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Advocard [28]
1 year ago
14

This is the third time I’m asking, please, On a wet road, is a higher coefficient of friction on the tires safer or a lower one

and pls explain I don’t get it
Physics
1 answer:
jenyasd209 [6]1 year ago
4 0

Answer:

higher is safer

Explanation:

because it is a wet slippery surface, you would need more friction on the tires, to get more traction in the slippery wet road, if you had low friction you would not move anywhere and or could swerve off somewhere

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Astronauts often undergo special training in which they are subjected to extremely high centripetal accelerations. One device ha
egoroff_w [7]

The centripetal acceleration of an object is given by the relation,

Ac =V^2/R

where Ac = centripetal acceleration = 98 m/s^2

R = radius of rotation = 15 m

V = speed of astronaut

Hence, \frac{V^2}{15} =98

solving this we get, V = 38.34 m/s

3 0
3 years ago
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1. The speed limit on a particular freeway is 28.0 m/s (about 101 km/hour). A car
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Answer:

the speed limit 6 seconds and the car will travel in 90m

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3 years ago
A comet of mass 1.20 × 10¹⁰kg moves in an elliptical orbit around the Sun. Its distance from the Sun ranges between 0.500 AU and
irina [24]

The eccentricity of its orbit is $$U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

<h3>What is mass?</h3>
  • Mass is a physical body's total amount of matter. It also serves as a gauge for the body's inertia or resistance to acceleration (change in velocity) in the presence of a net force. The strength of an object's gravitational pull to other bodies is also influenced by its mass.
  • The kilogram is the SI unit of mass (kg). In science and technology, a body's weight in a given reference frame is the force that causes it to accelerate at a rate equal to the local acceleration of free fall in that frame.
  • For instance, a kilogram mass weighs around 2.2 pounds at the surface of the planet. However, the same kilogram mass would weigh just about 0.8 pounds on Mars and about 5.5 pounds on Jupiter.
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The length of the semi-major axis is calculated as follows:

where, $G=6.67 \times 10^{-1} \mathrm{~m}^3 / \mathrm{kgs}$

$M=1.99 \times 10^{30} \mathrm{~kg}=$mass of sur

$m=1.20 \times 10^{10} \mathrm{~kg}$ - a mass of the comet

$$\begin{aligned}\therefore \quad \text { At aphelion, } r &=50 \times U \\&=50 \times 1.496 \times 10^{11} \mathrm{~m} . \\U=-\frac{6.67 \times 10^{-11} \times \mathrm{m} 1.99 \times 10^{30} \times 1.20 \times 10^{10}}{50 \times 1.496 \times 10^{11}} \\U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

$$U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

To learn more about mass, refer to:

brainly.com/question/3187640

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4 0
2 years ago
The cable of the 1800kg elevator cab in Fig. 8−56 snaps when the cab is at rest at the first floor, where the cab bottom is a di
drek231 [11]

a ) The speed of the cab just before it hits the spring = 7.4 m / s

b ) The maximum distance x that the spring is compressed = 0.9 m

c ) The distance that the cab will bounce back up the shaft = 2.8 m

a ) The speed of the cab just before it hits the spring,

Ki + Pi = K final + P final + W

1 / 2 m vo² + m g hi = 1 / 2 m v² + m g h + f d

0 + ( 1800 * 9.8 * 3.7 m ) = ( 1 / 2 * 1800 * v² ) + 0 + ( 4400 * 3.7 )

v = 7.4 m / s

b ) The maximum distance x that the spring is compressed,

Ki + Pi = K final + P final + W + Fs

1 / 2 m vo² + m g x = 1 / 2 m v² + m g h + f d + 1 /2 k x²

( 1/2 * 1800 * 7.4² ) + ( 1800 * 9.8 * x ) =0 + 0+ ( 4400 * x ) + ( 1/2 * 1800 * x² )

75000 x² + 13420 x - 50625 = 0

x = 0.9 m

c ) The distance that the cab will bounce back up the shaft,

Ki + Pi + Fs = K final + P final + W

1 / 2 m vo² + m g x + 1 /2 k x² = 1 / 2 m v² + m g h + f d

0 + 0 + ( 1 / 2 * 0.15 * 0.9² ) = 0 + ( 1800 * 9.8 * h ) + ( 4400 * h )

h = 2.8 m

Therefore,

a ) The speed of the cab just before it hits the spring = 7.4 m / s

b ) The maximum distance x that the spring is compressed = 0.9 m

c ) The distance that the cab will bounce back up the shaft = 2.8 m

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3 0
1 year ago
The value of acceleration due to gravity (g) on Pluto is about . 0.61 meters/second2. How much will an object that weighs 250 ne
Annette [7]
First, determine the mass of the object by dividing its weight on Earth by 9.8 m/s² as shown below,
                                   m = 250 N / 9.8 m/s² = 25.51 kg
Then, multiply the obtained mass by the acceleration due to gravity (g) on Pluto. 
                                  W (in Pluto) = (25.51 kg) x (0.61 m/s²)  = 15.56 N
Therefore, the object will only weigh 15.56 N. 
8 0
3 years ago
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