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3 years ago

What can we learn from space images?

Physics

1 answer:

Arte-miy333 [17]3 years ago

4 0

How planets look like,and maybe How the planets were created

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g A 5.60-kilogram hoop starts from rest at a height 1.80 m above the base of an inclined plane and rolls down under the influenc

swat32

Answer:

4.24m/s

Explanation:

Potential energy at the top= kinetic energy at the button

But kinetic energy= sum of linear and rotational kinetic energy of the hoop

PE= mgh

KE= 1/2 mv^2

RE= 1/2 I ω^2

Where

m= mass of the hoop

v= linear velocity

g= acceleration due to gravity

h= height

I= moment of inertia

ω= angular velocity of the hoop.

But

I = m r^2 for hoop and ω = v/r

giving

m g h = 1/2 m v^2 + 1/2 (m r^2) (v^2/r^2) = 1/2 m v^2 + 1/2 m v^2 = m v^2

and m's cancel

g h = v^2

Hence

v= √gh

v= √10×1.8

v= 4.24m/s

4 0

3 years ago

Read 2 more answers

An object begins with a speed of 20 meters per second and slows down to a speed of 10 meters per second over a time of 5 seconds

Andreyy89

Answer:

V = 6 m/s

Explanation:

Given that,

Initial speed of an object is 20 m/s

Final speed of an object is 10 m/s

Time, t = 5 s

We need to find the average speed of the object during these 5 seconds. Let it is equal to V. Here, time is same. The average speed is given by :

$V=\dfrac{x+y}{2}\\\\\text{Putting values}\\\\V=\dfrac{20+10}{5}\\\\V=6\ m/s$

So, the average speed of the object is 6 m/s.

3 0

3 years ago

A 16 kg object cuts 4 km in 25 minutes, find the applied force on the object?

GenaCL600 [577]

Answer:

14min i think im not quite sure

Explanation:

7 0

2 years ago

Salmon often jump waterfalls to reach their

PilotLPTM [1.2K]

The minimum velocity of the Salmon jumping at the given angle is 12.3 m/s.

The given parameters;

height of the waterfall, h = 0.432 m
distance of the Salmon from the waterfall, s = 3.17 m
angle of projection of the Salmon, = 30.8º

The time of motion to fall from 0.432 m is calculated as;

$h = v_0_y + \frac{1}{2} gt^2\\\\0.432 = 0 + (0.5\times 9.8)t^2\\\\0.432 = 4.9t^2\\\\t^2 = \frac{0.432}{4.9} \\\\t^2 = 0.088\\\\t = \sqrt{0.088} \\\\t = 0.3 \ s$

The minimum velocity of the Salmon jumping at the given angle is calculated as;

$X = v_0_x t\\\\3.17 = (v_0\times cos(30.8)) \times 0.3\\\\10.567 = v_0\times cos(30.8)\\\\v_0 = \frac{10.567}{cos(30.8)} \\\\v_0 = 12.3 \ m/s$

Thus, the minimum velocity of the Salmon jumping at the given angle is 12.3 m/s.

Learn more here: brainly.com/question/20064545

8 0

2 years ago

Superman does an exhibition run at a track meet. When he runs the 200 m

Ber [7]

Answer:

6.32s

Explanation:

Given parameters:

Length of track and distance covered = 200m

Acceleration = 10m/s²

Unknown:

Time taken to cover the track = ?

Solution:

To solve this problem, we apply one of the motion equations as shown below:

S = ut + $\frac{1}{2}$ at²

S is the distance covered

t is the time taken

a the acceleration

u is the initial velocity

The initial velocity of Superman is 0;

So;

S = $\frac{1}{2}$ at²

200 = $\frac{1}{2}$ x 10 x t²

200 = 5t²

t² = 40

t = 6.32s

5 0

3 years ago