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3 years ago

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A guy wire helps to hold a television tower in place. The wire forms an angle of 40 degrees with the tower. It is under 4000N te

nsion. (a) What force twnds to support the tower. (b) What force tends to pull the tower over.

1 answer:

Paul [167]3 years ago

5 0

A) Gravity
b) Tensional force

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I will mark brainlist

tensa zangetsu [6.8K]

Answer:

False

Explanation:

A wave is a disturbance that transfers energy from one place to another without transferring matter.

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2 years ago

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The kinetic friction force between a 60.0-kg object and a horizontal surface is 50.0 N. If the initial speed of the object is 25

Vikki [24]

Answer:

375 m.

Explanation:

From the question,

Work done by the frictional force = Kinetic energy of the object

F×d = 1/2m(v²-u²)..................... Equation 1

Where F = Force of friction, d = distance it slide before coming to rest, m = mass of the object, u = initial speed of the object, v = final speed of the object.

Make d the subject of the equation.

d = 1/2m(v²-u²)/F.................. Equation 2

Given: m = 60.0 kg, v = 0 m/s(coming to rest), u = 25 m/s, F = -50 N.

Note: If is negative because it tends to oppose the motion of the object.

Substitute into equation 2

d = 1/2(60)(0²-25²)/-50

d = 30(-625)/-50

d = -18750/-50

d = 375 m.

Hence the it will slide before coming to rest = 375 m

6 0

2 years ago

A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.76 m/s2 for t1 = 20 s. At that point the

alex41 [277]

Answer:

(a) $v_1 = a_1t_1 = 1.76 t_1$

(b) It won't hit

(c) 110 m

Explanation:

(a) the car velocity is the initial velocity (at rest so 0) plus product of acceleration and time t1

$v_1 = v_0 + a_1t_1 = 0 +1.76t_1 = 1.76t_1$

(b) The velocity of the car before the driver begins braking is

$v_1 = 1.76*20 = 35.2m/s$

The driver brakes hard and come to rest for t2 = 5s. This means the deceleration of the driver during braking process is

$a_2 = \frac{\Delta v_2}{\Delta t_2} = \frac{v_2 - v_1}{t_2} = \frac{0 - 35.2}{5} = -7.04 m/s^2$

We can use the following equation of motion to calculate how far the car has travel since braking to stop

$s_2 = v_1t_2 + a_2t_2^2/2$

$s_2 = 35.2*5 - 7.04*5^2/2 = 88 m$

Also the distance from start to where the driver starts braking is

$s_1 = a_1t_1^2/2 = 1.76*20^2/2 = 352$

So the total distance from rest to stop is 352 + 88 = 440 m < 550 m so the car won't hit the limb

(c) The distance from the limb to where the car stops is 550 - 440 = 110 m

8 0

2 years ago

For 0.37 moles of oxygen (02) gas at room temperature with active translational and rotational degrees of freedom.

Lelechka [254]

Answer:

B. 161.5 J

Explanation:

$n$ = Number of moles = 0.37

$\Delta T$ = Rise in the temperature of the oxygen gas = 15 K

$Q$ = heat added in order to raise the temperature

$c_{p}$ = specific heat at constant pressure = 3.5

At constant pressure, heat is given as

$Q = n c_{p} R \Delta T\\Q = (3.5) (0.37) (8.314) (15)\\Q = 161.5 J$

6 0

2 years ago

· A car moves at 12 m/s and coasts up a hill with a

Zigmanuir [339]

Explanation:

Given that,

Initial speed of a car, u = 12 m/s

Aceleration, a = -1.6 m/s²

(a) The displacement of car after 6 seconds is :

$d=ut+\dfrac{1}{2}at^2\\\\d=12\times 6+\dfrac{1}{2}\times (-1.6)\times 6^2\\\\d=43.2\ m$

(b) The displacement of the car after 9 seconds is :

$d=ut+\dfrac{1}{2}at^2\\\\d=12\times 9+\dfrac{1}{2}\times (-1.6)\times 9^2\\\\d=43.2\ m$

Hence, this is the required solution.

7 0

3 years ago