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3 years ago

15

a car travelling at 50m/h on a horizontal highway (a) if the coefficient of static friction between road and tyres on a rainy da

y is 0.100, what is the minimum distance in which the car will stop?

1 answer:

aleksandrvk [35]3 years ago

3 0

Hope this helps!

-Lilly

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If an object is thrown in an upward direction from the top of a building 160 ft high at an initial speed of 30 mi/h, what is

expeople1 [14]

Answer:

We can use 2 g H = v2^2 - v1^2 or

v2^2 = 2 g H + v1^2

Since 88 ft/sec = 60mph we have 30 mph = 44 ft/sec

The object will return with the same speed that it had initially so the object

starts out with a downward speed of 44 ft/sec

Then v2^2 = 2 * 32 ft/sec^2 * 160 ft + 44 (ft/sec)^2

v2^2 = (2 * 32 * 160 + 44^2) ft^2 / sec^2 = 12180 ft^2/sec^2

v2 = 110 ft/sec

8 0

2 years ago

A migrating robin flies due north with a speed of 12 m/s relative to the air. The air moves due east with a speed of 6.3 m/s rel

Sonja [21]

consider east-west direction along X-axis and north-south direction along Y-axis

$V_{ra}$ = velocity of migrating robin relative to air = 12 j m/s

(where "j" is unit vector in Y-direction)

$V_{ag}$ = velocity of air relative to ground = 6.3 i m/s

(where "i" is unit vector in X-direction)

$V_{rg}$ = velocity of migrating robin relative to ground = ?

using the equation

$V_{rg}$ = $V_{ra}$ + $V_{ag}$

$V_{rg}$ = 12 j + 6.3 i

$V_{rg}$ = 6.3 i + 12 j

magnitude : sqrt((6.3)² + (12)²) = 13.6 m/s

direction : tan⁻¹(12/6.3) = 62.3 deg north of east

4 0

3 years ago

Of all the planets in our solar system, Jupiter has the greatest gravitational strength. If a 1.5 kg pair of running shoes would

Andre45 [30]

Answer:

gₓ = 23.1 m/s²

Explanation:

The weight of an object is on the surface of earth is given by the following formula:

$W = mg$

where,

W = Weight of the object on surface of earth

m = mass of object

g = acceleration due to gravity on the surface of earth = strength of gravity on the surface of earth

Similarly, the weight of the object on Jupiter will be given as:

$W_{x} = mg_{x}$

where,

Wₓ = Weight of the object on surface of Jupiter = 34.665 N

m = mass of object = 1.5 kg

gₓ = acceleration due to gravity on the surface of Jupiter = strength of gravity on the surface of Jupiter = ?

Therefore,

$34.65 N = (1.5 kg)g_{x}$

$g_{x} = \frac{34.65 N}{1.5 kg}$

<u>gₓ = 23.1 m/s²</u>

7 0

3 years ago

The energy of random atomic and molecular motion is called.

Vanyuwa [196]

Energy of a random atomic and molecular is called molecules energy

5 0

2 years ago

Eating 2500 Cal every day a friend of mine maintains a stable weight of 70 kg. One day, after eating 3500 Cal, he decided to do

Kaylis [27]

Answer:

Explanation:

Calories to be burnt = 3500 - 2500 = 1000 Cals .

Efficiency of conversion to mechanical work is 25 % .

Work needed to burn this much of Cals = 1000 x 100 / 25 = 4000 Cals.

4000 Cals = 4.2 x 4000 = 16800 J .

Work done in one jump = kinetic energy while jumping

= 1/2 m v²

= .5 x 70 x 3.3²

= 381.15 J .

Number of jumps required = 16800 / 381.15

= 44 .

4 0

2 years ago