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fredd [130]
3 years ago
10

If a car skids 66 ft on wet concrete, how fast was it moving when the brakes were applied? (Round your answer to the nearest who

le number.)
Physics
1 answer:
Sonbull [250]3 years ago
8 0

Answer:

If a car skids 66 ft on wet concrete, it will move at 243 ft/s when the brake is applied.

Explanation:

To determine how fast the car was moving, after skidding, the formula below is used:

V = √32*fd

V is the car's speed (ft/s)

d is skid length (ft) = 66 ft

f is the coefficient of friction determined by the material the car was skidding on.

Coefficient of friction for wet concrete is 0.65

V = √32*fd

V = √32 *0.65* 66

V = 242.679 ft/s ≅ 243 ft/s (nearest whole number)

If a car skids 66 ft on wet concrete, it will move at 243 ft/s when the brake is applied.

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The Earth revolves around the Sun once a year at an average distance of 1.50×1011m. Find the orbital radius that corresponds to
DedPeter [7]

Answer:

9.4\cdot 10^{10} m

Explanation:

We can solve the problem by using Kepler's third law, which states that the ratio between the cube of the orbital radius and the square of the orbital period is constant for every object orbiting the Sun. So we can write

\frac{r_a^3}{T_a^2}=\frac{r_e^3}{T_e^2}

where

r_o is the distance of the new object from the sun (orbital radius)

T_o=180 d is the orbital period of the object

r_e = 1.50\cdot 10^{11} m is the orbital radius of the Earth

T_e=365 d is the orbital period the Earth

Solving the equation for r_o, we find

r_o = \sqrt[3]{\frac{r_e^3}{T_e^2}T_o^2} =\sqrt[3]{\frac{(1.50\cdot 10^{11}m)^3}{(365 d)^2}(180 d)^2}=9.4\cdot 10^{10} m

3 0
3 years ago
Newton’s second law of motion addresses the relationship between what two variables that influence the force on a body?
IgorC [24]

Newton’s Second Law concerns the generation of force based on an object’s mass and acceleration, as described by the equation F=ma.

Hope this helps!

8 0
3 years ago
Read 2 more answers
Two forces, F₁ and F₂, act at a point. F₁ has a magnitude of 8.00 N and is directed at an angle of 61.0° above the negative x ax
kirill115 [55]

1) -7.14 N

2) +2.70 N

3) 7.63 N

Explanation:

1)

In order to find the x-component of the resultant force, we have to resolve each force along the x-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: this means that the angle with respect to the positive x axis is

180^{\circ}-61^{\circ}

so its x-component is

F_{1x}=(8.00)(cos (180^{\circ}-61^{\circ}))=-3.88 N

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: so, its angle with respect to the positive x-axis is

180^{\circ}+52.8^{\circ}

Therefore its x-component is

F_{2x}=(5.40)(cos (180^{\circ}+52.8^{\circ}))=-3.26 N

So, the x-component of the resultant force is

F_x=F_{1x}+F_{2x}=-3.88+(-3.26)=-7.14 N

2)

In order to find the y-component of the resultant force, we have to resolve each force along the y-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: as we said previously, the angle with respect to the positive x axis is

180^{\circ}-61^{\circ}

so its y-component is

F_{1y}=(8.00)(sin (180^{\circ}-61^{\circ}))=7.00 N

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: as we said previously, its angle with respect to the positive x-axis is

180^{\circ}+52.8^{\circ}

Therefore its y-component is

F_{2y}=(5.40)(sin (180^{\circ}+52.8^{\circ}))=-4.30 N

So, the y-component of the resultant force is

F_y=F_{1y}+F_{2y}=7.00+(-4.30)=2.70 N

3)

The two components of the resultant force representent the sides of a right triangle, of which the resultant force corresponds tot he hypothenuse.

Therefore, we can find the magnitude of the resultant force by using Pythagorean's theorem:

F=\sqrt{F_x^2+F_y^2}

Where in this problem, we have:

F_x=-7.14 N is the x-component

F_y=2.70 N is the y-component

And substituting, we find:

F=\sqrt{(-7.14)^2+(2.70)^2}=7.63 N

6 0
3 years ago
Read 2 more answers
Three collinear forces,F1=45N west,F2=63N east and an unknown force F3 are applied to an object.The resultant force of the three
Greeley [361]

Take east to be the positive direction. Then the resultant force from adding <em>F</em>₁ and <em>F</em>₂ is

<em>F</em>₁ + <em>F</em>₂ = (-45 N) + 63 N = 18 N

which is positive, so it's directed east.

To this we add a third force <em>F</em>₃ such that the resultant is 12 N pointing west, making it negative, so that

18 N + <em>F</em>₃ = -12 N

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So <em>F</em>₃ has a magnitude of 30 N and points west.

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3 years ago
if an average cloud has a density of 0.5 g/m3 and has a volume of 1,000,000,000 m3,what is the weight of an average cloud
sveta [45]

If you have (1 x 10⁹) cubic meters of volume, and 1/2 gram of mass
in each cubic meter, then you have

          (1/2 x 10⁹) grams = (5 x 10⁸) grams = (5 x 10⁵) kilograms of mass . 

On Earth, that mass weighs  4,900,000 newtons .

                                            (about  1,102,300 pounds)

                                            (about  551 tons)

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3 years ago
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