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3 years ago

you arrive in my class 45 seconds after leaving math which is 90 meters away how fast did you travel?

Physics

2 answers:

Alekssandra [29.7K]3 years ago

3 0

Answer:

2m/s

Explanation:

90/45

Yanka [14]3 years ago

3 0

Answer:

Explanation:

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We want to construct a solenoid with a resistance of 4.30 Ω and generate a magnetic field of 3.70 × 10−2 T at its center when ap

marshall27 [118]

Answer with Explanation:

We are given that

Resistance of solenoid,R=4.3 ohm

Magnetic field,B= $3.7\times 10^{-2} T$

Current,I=4.6 A

Diameter of wire,d=0.5 mm= $0.5\times 10^{-3} m$

Radius of wire,r= $\frac{d}{2}=\frac{0.5\times 10^{-3}}{2}=0.25\times 10^{-3} m$

$1mm=10^{-3} m$

Radius of solenoid,r'=1 cm= $1\times 10^{-2} m$

$1 cm=10^{-2} m$

Resistivity of copper, $\rho=1.68\times 10^{-8}\Omega m$

We know that

$R=\frac{\rho l}{A}$

Where $A=\pi r^2$

Using the formula

$4.3=\frac{1.68\times 10^{-8}\times l}{\pi(0.25\times 10^{-3})^2}$

$l=\frac{4.3\times \pi(0.25\times 10^{-3})^2}{1.68\times 10^{-8}}=50.23 m$

Number of turns of wire= $\frac{l}{2\pi r'}$

Number of turns of wire= $\frac{50.26}{2\pi(1\times 10^{-2}}=800$

Hence, the number of turns of the solenoid,N=799

Magnetic field in solenoid,B= $\mu_0 nI$

$3.7\times 10^{-2}=4\pi\times 10^{-7} n\times 4.6$

$n=\frac{3.7\times 10^{-2}}{4\times 3.14\times 10^{-7}\times 4.6}$

$n=6404 turns/m$

$n=\frac{N}{L}$

$L=\frac{N}{n}$

$L=\frac{799}{6404}$

$L=0.125 m=0.125\times 100=12.5 cm$

Length of solenoid=12.5 cm

1m=100 cm

8 0

3 years ago

HELP ASAP

amid [387]

Answer:

3200 has the fewest number of sig figs

Explanation:

sorry i was late

7 0

3 years ago

Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103

son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

$PE = -\frac{GMm}{R}$

As M=m, then

$PE = -\frac{Gm^2}{R}$

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

$U = -\frac{2Gm^2}{R}$

$dU = -\frac{2Gm^2}{R}$

$dKE = -dU$

Therefore replacing we have that,

$\frac{1}{2}mv^2 =\frac{Gm^2}{2R}$

Re-arrange to find v,

$v= \sqrt{\frac{Gm}{R}}$

$v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}$

$v = 816.7m/s$

Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

$2r = 2*10^3m$

Therefore

$dU = Gm^2(\frac{1}{R}-\frac{1}{2r})$

$v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}$

$v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}$

$v = 1.83*10^7m/s$

Therefore the velocity when they are about to collide is $1.83*10^7m/s$

7 0

3 years ago

Determine the magnitude of the effective value of g⃗ at a latitude of 60 ∘ on the earth. assume the earth is a rotating sphere.

dezoksy [38]

In addition to acceleration of gravity we experience centrifugal acceleration away from the axis of rotation of the earth. this additional acceleration has value ac = r w^2 where w = angular velocity and r is distance from your spot on earth to the earth's axis of rotation so r = R cos(l) where l = 60 deg is the lattitude and R the earth's radius and w = 1 / (24hr x 3600sec/hr)
now you look up R and calculate ac then you combine the centrifugal acc. vector ac with the gravitational acceleration vector ag = G Me/R^2 to get effective ag' = ag -

5 0

3 years ago

Which of these will be the correct relationship between work input and work output?

dsp73

Answer:

Work input = Work output * Work against friction is your answer so C

Explanation:

I hope this helps you :)

8 0

2 years ago