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3 years ago

How Far East have you traveled if you ran 80 meters northeast

Physics

1 answer:

CaHeK987 [17]3 years ago

6 0

Explanation:

Given parameters:

Displacement = 80m

Unknown:

Displacement eastward = ?

Solution:

Displacement is the distance traveled in a specific direction.

To find the distance traveled eastward, we need to find the horizontal component of his displacement using the cosine rule:

Cos Ф = $\frac{adjacent }{hypotenuse}$

The unknown is the adjacent;

Adjacent = hypotenuse CosФ = 80 cos 45 = 56.57m

His displacement is 56.57m due east

learn more:

Displacement brainly.com/question/5461768

#learnwithBrainly

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Chứng minh V=kq/r từ mối liên hệ giữ E và V

Marat540 [252]

Answer:V=Aq=KQr

Explanation:

Không biết V bạn kí hiệu ở đây là gì nhỉ? Có phải là điện thế?

Điện thế tại 1 điểm trong điện trường được định nghĩa là công làm vật dịch chuyển từ vị trí đó đến vô cùng. V = A/q

Chứng minh thì được, nhưng chỉ e bạn không có hiểu biết về nguyên hàm, tích phân nên không hiểu.

- Xét tại vị trí cách điện tích Q một đoạn x, khi đó điện tích q sẽ chịu 1 lực: dF=KQ.qx2

Điện tích q dịch chuyển 1 khoảng dx rất nhỏ. Khi đó công do lực điện trường gây ra là:

dA=dF.dx=KQ.qx2dx

Công để dịch chuyển điện tích q từ vị trí r đến vô cùng là:

A=∫∞rdA=∫∞rKQ.qx2dx=KQ.qr−KQ.q∞=KQ.qr

Theo đúng định nghĩa: V=Aq=KQr

4 0

3 years ago

A 420N force act on a 400N object, and the force is from the north.

belka [17]

Answer:

I believe its forward or south :D

Explanation:

7 0

3 years ago

7. Water flows trough a horizontal tube of diameter 2.5 cm that is joined to a second horizontal tube of diameter 1.2 cm. The pr

Usimov [2.4K]

To solve this problem it is necessary to apply the concepts related to the continuity of fluids in a pipeline and apply Bernoulli's balance on the given speeds.

Our values are given as

$d_1 = 2.5cm \rightarrow r_1=1.25cm=1.25*10^{-2}m$

$d_2 = 1.2cm \rightarrow r_2 = 0.6cm = 0.6*10^{-2}m$

From the continuity equations in pipes we have to

$A_1V_1 = A_2 V_2$

Where,

$A_{1,2}$ = Cross sectional Area at each section

$V_{1,2}$ = Flow Velocity at each section

Then replacing we have,

$(\pi r_1^2) v_1 = (\pi r_2^2) v_2$

$(1.25*10^{-2})^2 v_1 = ( 0.06*10^{-2})^2 v_2$

$v_2 = \frac{(1.25*10^{-2})^2 }{0.6*10^{-2})^2} v_1$

From Bernoulli equation we have that the change in the pressure is

$\Delta P = \frac{1}{2} \rho (v_2^2-v_1^2)$

$7.3*10^3 = \frac{1}{2} (1000)([ \frac{(1.25*10^{-2})^2 }{0.6*10^{-2})^2} v_1 ]^2-v_1^2)$

$7300= 8919.01 v_1^2$

$v_1 = 0.9m/s$

Therefore the speed of flow in the first tube is 0.9m/s

6 0

4 years ago

two vehicles have a head on collision. one vehicle has a mass of 3000 kg and moves at 25 m/s while the second vehicle has a mass

emmasim [6.3K]

Answer:

The speed of the combined vehicles is 6.82m/s

Explanation:

Using the law of conservation of momentum which stayed that the sum of momentum of bodies before collision is equal to their sum of momentum after collision. After collision, both object moves with the same velocity.

Momentum = mass×velocity

Before collision:

Momentum of vehicle or mass 3000kg moving with velocity 25m/s

= 3000×25

= 75000kgm/s

Pa = 75000kgm/s

Momentum of vehicle with mass 2500kg moving with velocity of -15m/s

= 2500×-15

= -37500kgm/s

After collision:

Momentum = (3000+2500)V

Where v is their common velocity

Momentum after collision = 5500V

Based on the law:

75000+(-37500) = 5500V

75000-37500 = 5500V

37500 = 5500V

V = 37500/5500

V = 6.82m/s

7 0

3 years ago

A glider of mass m1 on a frictionless horizontal track is connected to an object of mass m2 by a massless string. The glider acc

inysia [295]

Answer:

m2g -> T2 -> T1

Explanation:

m2g -> T2 -> T1

5 0

4 years ago