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3 years ago

Can some help me with this question

Physics

2 answers:

Assoli18 [71]3 years ago

4 0

Answer:

It maybe 1600

Explanation:

Because it's near to 1500 in the line.

Luba_88 [7]3 years ago

3 0

The answer is 2000 I need 20 characters do djrjrkrir

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When the inferior angle of the scapula moves in the superior and lateral direction, the motion is called:

kkurt [141]

The motion of the inferior angle of the scapula in the superior and lateral direction is called upward rotation.

<h3>What are the possible motions of the scapula?</h3>

The scapula or the shoulder blade has about six different types of motion it undergoes.

The six ways of movement of the scapula are:

protraction,
retraction,
elevation,
depression,
upward rotation, and
downward rotation

When the inferior angle of the scapula moves in the superior and lateral direction, the motion is called upward rotation.

Learn more about scapula motion at: brainly.com/question/16868917

#SPJ12

8 0

2 years ago

An object moving at 36 km/h reduces its speed to 18 km/h in 2sec. If the mass of the object is 20 kg, find the average force act

laiz [17]

Answer:

-50 N

Explanation:

Givens:

V_i = 36 km/h

V_f = 18 km/h

t = 2 s

m = 20 kg

First we have to convert our km/h into m/s:

(36 km*(1000 m/1 km)) / (60 min *(60 s/1 min)) = 10 m/s

(18 km*(1000 m/1 km)) / (60 min *(60 s/1 min)) = 5 m/s

a = (V_f - V_i)/t

a = (5 m/s - 10 m/s) / 2 s

a = -2.5 m/s^2

F = m(a)

F = 20 kg(-2.5 m/s^2)

F = -50 N

It's a negative force meaning its acting on it opposite its current direction of movement.

7 0

3 years ago

Question 1 of 4 Attempt 4 The acceleration due to gravity, ???? , is constant at sea level on the Earth's surface. However, the

Evgen [1.6K]

Answer:

g(h) = g ( 1 - 2(h/R) )

<em>*At first order on h/R*</em>

Explanation:

Hi!

We can derive this expression for distances h small compared to the earth's radius R.

In order to do this, we must expand the newton's law of universal gravitation around r=R

Remember that this law is:

$F = G \frac{m_1m_2}{r^2}$

In the present case m1 will be the mass of the earth.

Additionally, if we remember Newton's second law for the mass m2 (with m2 constant):

$F = m_2a$

Therefore, we can see that

$a(r) = G \frac{m_1}{r^2}$

With a the acceleration due to the earth's mass.

Now, the taylor series is going to be (at first order in h/R):

$a(R+h) \approx a(R) + h \frac{da(r)}{dr}_{r=R}$

a(R) is actually the constant acceleration at sea level

and

$a(R) =G \frac{m_1}{R^2} \\ \frac{da(r)}{dr}_{r=R} = -2 G\frac{m_1}{R^3}$

Therefore:

$a(R+h) \approx G\frac{m_1}{R^2} -2G\frac{m_1}{R^2} \frac{h}{R} = g(1-2\frac{h}{R})$

Consider that the error in this expresion is quadratic in (h/R), and to consider quadratic correctiosn you must expand the taylor series to the next power:

$a(R+h) \approx a(R) + h \frac{da(r)}{dr}_{r=R} + \frac{h^2}{2!} \frac{d^2a(r)}{dr^2}_{r=R}$

6 0

3 years ago

Read 2 more answers

A 2530-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force so that

gavmur [86]

Answer:

A = 1.4 m/s²

B = -0.10493 m/s³

a = 1.29507 m/s²

T = 28095.8271 N

T = 1.13198 W

Explanation:

t = Time taken

g = Acceleration due to gravity = 9.81 m/s²

The equation

$v(t)=At+Bt^2$

Differentiating with respect to time

$\frac{dv}{dt}=\frac{d(At+Bt^2)}{dt}\\\Rightarrow 1.4=A+2Bt$

At t = 0

$1.4=A$

Hence, A = 1.4 m/s²

$B=\frac{v-At}{t^2}\\\Rightarrow B=\frac{2.18-1.4\times 1.8}{1.8^2}\\\Rightarrow B=-0.10493\ m/s^3$

B = -0.10493 m/s³

At t = 5 seconds

$a=1.4+2\times -0.010493\times 5=1.29507\ m/s^2$

a = 1.29507 m/s²

$T=m(a+g)\\\Rightarrow T=2530(1.29507+9.81)\\\Rightarrow T=28095.8271\ N$

T = 28095.8271 N

Weight of rocket

$W=2530\times 9.81=24819.9\ N$

$\frac{T}{W}=\frac{28095.8271}{24819.9}\\\Rightarrow \frac{T}{W}=1.13198\\\Rightarrow T=1.13198W$

T = 1.13198 W

3 0

3 years ago

Contrast radiation with conduction<br> ( keep it short and simple , make one sentence . )

TEA [102]

Answer:

In radiation, heat transfer occurs through electromagnetic waves without involving particles. In conduction, heat transfer takes place between objects by direct contact.

7 0

3 years ago

Read 2 more answers

Can some help me with this question​

Can some help me with this question