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alukav5142 [94]
3 years ago
13

Which of these best explains the ability of small insects to walk on the surface of still water?

Physics
1 answer:
Furkat [3]3 years ago
7 0
Well, the surface of still water has surface tension. If there isn't enough mass or weight to break the surface tension, the object will float.
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A rocket carrying a satellite is accelerating straight up from the earth’s surface. At 1.15 s after liftoff, the rocket clears t
yarga [219]

Answer:

197.263157895 m/s

169.491525424 m/s

Explanation:

x Denotes position

t Denotes time

Average velocity is given by

v_a=\dfrac{x_2-x_1}{t_2}\\\Rightarrow v_a=\dfrac{1000-63}{4.75}\\\Rightarrow v_a=197.263157895\ m/s

The average velocity is 197.263157895 m/s

v_a=\dfrac{x_2-x_1}{t_2}\\\Rightarrow v_a=\dfrac{1000-0}{5.9}\\\Rightarrow v_a=169.491525424\ m/s

The average velocity is 169.491525424 m/s

8 0
3 years ago
A force AB of length 100m and weight 600N has its centre of gravity 4.0 m from the end, and lies on horizontal ground calculate
Stells [14]

Answer:

F = 24 N

Explanation:

In this exercise we have a bar l = 100 m with a center of gravity x = 4 m, which force is needed to lift it from the other end

Let's use the rotational equilibrium relationship, where we consider the counterclockwise rotations as positive and fix the reference system at the point closest to the center of gravity

           ∑ τ = 0

           F l -x W = 0

           F = \frac{x}{l} \  W

let's calculate

          F = \frac{4}{100}4/100 600

          F = 24 N

8 0
3 years ago
A laser with a power of 1.0 mW has a beam radius of 1.0 mm. What is the peak value of the electric field in that beam
Fynjy0 [20]

Answer:

The peak value of the electric field is 489.64 V/m

Explanation:

Given;

power of the laser, P = 1.0 mW = 1 x 10⁻³ W

Radius of the beam, R = 1.0 mm = 1 x 10⁻³ m

Area of the beam = πr² = π(1 x 10⁻³ )² = 3.142 x 10⁻⁶ m²

The average intensity of the light = P / A

The average intensity of the light = ( 1 x 10⁻³) / (3.142 x 10⁻⁶)

The average intensity of the light = 318.27 W/m²

The peak value of the electric field is given by;

E_o = \sqrt{\frac{2I_{avg}}{c\epsilon_o}}\\\\E_o = \sqrt{\frac{2(318.27)}{(3*10^8)(8.85*10^{-12})}}\\\\E_o = 489.64 \ V/m

Therefore, the peak value of the electric field is 489.64 V/m.

4 0
3 years ago
Suppose that a spring with a larger spring constant was used in this same apparatus. If a given mass were rotated at the same ra
malfutka [58]

Answer:

The new period of rotation using the new spring would be less than the period of rotation using the original spring

Explanation:

Generally the  period of rotation of the mass is mathematically represented as

       T = 2 \pi \sqrt{\frac{I}{k} }

Here I is the moment of inertia of the mass about the rotation axis and  k is the spring constant

Now looking at the equation we can tell that T is inversely proportional to the square root of the spring constant which means that for a larger spring constant the time period would be lesser

6 0
3 years ago
HELP!! ALL MY POINTS WILL BE GIVEN
Marta_Voda [28]

Answer:

-6 m/s^2

Explanation:

30 - 90 = -60

-60 / 10 = -6

If acceleration was constant, it will be -6 m/s^2

5 0
2 years ago
Read 2 more answers
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