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3 years ago

15

A car increases its velocity from 0 m/s to 15 m/s in 6 seconds. Calculate its acceleration in n m/s/s. Round your answer to one

decimal place

1 answer:

iren [92.7K]3 years ago

7 0

Answer:

i cant find answer sorry :((( wish you best of luck <333

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If you attach a 100 g mass to the spring whose data are shown in the graph, what will be the period of its oscillations?

ycow [4]

if for a force of 0.5 N we have a displacement of -0.02m we can calculate the elastic constant(k) with the formula F=-kx(F=0.5N x=-0.02)

k=F/x k=0.5/0.02=25N/m

now we can calculate the period by the formula

T=2π√(m/k)

in the mass we convert grams to kilograms so 100g=0.1kg

T=6.28√(0.1/25)⇒T=0.39seconds

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3 years ago

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You find a measurement or 34 kg written in a laboratory notebook. What physical characteristic is this a measurement of?

Elis [28]

Planck's constant. A physical constant adopted in 2011 by the CGPM.

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4 years ago

You were driving your car to UTD at a speed of 35 miles per hour. You stopped at the FloydCampbell intersection with the signal

ser-zykov [4K]

Answer:

green light have high energy

Explanation:

We have given the wavelength of the red light $\lambda =6.45\times 10^{-5}cm=6.45\times 10^{-7}m$

Speed of the light $c=3\times 106{8}m/sec$

The energy of the signal is given by $E=h\nu =h\frac{c}{\lambda }=\frac{6.67\times 10^{-34}\times 3\times 10^{8}}{6.45\times 10^{-7}}=3.1023\times 10^{-15}j$

The frequency of the green light is given by:

$f=5.80\times 10^{14}s^{-1}$

So energy $E=h\nu =6.67\times 10^{-34}\times 5.80\times 10^{14}=3.8686\times 10^{-19}j$

So green light have high energy

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3 years ago

One joule of work is needed to move one coulomb of charge from one point to another with no change in velocity. Which of the fol

podryga [215]

It B the current is one ampere

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4 years ago

A car with speed v and an identical car with speed 2v both travel the same circular section of an unbanked road. If the friction

yawa3891 [41]

Answer:

$F'=\dfrac{F}{4}$

Explanation:

Let m is the mass of both cars. The first car is moving with speed v and the other car is moving with speed 2v. The only force acting on both cars is the centripetal force.

For faster car on the road,

$F=\dfrac{mv^2}{r}$

v = 2v

$F=\dfrac{m(2v)^2}{r}$

$F=4\dfrac{m(v)^2}{r}$ ..........(1)

For the slower car on the road,

$F'=\dfrac{mv^2}{r}$ ............(2)

Equation (1) becomes,

$F=4F'$

$F'=\dfrac{F}{4}$

So, the frictional force required to keep the slower car on the road without skidding is one fourth of the faster car.

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4 years ago