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3 years ago

5

Any Help??? Dued before 2:30pm and it's 9:30am by me... so please take yuh time and ans.... jus ans 1 if you want... This is for

course marks.... I appreciate all efforts❤

1 answer:

Kryger [21]3 years ago

5 0

Answer:

The methodology employed by Galileo contributed to the development of Physics by find moons of Jupiter. (I think)

sorry if it's wrong

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If the weight of a person is 500 newton what is his mass on the earth ?

miskamm [114]

Answer:

The person is on the Moon having a weight of 500 N. The acceleration of gravity on the Moon is approximately 1.6 m/s2. What is your his, which includes his space suit?

f= Force (of gravity)=500N

g=acceleration of gravity=1.6m/s^2

m=mass=312kg

m=f/a= 500N/1.6 m/s^2 = 500 (kg-m/1.6m/s^2) = 500/1.6kg = 312kg

his mass is 312kg

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2 years ago

Two uniform solid cylinders, each rotating about its central (longitudinal) axis, have the same mass of 2.88 kg and rotate with

Tanya [424]

Answer:

(a) $K_{small}=4839.3J$

(b) $K_{larger}=17406.4J$

Explanation:

Given data

The angular velocity of two cylinders ω=257 rad/s

The mass of the two cylinders m=2.88 kg

The radius of small cylinder r₁=0.319 m

The radius of larger cylinder r₂=0.605 m

For Part (a)

The rotational kinetic energy of the cylinder is given by:

$K=\frac{1}{2}Iw^2$

Where I is rotational of inertia of solid cylinder about its central axis.

So

$K=\frac{1}{2}Iw^2\\ K=\frac{1}{2}(1/2mr^2)w^2$

Substitute the given values

So

$K_{small}=\frac{1}{4}(2.88kg)(0.319)^2(257rad/s)^2 \\K_{small}=4839.3J$

For Part (b)

$K=\frac{1}{2}Iw^2\\ K=\frac{1}{2}(1/2mr_{2}^2)w^2$

Substitute the given values

$K_{larger}=\frac{1}{4}mr_{2}^2w^2\\ K_{larger}=\frac{1}{4}(2.88kg)(0.605m)^2(257rad/s)^2\\ K_{larger}=17406.4J$

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3 years ago

Que significa gobernar

kow [346]

Answer: conducir la política, acciones y asuntos de (un estado, organización o personas).

7 0

2 years ago

Find the magnitude and direction of a force between a 25.0 coulomb charge and a 40.0coulomb charge when they are separated by a

deff fn [24]

Answer:

95.0 colomb

Explanation:

Make sure to understand the concept

6 0

2 years ago

А A pool of water of refractive index

babymother [125]

Answer:

Apparent depth = 45 cm

Explanation:

The refractive index of water in a pool, n = 4/3

Real depth, d = 60 cm

We need to find its apparent depth when viewed vertically through air. The ratio of real depth to the apparent depth is equal to the refractive index of the material. Let the apparent depth is d'. So,

$n=\dfrac{d}{d'}\\\\d'=\dfrac{d}{n}\\\\d'=\dfrac{60}{\dfrac{4}{3}}\\\\d'=45\ cm$

So, the apparent depth is 45 cm.

3 0

2 years ago