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Bezzdna [24]
4 years ago
7

About how much time elapses between high tides on the same day?

Physics
1 answer:
Pani-rosa [81]4 years ago
6 0
There is about 12 hours
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A contestant in a winter games event pushes a rock across a frozen lake (frictionless) with a force of 25 N at 60° below the hor
Orlov [11]
Text me! My number is (856)449-5259
8 0
3 years ago
While in a car at 4.47 meters per second a passenger drops a ball from a height of 0.70 meters above the top of a bucket how far
viktelen [127]

Answer:

1.7 m

Explanation:

v_x = Velocity of ball in x direction = 4.47 m/s

u_y = Velocity of ball in y direction = 0

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

t = Time taken

s_y = Vertical displacement = 0.7 m

s_y=u_yt+\dfrac{1}{2}gt^2\\\Rightarrow 0.7=0+\dfrac{1}{2}\times 9.81t^2\\\Rightarrow t=\sqrt{\dfrac{0.7\times 2}{9.81}}\\\Rightarrow t=0.38\ \text{s}

Horizontal displacement is given by

s_x=v_xt\\\Rightarrow s_x=4.47\times 0.38\\\Rightarrow s_x=1.7\ \text{m}

The passenger should throw the ball 1.7 m in front of the bucket.

5 0
3 years ago
VERY URGENT GIVING BRAINLY​
lakkis [162]

Answer:

A

Explanation:

There is a mechanical advantage in this system....

    she only needed 100 N to lift 500 N

        1:5      she lifted it 2 meters, so if there WAS NO friction, she would have to pull in 5 x 2 = 10 meters

   but there IS friction so she must have pulled in MORE than 10 m  

8 0
2 years ago
What is the approximate weight of the air inside the tire in English Engineering Units (tire outside diameter = 49", rim diamete
Mademuasel [1]

Answer:

1.265 Pounds

Explanation:

Data provided:

Tire outside diameter = 49"

Rim diameter = 22"

Tire width = 19"

Now,

1" = 0.0254 m

thus,

Tire outside radius, r₁ = 49"/2 = 24.5" = 24.5 × 0.0254 = 0.6223 m

Rim radius, r₂   = 22" / 2 = 11" = 0.2794 m

Tire width, d = 19" = 0.4826 m

Now,

Volume of the tire = π ( r₁² - r₂² ) × d

on substituting the values, we get

Volume of air in the tire = π (  0.6223² - 0.2794² ) × 0.4826 = 0.46877 m³

Also,

Density of air = 1.225 kg/m³

thus,

weight of the air in the tire = Density of air × Volume air in the tire

or

weight of the air in the tire = 1.225 × 0.46877 = 0.5742 kg

also,

1 kg = 2.204 pounds

Hence,

0.5742 kg = 0.5742 × 2.204  = 1.265 Pounds

3 0
3 years ago
An electric field of 6.1x 10^5 N/C is directed along a charge 2.9 x 10^-19 C. What
Aloiza [94]

Answer:

Electric field is a function 1/r^2

Explanation:

4 0
3 years ago
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