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Citrus2011 [14]
1 year ago
15

Astronauts often undergo special training in which they are subjected to extremely high centripetal accelerations. One device ha

s a radius of 15 m and can accelerate a person at 98 m/s2. What is the speed of the astronaut in this device?
Physics
2 answers:
qwelly [4]1 year ago
5 0

Answer: The speed of the astronaut in this device 38.34 m/s.

Explanation:

Radius of the device ,r = 15 m

Acceleration of the device = =a_c=98m/s^2

Velocity of the astronaut in a device ,v = ?

a_c=\frac{v^2}{r}

v^2=98 m/s^2\times 15 m

v=38.34 m/s

Hence,the speed of the astronaut in this device 38.34 m/s.

egoroff_w [7]1 year ago
3 0

The centripetal acceleration of an object is given by the relation,

Ac =V^2/R

where Ac = centripetal acceleration = 98 m/s^2

R = radius of rotation = 15 m

V = speed of astronaut

Hence, \frac{V^2}{15} =98

solving this we get, V = 38.34 m/s

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Oksi-84 [34.3K]

Answer:

The horizontal distance d does the ball travel before landing is 1.72 m.

Explanation:

Given that,

Height of ramp h_{1}=2.30\ m

Height of bottom of ramp h_{2}=1.69\ m

Diameter = 0.17 m

Suppose we need to calculate the horizontal distance d does the ball travel before landing?

We need to calculate the time

Using equation of motion

h_{2}=ut+\dfrac{1}{2}gt^2

t=\sqrt{\dfrac{2h_{2}}{g}}

t=\sqrt{\dfrac{2\times1.69}{9.8}}

t=0.587\ sec

We need to calculate the velocity of the ball

Using formula of kinetic energy

K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2

K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times(\dfrac{2}{5}mr^2)\times(\dfrac{v}{r})^2

K.E=\dfrac{7}{10}mv^2

Using conservation of energy

K.E=mg(h_{1}-h_{2})

\dfrac{7}{10}mv^2=mg(h_{1}-h_{2})

v^2=\dfrac{10}{7}\times g(h_{1}-h_{2})

Put the value into the formula

v=\sqrt{\dfrac{10\times9.8\times(2.30-1.69)}{7}}

v=2.922\ m/s

We need to calculate the horizontal distance d does the ball travel before landing

Using formula of distance

d =vt

Where. d = distance

t = time

v = velocity

Put the value into the formula

d=2.922\times 0.587

d=1.72\ m

Hence, The horizontal distance d does the ball travel before landing is 1.72 m.

8 0
2 years ago
An ant travels 2.78 cm [W] and then turns and travels 6.25 cm [S 40 degrees E]. What is the ant's total displacement?
Sunny_sXe [5.5K]
Answer is 6.84 approx
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4 0
2 years ago
When the wind kicks up dust and sand, the dust grains are charged. The small grains tend to get a negative charge, and the large
diamong [38]

Answer:

Explanation:

Small grains are negatively charged by the wind while big grains is positively charged and remains at the ground . This process creates an electric field due to the presence of oppositely charged particles.

When ever electric field exists it is directed from a positive charge to a negative charge so the here electric field is towards an upwards direction.                  

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2 years ago
A 1 200-kg car traveling initially at vCi 5 25.0 m/s in an easterly direction crashes into the back of a 9 000-kg truck moving i
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Answer:

The velocity of the truck after the collision is 20.93 m/s

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Initial velocity of the truck, v_{Ti}=20\ m/s

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Explanation:

The complete question is

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check all that apply to the above

They have different wavelengths.

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