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3 years ago

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a car accelerates along a straight road from rest to 75km/h in 5.0s. What is the magnitude of its average acceleration? show up

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1 answer:

kykrilka [37]3 years ago

5 0

Answer:

4.186 m/s^2

Explanation:

First, convert km/hr to m/s:

(75 km/hr)(1000m/1km)(1hr/60min)(1min/60s) = 20.83 m/s

Then, divide 20.93 m/s by 5.0s

(20.93 m/s) / (5.0s) = 4.186 m/s^2

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If a proton emits a photon in a spin-flipping event in a magnetic field, which is true?

vodomira [7]

Answer:

C.

Explanation:

We know that the magnetic moment of electron is in same direction as it's spin so it's spin flip from down to up.

The correct answer is C that is . It flips from spin down to spin up.

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3 years ago

If a pitcher throws a baseball 10 m in 30 seconds - what would be the average speed?

allsm [11]

Answer:

0.333 m/s

Explanation:

avg speed= (total distance)/(total time)

10m/30s

0.333 m/s

7 0

3 years ago

Read 2 more answers

An emf is induced by rotating a 1060 turn, 20.0 cm diameter coil in the Earth's 5.25 ✕ 10−5 T magnetic field. What average emf (

lara31 [8.8K]

Answer:

The average emf induced in the coil is 175 mV

Explanation:

Given;

number of turns of the coil, N = 1060 turns

diameter of the coil, d = 20.0 cm = 0.2 m

magnitude of the magnetic field, B = 5.25 x 10⁻⁵ T

duration of change in field, t = 10 ms = 10 x 10⁻³ s

The average emf induced in the coil is given by;

$E = N\frac{\delta \phi}{dt} \\\\E = N\frac{\delta B}{\delta t}A$

where;

A is the area of the coil

A = πr²

r is the radius of the coil = 0.2 /2 = 0.1 m

A = π(0.1)² = 0.03142 m²

$E = \frac{NBA}{t} \\\\E = \frac{1060*5.25*10^{-5}*0.03142}{10*10^{-3}} \\\\E = 0.175 \ V\\\\E = 175 \ mV$

Therefore, the average emf induced in the coil is 175 mV

3 0

3 years ago

Need help please... thank u

tangare [24]

Answer:

the answer from my side is both vertical and horizontal

4 0

3 years ago

Read 2 more answers

A circular loop of wire 75 mm in radius carries a current of 113 A. Find the (a) magnetic field strength and (b) energy density

Roman55 [17]

The magnetic field strength is 9.47 ×10⁻⁴ T

The energy density at the center of the loop is 0.36 J/m³

<h3>Calculating Magnetic field strength & Energy density </h3>

From the question, we are to find the magnetic field strength

The magnetic field strength of a loop can be calculated by using the formula,

$B = \frac{\mu_{0} I}{2R}$

Where B is the magnetic field strength

$\mu_{0}$ is the permeability of free space $(\mu_{0}=4\pi \times 10^{-7} \ N/A^{2})$

$I$ is the current

and R is the radius

From the give information,

$R = 75 \ mm= 75 \times 10^{-3} \ m$

and $I = 113 \ A$

Putting the parameters into the formula, we get

$B = \frac{4\pi \times 10^{-7} \times 113}{2 \times 75 \times 10^{-3} }$

$B = 9.47 \times 10^{-4} \ T$

Hence, the magnetic field strength is 9.47 ×10⁻⁴ T

Now, for the energy density

Energy density can be calculated by using the formula,

$u_{B} = \frac{B^{2} }{2\mu_{0} }$

Where $u_{B}$ is the energy density

Then,

$u_{B}= \frac{(9.47\times 10^{-4} )^{2} }{2 \times 4\pi \times 10^{-7} }$

$u_{B} = 0.36 \ J/m^{3}$

Hence, the energy density at the center of the loop is 0.36 J/m³

Learn more on Magnetic field stregth & Energy density here: brainly.com/question/13035557

7 0

2 years ago