightly less than 243 N

Explanation:

Road friction and air resistance aren't that much on a force. Try pushing something and see how much friction there is. Not that much.

a car moving at a speed shows that the force applied to the car is greater than the frictional force and air resistance

c. Slightly less than 243 N.

three charged particals are located at the corners of an equil triangle shown in the figure showing let (q 2.20 Uc) and L 0.650

Answer:

V2 = 15.53 [m/s]]

Explanation:

In order to solve this problem we must use the principle of energy conservation, where potential energy is transformed into kinetic energy. At the bottom is taken as a reference level of potential energy, where the value of this energy is equal to zero.

Above the inclined plane we have two energies, kinetics and potential. While when the sled is at the reference level all this energy will have been transformed into kinetic energy.

[tex]E_{1}=E_{2}\\ m*g*h+(\frac{1}{2} )*m*v_{1} ^{2}=\frac{1}{2}*m*v_{2} ^{2} \\(9.81*8.21)+(0.5*8.96^{2} )=(0.5*v_{2}^{2} )\\(0.5*v_{2}^{2} )=120.68\\v_{2} ^{2}=241.36\\v_{2} =\sqrt{241.36}\\ v_{2} =15.53[m/s][/tex]

Answer:

Water isn't wet by itself, but it makes other materials wet when it sticks to the surface of them.

Explanation:

Answer:

water is wet

it is a liquid

Explanation:

Answer:

changes by 9.8 m/s each second.

R=A2+B2+2ABcosβ−−−−−−−−−−−−−−−−√R=A2+B2+2ABcosβ

A=4NA=4N , B=3NB=3N , β=90°β=90° , cosβ=0cosβ=0

R=A2+B2−−−−−−−√R=A2+B2

R=42+32−−−−−−√=25−−√=5NR=42+32=25=5N

tanα=Bsin90°A+Bcos90°=34tanα=Bsin90°A+Bcos90°=34

α=37°α=37°

Therefore the resultant of the two forces has a magnitude of 5N5N and is at an angle of 37°37° with respect to

Take the moment Allen sees the moose to be the origin.

First, convert his speed to m/s.

90 km/h = (90 km/h) • (1000 m/km) • (1/3600 h/s) = 25 m/s

(a) For the time it takes him to react (0.85 s), Allen is moving at a constant speed of 25 m/s, so that before he actually does anything, he covers a distance of

(25 m/s) • (0.85 s) = 21.25 m

(b) Once he presses the brakes, Allen's vehicle covers a distance x in time t of

x = 21.25 m + (25 m/s) t + 1/2 a t²

and has a speed v of

v = 25 m/s + a t

It takes him 3.5 s to come to a full stop. Use this to find the acceleration:

0 = 25 m/s + a (3.5 s)

a = - (25 m/s) / (3.5 s)

a ≈ - 7.1 m/s²

After 3.5 s, he will have traveled a total distance of

x = 21.25 m + (25 m/s) (3.5 s) + 1/2 (- 7.1 m/s²) (3.5 s)²

x = 152.5 m ≈ 150 m

(c) This one is worded a bit strangely, specifically "once he presses the brake" seems to suggest instantaneous acceleration, not average. Average acceleration is defined for some duration of time. You're probably expected to report the acceleration of the car as it comes to a stop, which we found earlier to be

a ≈ - 7.1 m/s²

Answer:

Higher denisty

Explanation:

High pressure=high denisty

Answer: D

Explanation: it seem right to me I really don't know if this right but I hope this helps

Answer:

i think its the top one

Explanation:

pls tell me if im wrong  

Answer:

A. right before it hits the ground.

Explanation:

because all the way down, it is building kinetic energy.

Answer:

all of the above I think??

Answer: 6 J

Explanation:

Total force applied = 47 N Assuming that direction of movement of pencil and applied force is same. Work done by force in moving the pencil W = Force × Distance through which force moves ⇒ W = 47 × 0.25 = 11.75 J .-.-.-.-.-.-.-.-. In case we are asked useful work done then we calculate net force used for pushing the pencil: Net force used for pushing the pencil = 47 − 23 = 24 N Assuming that direction of movement of pencil and net force is same. Useful Work done by force in moving the pencil W u = Force × Distance ⇒ W u = 24 × 0.25 = 6 J

The work done the pushing force is required.

The work done by the pushing force is [tex]6\ \text{J}[/tex]

[tex]F_1[/tex] = Pushing force = 47 N

[tex]F_2[/tex] = Opposing force = 23 N

m = Mass of object = 0.025 kg

s = Displacement = 0.25 m

Work done is given by

[tex]W=F_ns[/tex]

[tex]\Rightarrow W=(47-23)\times 0.25[/tex]

[tex]\Rightarrow W=6\ \text{J}[/tex]

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Explanation:

Glomar Challenger studies about the "age of rocks in various places in the ocean" in 1968. EXPLANATION: Glomar Challenger was a "deep sea research vessel" for marine geology and oceanography studies.

