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4 years ago

6

Formulating a Hypothesis: Part I Since the investigative question has two variables, you need to focus on each one separately. T

hinking only about the first part of the question, mass, what might be a hypothesis that would illustrate the relationship between mass and kinetic energy? Use the format of "if...then... because when writing your hypothesis
Help?

2 answers:

Bad White [126]4 years ago

4 0

Answer:

If the mass of an object increases, then its kinetic energy will increase proportionally because mass and kinetic energy have a linear relationship when graphed.

Explanation:

Westkost [7]4 years ago

3 0

Answer : The kinetic energy depends directly on the mass of a particle.

Explanation :

We know that the kinetic energy of any particle is given by :

$KE=\dfrac{1}{2}mv^2$

Where,

m is the mass of an object.

v is the velocity with which it is moving

Kinetic energy is due to the motion of the particle.

So, the kinetic energy of a particle is directly proportional to its mass.

Hence, the conclusion of the question is if the mass of a particle is increases then its kinetic energy also increase.

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A solenoid is built with 870 turns uniformly distributed over a length of 0.390 m to produce a magnetic field of magnitude 1.00

Y_Kistochka [10]

Answer:

35.7 mA

Explanation:

The magnetic field inside a solenoid is given by:

$B=\mu_0 I n$ (1)

where

$\mu_0 = 4\pi \cdot 10^{-7} H/m$ is the vacuum permeability

I is the current

n is the number of turns per unit length

Since we have

N = 870 turns

L = 0.390 (length of the solenoid)

we can calculate n

$n=\frac{N}{L}=\frac{870}{0.390}=2230.8$

And now we can re-arrange eq.(1) to find the current, I:

$I=\frac{B}{\mu_0 n}=\frac{1.00\cdot 10^{-4} T}{(4\pi\cdot 10^{-7} H/m)(2230.8m^{-1})}=0.0357 A = 35.7 mA$

6 0

3 years ago

In regards to Pressure ( Ch. Static Fluids - Introductory Physics)

ehidna [41]

Explanation:

P₁ = P₂ + ρgh

g is the acceleration due to gravity

ρ is the density of the fluid

h is the depth of the fluid

P₁ is the pressure at that depth

P₂ is the pressure at the surface

P₁ and P₂ can either be absolute pressures or gauge pressures, but they must match.

For example, if you wanted to find the <em>absolute</em> pressure at the bottom of an <em>open</em> tank, you would use P₂ = Patm = 14.7 psi or 101.3 kPa.

If instead you wanted to find the <em>gauge</em> pressure, you would use P₂ − Patm = 0 psi or 0 kPa.

If the tank is sealed and pressurized, you would use the P₂ of the tank.

3 0

3 years ago

A magnetic field is aligned perpendicular to the plane of a circular loop of wire with radius 5.0 cm and 20 turns. The magnetic

Mariana [72]

Answer:

$V = 0.157(-36 e^{-3t} + 1.5)$

$t = 1.24 s$

Part b)

$EMF = 0.136(-36e^{-3t} + 1.5)$

Explanation:

magnetic field due to external source is given as

$B = 12 e^{-3t} + 1.5 t + 6$

area of the loop is given as

$A = \pi r^2$

$A = \pi(0.05)^2$

$A = 7.85 \times 10^{-3} m^2$

Now we have

$\phi = NBAcos0$

$\phi = (20)(12e^{-3t} + 1.5t + 6)(7.85 \times 10^{-3})$

$V = \frac{d\phi}{dt}$

$V = 0.157\frac{d}{dt}(12e^{-3t} + 1.5t + 6)$

$V = 0.157(-36 e^{-3t} + 1.5)$

now we need to find the time at which voltage is 0.1 Volts so we have

$0.1 V = 0.157(-36e^{-3t} + 1.5)$

$t = 1.24 s$

Part b)

If magnetic field is inclined at an angle of 30 degree with the normal of the loop then

$\phi = NBAcos30$

now we know that induced EMF is given as

$EMF = \frac{d\phi}{dt}$

$EMF = NAcos30\frac{dB}{dt}$

$EMF = (20)(7.85 \times 10^{-3})cos30(\frac{d}{dt}(12e^{-3t} + 1.5t + 6))$

$EMF = 0.136(-36e^{-3t} + 1.5)$

7 0

3 years ago

The breaks on a 15,680 N car exert a stopping force of 640 N.

BartSMP [9]

The brakes on a 15,680 N car exert a stopping force of 640 N. The car's velocity changes from 20.0 m/s to 0 m/s.

8 0

3 years ago

An 800-kHz radio signal is detected at a point 4.5 km distant from a transmitter tower. The electric field amplitude of the sign

saul85 [17]

Answer:

$2.1\times 10^{-9} T$

Explanation:

We are given that

Frequency,f=800KHz= $800\times 10^{3} Hz$

$1kHz=10^{3} Hz$

Distance,d=4.5 km= $4.5\times 10^{3} m$

1 km=1000 m

Electric field,E=0.63V/m

We have to find the magnetic field amplitude of the signal at that point.

$c=3\times 10^8 m/s$

We know that

$B=\frac{E}{c}$

$B=\frac{0.63}{3\times 10^8}=0.21\times 10^{-8} T$

$B=2.1\times 10^{-9} T$

8 0

4 years ago