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motikmotik
1 year ago
5

A car is traveling along a straight road at a velocity of 30m/s when its engine cuts off. For the next ten seconds, the car slow

s down, and its average acceleration is a₁. For the next five seconds, the car slows down further, and its average acceleration is a₂. The ratio of the average acceleration values is a₁/a₂=1.5. Find the velocity of the car at the end of the initial ten-second interval.​
Physics
1 answer:
nasty-shy [4]1 year ago
4 0

The velocity of the car at the end of the initial ten-second interval is 15a₂ + 30.

<h3>Initial velocity of the car</h3>

v = u + at

after 10 seconds;

v = 30 + 10a₁  --(1)

after 5 seconds

v₂ = v + 5a₂ ---- (2)

solve (1) and (2) together

v₂ = 30 + 10a₁ + 5a₂

but a₁/a₂=1.5

a₁ = 1.5a₂

v₂ = 30 + 10(1.5a₂) + 5a₂

v₂ = 30 + 15a₂ + 5a₂

v₂ = 30 + 20a₂

from equation (2), v₂ = v + 5a₂

v + 5a₂ = 30 + 20a₂

v = 15a₂ + 30

Thus, the velocity of the car at the end of the initial ten-second interval is 15a₂ + 30.

Learn more about velocity here: brainly.com/question/4931057

#SPJ1

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Force can alter its direction,slow or stop it you could say it can change its velocity

3 0
3 years ago
According to Kepler's Third Law, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about
NARA [144]

Answer:

Orbital period, T = 1.00074 years

Explanation:

It is given that,

Orbital radius of a solar system planet, r=4\ AU=1.496\times 10^{11}\ m

The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :

T^2=\dfrac{4\pi^2}{GM}r^3

M is the mass of the sun

T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times (1.496\times 10^{11})^3    

T^2=\sqrt{9.96\times 10^{14}}\ s

T = 31559467.6761 s

T = 1.00074 years

So, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about 1.00074 years.

6 0
3 years ago
A bird flies north 3 kilometers and then south 4 kilometers, what is the resultant displacement of the bird? Do not forget to in
Ber [7]

Answer: let north be+ and south be-

therefore, -1 km  is the resultant displacement of the bird

hope it helped u,

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8 0
2 years ago
The magnetic domains in a non-magnetized piece of iron are characterized by which orientation
mel-nik [20]
The unmagnetized pieces of iron would be randomly pointing to directions, this is true because although influenced with the magnetic domain, the direction of the unmagnetized iron field of attraction is not uniform or does not have preferred direction.
3 0
3 years ago
A physical pendulum consists of a meter stick that is pivoted at a small hole drilled through the stick a distance d from the 50
PSYCHO15rus [73]

Answer:

(a). The value of d is 0.056 cm and 1.496 cm.

(b). The time period is 1.35 sec.

Explanation:

Given that,

Length = 50.00 cm

Time period = 2.50 s

Time period of pendulum is defined as the time for one complete cycle.

The period depends on the length of the pendulum.

Using formula of time period

T=2\pi\sqrt{\dfrac{I}{mgh}}

Where, I = moment of inertia

We need to calculate the value of d

Using parallel theorem of moment of inertia

I=I_{cm}+md^2

For a meter stick mass m , the rotational inertia about it's center of mass

I_{cm}-\dfrac{mL^2}{12}

Where, L = 1 m

Put the value into the formula of time period

T=2\pi\sqrt{\dfrac{\dfrac{mL^2}{12}+md^2}{mgd}}

T=2\pi\sqrt{\dfrac{L^2}{12gd}+\dfrac{d}{g}}

T^2=4\pi^2(\dfrac{L^2}{12gd}+\dfrac{d}{g})

Multiplying both sides by d

tex]T^2d=4\pi^2(\dfrac{L^2}{12g}+\dfrac{d^2}{g})[/tex]

(\dfrac{4\pi^2}{g})d^2-T^2d+\dfrac{\pi^2L^2}{3g}=0

Put the value of T, L and g into the formula

4.028d^2-6.25d+0.336=0

d = 0.056\ m, 1.496\ m

The value of d is 0.056 cm and 1.496 cm.

(b). Given that,

L = 50-5 = 45 cm

We need to calculate the time period

Using formula of period

T=2\pi\sqrt{\dfrac{l}{g}}

Put the value into the formula

T=2\pi\sqrt{\dfrac{45\times10^{-2}}{9.8}}

T=1.35\ sec

Hence, (a). The value of d is 0.056 cm and 1.496 cm.

(b). The time period is 1.35 sec.

7 0
3 years ago
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