I hope this helps you :)

Answer:

c no cappp :)

Explanation:

Answer:

False

Explanation:

in my point of view the human race has adapted to being greedy in some ways we can save resources but its very rare

hope this helps! : )

Answer:

Since the ball is travelling 32 degrees above the horizontal, the value of Θ is 32

In the figure, v vector is the vertical component whereas h vector is the horizontal component

Using trigonometry:

CosΘ = h /24

Cos 32 = h/  24

0.85 = h / 24

h = 24*0.85

h = 20.4 m/s

Answer:

h = 2.49 [m]

Explanation:

In order to solve this problem we must use the definition of potential energy, which tells us that energy is equal to the product of mass by gravity by height.

The potential energy can be calculated by means of this equation:

Ep = m*g*h

where:

Ep = potential energy = 980 [J]

m = mass = 40 [kg]

g = gravity acceleration = 9.81 [m/s^2]

h = elevation [m]

Now replacing:

980 = 40*9.81*h

h = 2.49 [m]

Answer:

8.8 kN

Explanation:

V = 2 m³, W = 40 kN, SG = 1.59

Bouyant force N = 1.59 * 1000 kg/m³ * 9.81 N/kg * 2 m³ = 31.2 kN

So the weight becomes 40 - 31.2 = 8.8 kN

Answer:

  orbiting the Sun in an elliptical orbit

Explanation:

When orbits are circular, the center and the focus are the same point. For an elliptical orbit, the orbited object is at one focus of the ellipse. Kepler found planetary orbits to be elliptical with the sun at one focus.

__

As with a lot of scientific theories, it only explained some of the motion. As with a lot of scientific theories, explaining the discrepancies resulted in new discoveries.

Answer:

at one focus of the elliptical orbit of the planets.

The order of magnitude of the final velocity of an object that begins from rest and accelerates at a rate of 20m/s² for 5.0 seconds is 100m/s

​

The acceleration of a body is the change in velocity with respect to time as shown:

[tex]a=\frac{v-u}{t}[/tex]

Given the following parameters

a is the acceleration = 20m/s²

u is the initial velocity = 0m/s

t is the time taken = 5.0seconds

Required

Final velocity "v"

Substitute the given parameters into the formula:

[tex]20=\frac{v-0}{5} \\20 =\frac{v}{5}\\v = 20 \times 5\\v =100m/s[/tex]

Hence the order of magnitude of the final velocity of an object that begins from rest and accelerates at a rate of 20m/s² for 5.0 seconds is 100m/s

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Answer:

Motion with constant velocity of magnitude 1 m/s (uniform motion) for 4 seconds in a positive direction and then for 2 seconds uniform motion with constant velocity of magnitude 3 m/s in reverse direction .

Explanation:

The graph shows a constant velocity of 1 m/s for 4 seconds in the positive direction. After that, between 4 seconds and 6 seconds, the object reverses its motion with constant velocity of magnitude 3m/s.

Answer:

Archimedes' principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces.

Answer: A : 250 is the answer

B; The frequency of a wave is the number of complete oscillations (cycles) made by the wave in one second.

Instead, the wavelength is the distance between two consecutive crests (highest position) or 2 troughs (lowest position) of the wave.

In this problem, we are told that the leaf does two full up and down bobs: this means that it completes 2 full cycles in one second. Therefore, its frequency is

where  is called Hertz (Hz). So, the correct answer is

Explanation:

#Wavespeed

#1

[tex]\\ \rm\Rrightarrow v=\nu\lambda=500(3)=1500m/s[/tex]

#2

[tex]\\ \rm\Rrightarrow v=2(0.4)=0.8m/s[/tex]

#Wavelength

#1

[tex]\\ \rm\Rrightarrow \lambda=\dfrac{v}{\nu}=\dfrac{3}{2}=1.5m[/tex]

#2

[tex]\\ \rm\Rrightarrow \lambda= \dfrac{1450}{40000}=0.03625m[/tex]

#Frequency

[tex]\\ \rm\Rrightarrow \nu=\dfrac{v}{\lambda}=\dfrac{8}{20}=0.4Hz[/tex]

#2

[tex]\\ \rm\Rrightarrow \nu=\dfrac{3\times 10^8}{15\times 10^{-2}}=0.2\timee 10^{10}=2\times 10^9Hz[/tex]

The periodic time of the campus shuttle is = 1 sec

The periodic time is also known as  revolution per second is the time it takes object in motion ( revolving ) to pass through a point twice ( usually the starting position ) of the object and it calculated as :

For a simple pendulum = 2π√L/g

But a campus shuttle bus in motion

since it takes

1 revolution = 1 sec therefore the time period is = 1 sec

Hence we can conclude that the periodic time of the campus shuttle = 1 second.

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Answer:

be in front of the club

Explanation:

The ball should be highest off the ground for a driver. The general recommendation is that the bottom of the golf ball on a tee should be level with the top of the driver; for long and mid-irons, push the tee into the ground so that only about a quarter-inch is above ground.