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Gwar [14]
4 years ago
9

Solve 13.004+3.09+112.947

Physics
1 answer:
Mashcka [7]4 years ago
4 0

Answer:

The answer is 129.041

Explanation:

Because when you add desimals you need to keep them lined up and not uneven

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What is the work done by a car's braking system when it slows the 1500-kg car from an initial speed of 96 km/h down to 56 km/h i
kondor19780726 [428]

Answer:

-352.275KJ

Explanation:

We are given that

Mass of car=1500kg

Initial speed of car =u=96 km/h=96\times \frac{5}{18}=26.67m/s

1km/h=\frac{5}{18}m/s

Final speed of car=v=56km/h=56\times \frac{5}{18}=15.56m/s

Distance traveled by car=s=55m

We have to find the work done by the car's braking system.

Using third equation of motion

v^2-u^2=2as

(15.56)^2-(26.67)^2=2a(55)

-469.18=110a

a=\frac{-469.18}{110}=-4.27m/s^2

Where negative sign indicates that velocity of car decreases.

Work done by a car's barking system=w=F\times s=ma\times s

Work done by a car's barking system=1500\times -4.27\times 55=352275J

Work done by a car's barking system=-\frac{352275}{1000}=-352.275KJ

1KJ=1000J

Where negative sign indicates that work done in opposite direction of motion.

7 0
4 years ago
Read 2 more answers
What is the difference between a 4x100 and a 4x400 relay
qaws [65]
A 4x100 relay is where 4 people run 100 meters and a 4x400 relay is where 4 people run 400 meters
6 0
3 years ago
By using kepler's 3rd law we find that ___________.
tangare [24]

Answer;

By using kepler's 3rd law we find that;

-A year on Earth is shorter than a year on Saturn.

Explanation;

-Kepler’s 3rd law states that the square of a planet’s orbital period is proportional to the cube of its average distance from the Sun (semi-major axis), which tells us that more distant planets move more slowly in their orbits.

-In other words, if you square the 'year' of each planet, and divide it by the cube of its distance to the Sun, you get the same number, for all planets. The law captures the relationship between the distance of planets from the Sun, and their orbital periods.

8 0
3 years ago
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The planet Uranus has a radius of 25,360 km and a surface acceleration due to gravity of 9.0 m/s^2 at its poles. Its moon Mirand
AlexFokin [52]

Answer:

8.67791\times 10^{25}\ kg

0.34589\ m/s^2

0.07903\ m/s^2

Explanation:

M = Mass of Uranus

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Uranus = 25360 km

h = Altitude = 104000 km

r_m = Radius of Miranda = 236 km

m = Mass of Miranda = 6.6\times 10^{19}\ kg

Acceleration due to gravity is given by

g=\dfrac{GM}{r^2}\\\Rightarrow M=\dfrac{gr^2}{G}\\\Rightarrow M=\dfrac{9\times 25360000^2}{6.67\times 10^{-11}}\\\Rightarrow M=8.67791\times 10^{25}\ kg

The mass of Uranus is 8.67791\times 10^{25}\ kg

Acceleration is given by

a_m=\dfrac{GM}{(r+h)^2}\\\Rightarrow a_m=\dfrac{6.67\times 10^{-11}\times 8.67791\times 10^{25}}{(25360000+104000000)^2}\\\Rightarrow a_m=0.34589\ m/s^2

Miranda's acceleration due to its orbital motion about Uranus is 0.34589\ m/s^2

On Miranda

g_m=\dfrac{Gm}{r_m^2}\\\Rightarrow g_m=\dfrac{6.67\times 10^{-11}\times 6.6\times 10^{19}}{236000^2}\\\Rightarrow g_m=0.07903\ m/s^2

Acceleration due to Miranda's gravity at the surface of Miranda is 0.07903\ m/s^2

No, both the objects will fall towards Uranus. Also, they are not stationary.

6 0
3 years ago
If I spacecraft seems to be Motionless in deep space is given some type of quick push what will happen
lapo4ka [179]
It will stay in motion unless an outside force slows it down.
5 0
4 years ago
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