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3 years ago

A 1-kilogram object is thrown horizontally and a 2-kilogram object is dropped vertically at

Physics

1 answer:

Semenov [28]3 years ago

4 0

Their vertical position will be the same for both at any instant. They will both hit the ground at the same time.

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A car races around a circular track. Friction on the tires is the what that acts toward the center of the circle and keeps the c

Ivanshal [37]

I would think force. Because friction has to have force to work ^-^

8 0

4 years ago

A truck is traveling east at 80 km/h. At an intersection 32 km ahead, a car is traveling north at 50 km/h. How long after this m

nirvana33 [79]

The time elapsed when the vehicles are closest to each other is 20 min.

The given parameters:

Speed of the truck, u = 80 km/h
Distance, d = 32 km
Speed of the car, v = 50 km/h

<h3>Principles of relative speed</h3>

The time elapsed when the cars are close to each other is calculated by applying the principles of relative speed.

$(V_r) t = d\\\\V_r^2 = 50^2 + 80^2\\\\V_r =\sqrt{50^2 + 80^2} \\\\V_r = 94.34 \ km/h$

$94.34 t = 32\\\\t = \frac{32}{94.34} \\\\t = 0.34 \ hr\\\\t \approx 20 \min$

Thus, the time elapsed when the vehicles are closest to each other is 20 min.

Learn more about relative velocity here: brainly.com/question/24430414

3 0

3 years ago

A pressure cylinder has a diameter of 150-mm and has a 6-mm wall thickness. What pressure can this vessel carry if the maximum s

myrzilka [38]

Answer:

p = 8N/mm2

Explanation:

given data ;

diameter of cylinder = 150 mm

thickness of cylinder = 6 mm

maximum shear stress = 25 MPa

we know that

hoop stress is given as = $\frac{pd}{2t}$

axial stress is given as = $\frac{pd}{4t}$

maximum shear stress = (hoop stress - axial stress)/2

putting both stress value to get required pressure

$25 = \frac{ \frac{pd}{2t} -\frac{pd}{4t}}{2}$

$25 = \frac{pd}{8t}$

t = 6 mm

d = 150 mm

therefore we have pressure

p = 8N/mm2

7 0

3 years ago

A dwarf planet discovered out beyond the orbit of Pluto is known to have an orbital period of 619.36 years. What is its average

Maksim231197 [3]

Answer: $72.66 AU=1.089(10)^{10} km$

Explanation:

Let's begin by explaining that according to Kepler’s Third Law of Planetary motion “The square of the orbital period $T$ of a planet is proportional to the cube of the semi-major axis $a$ of its orbit”:

$T^{2}\propto a^{3}$ (1)

Now, if $T$ is measured in years (Earth years), and $a$ is measured in astronomical units (equivalent to the distance between the Sun and the Earth: $1AU=1.5(10)^{8}km$ ), equation (1) becomes: