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3 years ago

What is the name of this muscle?

Physics

2 answers:

DaniilM [7]3 years ago

8 0

Shoulder
It’s on the shoulder

suter [353]3 years ago

7 0

Answer:

shoulder

it is at the shoulder region

Explanation:

shoulder

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A 2:2 kg toy train is con ned to roll along a straight, frictionless track parallel to the x-axis. The train starts at the origi

Liula [17]

Answer:

a) 10.51 J

b) 3.48 m/s

Explanation:

Given data :

mass of train ( M ) = 2.2 kg

Given initial velocity ( u ) = 1.6 m/s

a) calculating work done by the force over the journey of the train

F = mx + b ------ ( 1 )

m = slope = ( Δ f / Δ x ) = 2.8 / -7.5 = - 0.373 N/m

x = distance travelled on the x axis by the train = 7.5 m

F = force experienced by the train = 2.8 N

x = 0

∴ b = 2.8

hence equation 1 can be written as

F = ( -0.373) x + 2.8 ----- ( 2 )

hence to determine the work done by the force

W = $\int\limits^7_0 { ( -0.373) x + 2.8 )} \, dx$ Note: the limits are actually 7.5 and 0

∴ W ( work done ) = -10.49 + 21 = 10.51 J

b) calculate the speed of the train at the end of its journey

we will apply the work energy theorem

W = 1/2 m*v^2 - 1/2 m*u^2

∴ V^2 = 2 / M ( W + 1/2 M*u^2 ) ( input values into equation )

V^2 = 12.11

hence V = 3.48 m/s

6 0

3 years ago

A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and

Svetlanka [38]

Answer:

a = 1.152s

b = 0.817 m

c = 7.29m/s

Explanation: let the following

From the first equation of linear motion

V = u+at..........1

parameters be represented as :

t = Time taken

v = Final velocity

a = Acceleration due to gravity = 9.8m/s²

u = Initial velocity = 4 m/s

s = Displacement

V = 0

Substitute the values into equation 1

0 = 4-9.8(t)

-4 = -9.8t

t = 4/9.8

t = 0.408s

From : s = ut+1/2at^2.........2

S = 4×0.408+0.5(-9.8)×0.408^2

S= 1.632-4.9(0.166)

S = 1.632-0.815

S = 0.817m

Her highest height above the board is 0.817 m

Total height she would fall is 0.817+1.90 = 2.717 m

From equation 2

s = ut+1/2at^2

2.717 m = 0t+0.5(9.8)t^2

2.717 m = 0+4.9t^2

2.717 m = 4.9t^2

2.717/4.9 = t^2

0.554 =t^2

t =√0.554

t = 0.744s

Hence, her feet were in the air for 0.744+0.408seconds

= 1.152s

Also recall from equation 1

V= u+at

V = 0+9.8(0.744)

V = 7.29m/s

Hence, the velocity when she hits the water is 7.29m/s

Finally,

a = 1.152s

b = 0.817 m

c = 7.29m/s

4 0

3 years ago

Which geologic feature most likely be represented by contour lines drawn far apart from one another?

andrey2020 [161]

Answer;

A plain

Explanation;

A contour line connects points of the same elevation. Contour lines are usually curves. Closed contours represent hills.

Contour lines can not cross since they represent different elevations. A contour interval is the difference in elevation between one contour and an adjacent contour.

5 0

3 years ago

What happens to kinetic energy when you increase the mass?

Rus_ich [418]

Increasing mass increases kinetic energy. This can be seen in the equation KE = 1/2 (m) (v)^2

If you found this helpful, please brainliest me!

5 0

3 years ago

On level ground a shell is fired with an initial velocity of 49.0 m/s at 70.0 โ above the horizontal and feels no appreciable ai

xenn [34]

Answer:

Horizontal component = 16.8 m/s

Vertical component = 46.0 m/s

Explanation:

If we denote the initial velocity by v and the angle above the horizontal by θ,

the horizontal component of this initial velocity is given by

$v_x = v\cos \theta$

$v_x = (49.0\text{ m/s})\cos\,(70.0^\circ) = 16.8\text{ m/s}$

The vertical component is given by

$v_y = v\sin\theta$

$v_x = (49.0\text{ m/s})\sin\,(70.0^\circ) = 46.0\text{ m/s}$

3 0

3 years ago