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olya-2409 [2.1K]

3 years ago

15

In order to simulate weightlessness for astronauts in training, they are flown in a vertical circle. if the passengers are to ex

perience weightlessness, how fast should an airplane be moving at the top of a vertical circle with a radius of 2.50 km

1 answer:

sleet_krkn [62]3 years ago

8 0

The answer is "156.6 m/s".

This is how we calculate this;

-N + mg = ma = mv²/r

For "weightlessness" N = 0, so

0 = mg - mv²/r

g - v²/r = 0

v =√( gr)
g = 9.8 and r = 2.5km = 2500 m

v = √(9.8 x 2500)

= 156.6 m/s

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How to use heliocentric in your own sentence?

Alexus [3.1K]

According to the heliocentric theory, the sun is the center of everything in the universe.

8 0

3 years ago

What are the horizontal and vertical components of a lizard’s displacement if it has climbed 7m directly up a tree?

madam [21]

Answer:

The horizontal component is zero.

The vertical component is $7\sin\theta$

Explanation:

Given that,

The lizard climb 7m directly up on a tree.

We know that,

The horizontal component is

$x=\cos\theta$

The vertical component is

$y=\sin\theta$

If the lizard climb 7m directly up on a tree then,

We need to find the components

Using given data

The horizontal component of lizard is

$x=0$

The vertical component is

$y=7\sin\theta$

Hence, The horizontal component is zero.

The vertical component is $7\sin\theta$

7 0

3 years ago

What is the gravitational force between a 45 kg person, and the Earth at 5.98 x 1024 kg, with a distance of

Assoli18 [71]

Answer:

C. 441 N

Explanation:

Gravitational force between two objects can by calculated by the formula

= G m₁m₂ / r² , m₁ and m₂ are masses at distance r

= ( 6.67 x 10⁻¹¹ x 45 x 5.98 x 10²⁴) / ( 6.38 x 10⁶ )²

= 44.09 x 10

= 440.9 N

= 441 N .

7 0

3 years ago

Susan's 10.0kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30? above the f

elena-s [515]

Answer:

The speed after being pulled is 2.4123m/s

Explanation:

The work realize by the tension and the friction is equal to the change in the kinetic energy, so:

$W_T+W_F=K_f-K_i$ (1)

Where:

$W_T=T*x*cos(0)=32N*3.2m*cos(30)=88.6810J\\W_F=F_r*x*cos(180)=-0.190*mg*x =-0.190*10kg*9.8m/s^{2}*3.2m=59.584J\\ K_i=0\\K_f=\frac{1}{2}*m*v_f^{2}=5v_f^{2}$

Because the work made by any force is equal to the multiplication of the force, the displacement and the cosine of the angle between them.

Additionally, the kinetic energy is equal to $\frac{1}{2}mv^{2}$ , so if the initial velocity $v_i$ is equal to zero, the initial kinetic energy $K_i$ is equal to zero.

Then, replacing the values on the equation and solving for $v_f$ , we get:

$W_T+W_F=K_f-K_i\\88.6810-59.5840=5v_f^{2}\\29.097=5v_f^{2}$

$\frac{29.097}{5}=v_f^{2}\\\sqrt{5.8194}=v_f\\2.4123=v_f$

So, the speed after being pulled 3.2m is 2.4123 m/s

8 0

3 years ago

The height of a projectile t seconds after it is launched straight up in the air is given by f (t )equals negative 16 t squared

velikii [3]

Answer:

$\displaystyle a(5)=-32$

Explanation:

<u>Instant Acceleration</u>

The kinetic magnitudes are usually related as scalar or vector equations. By doing so, we are assuming the acceleration is constant over time. But when the acceleration is variable, the relations are in the form of calculus equations, specifically using derivatives and/or integrals.

Let f(t) be the distance traveled by an object as a function of the time t. The instant speed v(t) is defined as:

$\displaystyle v(t)=\frac{df}{dt}$

And the acceleration is

$\displaystyle a(t)=\frac{dv}{dt}$

Or equivalently

$\displaystyle a(t)=\frac{d^2f}{d^2t}$

The given height of a projectile is

$f(t)=-16t^2 +238t+3$

Let's compute the speed

$\displaystyle v(t)=-32t+238$

And the acceleration

$\displaystyle a(t)=-32$

It's a constant value regardless of the time t, thus

$\boxed{\displaystyle a(5)=-32}$

3 0

3 years ago