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3 years ago

The weight of air measured in units of force per area is called _____.

2 answers:

6 0

The weight of air resting on a surface, divided by the area
of the surface ... described in units of force per unit area ...
is called air pressure.

Mrac [35]3 years ago

5 0

Answer:

The weight of air measured in units of force per area is called Air pressure Hopefully this helps!

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A 100 kg cart on a roller coaster has 1800J of kinetic energy. How fast is it going? KE= 1/2mv2

34kurt

Answer:

the speed is equal to 6 m/s

Explanation:

7 0

2 years ago

g If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of th

Reil [10]

Answer:

Speed; v = 17 m/s

Explanation:

We are given;

Radius; r = 110m

Angle; θ = 15°

Now, we know that in circular motion,

v² = rg•tanθ

Thus,

v = √(rg•tanθ)

Where,

v is velocity

r is radius

g is acceleration due to gravity

θ is the angle

Thus,

v = √(rg•tanθ) = √(110 x 9.8•tan15)

v = √(288.85)

v = 17 m/s

8 0

3 years ago

Calculate the gravitational potential energy of a 1200 kg car at the top of a hill that is 42 m high.

GalinKa [24]

It IS PE = (1200 kg)(9.8 m/s²)(42 m) = 493,920 J

6 0

3 years ago

An object is dropped from a bridge. A second object is thrown downwards 1.00 s later. They both reach the water 20.0 m below at

Deffense [45]

Answer:

$v_{o}=-14.60m/s$

Explanation:

Kinematics equation for first Object:

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

but:

$v_{o}=0m/s$ The initial velocity is zero

$y_{o}=20m$

it reach the water at in instant, t1, y(t)=0:

$0=y_{o}-1/2*g*t_{1}^{2}$

$t_{1}=\sqrt{2y_{o}/g}=\sqrt{2*20/9.81}=2.02s$

Kinematics equation for the second Object:

The initial velocity is zero

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

but:

$y_{o}=20m$

it reach the water at in instant, t2, y(t)=0. If the second object is thrown 1s later, t2=t1-1=1.02s

$0=y_{o}+v_{o}t_{2}-1/2*g*t_{2}^{2}$

$v_{o}=1/(t_{2})*(1/2*g*t_{2}^{2}-y_{o})=(1/1.02)*(1/2*9.81*1.02^{2}-20)=-14.60m/s$

The velocity is negative, because the object is thrown downwards.

6 0

3 years ago

The Sun orbits the center of the Milky Way galaxy once each 2.60 × 108 years, with a roughly circular orbit averaging 3.00 × 104

Mamont248 [21]

To solve this problem it is necessary to apply the kinematic equations of linear and angular motion, as well as the given definitions of the period.

Centripetal acceleration can be found through the relationship

$a_c = \frac{v^2}{R}$

Where

v = Tangential Velocity

R = Radius

At the same time linear velocity can be expressed in terms of angular velocity as

$v = R\omega$

Where,

R = Radius

$\omega =$ Angular Velocity

PART A) From this point on, we can use the values used for the period given in the exercise because the angular velocity by definition is described as

$\omega = \frac{2\pi}{T}$

T = Period

So replacing we have to

$\omega = \frac{2\pi}{2.6*10^8years}\\\omega = 2.4166*10^{-8}rad/years\\\omega = 2.4166*10^{-8}rad/years(\frac{1years}{365days})(\frac{1day}{86400s})\\\omega = 7.663*10^{-16}rad/s$

Since $1 Light year = 9.48*10^{15}m$

Then the radius in meters would be

$R = (3*10^4ly)(\frac{9.48*10^{15}m}{1ly})$

$R = 2.844*10^{20}m$

Then the centripetal acceleration would be

$a_c = \frac{v^2}{R}\\a_c = \frac{(R\omega)^2}{R}\\a_c = R\omega^2 \\a_c = 2.844*10^{20}(7.663*10^{-16})^2\\a_c = 1.67*10^{-10}m/s^2$

From the result obtained, considering that it is an unimaginably low value of an order of less than $10^{-10}$ it is possible to conclude that it supports the assertion on the inertial reference frame.

8 0

3 years ago