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3 years ago

A car traveling at 27 m/s starts to decelerate steadily. It comes to a complete stop in 12 seconds. What is its acceleration?

1 answer:

5 0

Initial speed = 27 m/s
Final speed = zero

Acceleration = (final speed - initial speed) / (time)

= (-27 m/s) / (12 s)

= (-27/12) m/s²

= - 2.25 m/s²

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In 1996, NASA performed an experiment called the Tethered Satellite experiment. In this experiment a 4.32 x 104-m length of wire

elena55 [62]

Answer:

Explanation:

Length, l = 4.32 x 10^4 m

speed, v = 7.07 x 10^3 m/s

magnetic field, B = 5.81 x 10^-5 T

The formula for the motion emf is given by

e = B x v x l

e = 5.81 x 10^-5 x 7.07 x 10^3 x 4.32 x 10^4

e = 17745.1 V

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3 years ago

Which of the following is true about magnetic poles?

Vika [28.1K]

The answer is A. Have you ever heard the saying, opposites attract? That comes from magnetism. Like poles repel each other while opposites attract.

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3 years ago

Read 2 more answers

The earth’s magnetic field is located in the _____.

Natali [406]

Near Greenland in the northern hemisphere <span />

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3 years ago

A 41.0 g marble moving at 2.30 m/s strikes a 25.0 g marble at rest. What is the speed of each marble immediately after the colli

Gre4nikov [31]

Answer:

speed of each marble after collision will be 1.728 m/sec

Explanation:

We have given mass of the marble $m_1=41gram=0.041kg$

Velocity of marble $v_1=2.30m/sec$

Its collides with other marble of mass 25 gram

So mass of other marble $m_2=25gram=0.025kg$

Second marble is at so $v_2=0m/sec$

We have to find the velocity of second marble

From momentum conservation we know that

$m_1v_1+m_2v_2=(m_!+m_2)v$ , here v is common velocity of both marble after collision

So $0.041\times 2.30+0.025\times 0=(0.041+0.025)v$

v = 1.428 m /sec

So speed of each marble after collision will be 1.728 m/sec

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3 years ago

Charge is uniformly distributed around a ring of radius R and the resulting electric field magnitude E is measured along the rin

Arte-miy333 [17]

Answer:

$x=\dfrac{r}{\sqrt2}$

Explanation:

Given that

Radius =r

Electric filed =E

Q=Charge on the ring

The electric filed at distance x given as

$E=K\dfrac{Q}{(r^2+x^2)^{3/2}}$

For maximum condition

$\dfrac{dE}{dx}=0$

$E=K{Q}{(r^2+x^2)^{-3/2}}$

$\dfrac{dE}{dx}=K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}$

For maximum condition

$\dfrac{dE}{dx}=0$

$K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}=0$

$r^2+x^2-3x^2=0$

$x=\dfrac{r}{\sqrt2}$

At $x=\dfrac{r}{\sqrt2}$ the electric field will be maximum.

3 0

3 years ago