The pull of what is weaker when 2 objects are farther apart

2) Two railway tracks are parallel to west east direction. Along one track, train A moves with a speed of 45 m/s from

Natalija [7]

Answer:

105 m/s

Explanation:

Given that the speed of train A, $V_A$ = 45 m/s from west to east.

Speed of train B, $V_B$ = 60 m/s from east to west.

Train B is moving in the opposite direction with respect to the speed of train A. Assuming that the speed from east to west direction is positive.

So, the speed of train A from east to west= - 45 m/s

The speed of train B w.r.t train A $= V_B - V_A=60-(-45)=60+45=105$ m/s

Hence, the speed of train B w.r.t train A is 105 m/s from east to west.

5 0

2 years ago

A kangaroo jumps to a vertical height of 2.8 m. How long was it in the air before returning to earth

BaLLatris [955]

The answer would be 2.8m height on earth takes
2.8=1/2*9.8*t^2 => <span>s = ut +1/2at^2 </span>

8 0

2 years ago

Look at the circuit diagram. Which of these components is part of the circuit?

Roman55 [17]

Answer:

b ac power source

Explanation:

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7 0

2 years ago

If an object force of 50 N is used to move an object a distance of 20 m, what distance must the object be moved if the input for

steposvetlana [31]

Answer:

$\ d_{out} = 100 \ m.$

Explanation:

Given data:

$F_{in} = 50 \ \rm N$

$F_{out} = 10 \ \rm N$

$d_{in} = 20 \ m$

Let the distance traveled by the object in the second case be $d_{out}.$

In the given problem, work done by the forces are same in both the cases.

Thus,

$W_{in} = W_{out}$

$F_{in}.d_{in} = F_{out}.d_{out}$

$\Rightarrow \ d_{out} = \frac{F_{in}.d_{in}}{F_{out}}$

$\ d_{out} = \frac{50 \times 20}{10}$

$\ d_{out} = 100 \ m.$

5 0

2 years ago

During takeoff, an airplane climbs with a speed of 195 m/s at an angle of 15° above the horizontal. The speed and angle constitu

matrenka [14]

Answer:

The horizontal component of the velocity is 188 m/s

The vertical component of the velocity is 50 m/s.

Explanation:

Hi there!

Please, see the figure for a graphic description of the problem. Notice that the x-component of the vector velocity (vx), the y-component (vy) and the vector velocity form a right triangle. Then, we can use trigonometry to obtain the magnitude of vx and vy:

We can find vx using the following trigonometric rule of a right triangle:

cos α = adjacent / hypotenuse

cos 15° = vx / 195 m/s

195 m/s · cos 15° = vx

vx = 188 m/s

The horizontal component of the velocity is 188 m/s

To calculate the y-component we will use the following trigonometric rule:

sin α = opposite / hypotenuse

sin 15° = vy / 195 m/s

195 m/s · sin 15° = vy

vy = 50 m/s

The vertical component of the velocity is 50 m/s.

4 0

2 years ago