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2 years ago

Please help! It’s urgent

Physics

2 answers:

Elza [17]2 years ago

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kobusy [5.1K]2 years ago

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

Furkat [3]

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.

7 0

2 years ago

A volleyball player bumps a ball across a net with the velocity and angle shown below. What is the maximum height of the ball?

Marrrta [24]

Answer:

D. 12.4 m

Explanation:

Given that,

The initial velocity of the ball, u = 18 m/s

The angle at which the ball is projected, θ = 60°

The maximum height of the ball is given by the formula

h = u² sin²θ/2g m

Where,

g - acceleration due to gravity. (9.8 m/s)

Substituting the values in the above equation

h = 18² · sin²60 / 2 x 9.8

= 18² x 0.75 / 2 x 9.8

= 12.4 m

Hence, the maximum height of the ball attained, h = 12.4 m

6 0

2 years ago

If you were stuck in the center of the bridge when it was swaying, where would be the safest place to walk back to land?

Ludmilka [50]

Straight
You already have to momentum of walking forward, and going back and forth are the same distance. If you go back then you would have to stop, turn and walk, but if you go forward you just have to walk.

4 0

2 years ago

Calculate the pressure exerted on the heel of a boy’s foot if the boy weighs 80 N and he lands on one heel,which has an area of

cricket20 [7]

Pressure at a given surface is given as ratio of normal force and area

so here force due to heel of the shoes is given as 80 N

and the area of the heel is given as 16 cm^2

so we can say

$P = \frac{F}{A}$

here we have

F = 80 N

$A = 16 cm^2 = 16 * 10^{-4} m^2$

$P = \frac{80}{16 * 10^{-4}}$

$P = 5 * 10^4 N/m^2$

so pressure at the surface due to its heel will be 5 * 10^4 N/m^2

3 0

2 years ago

Calculate the force which will produce an extension of 0.30mm in a steel wire with a length of 4.0m and a cross section area of

Anna [14]

Given data:

* The extension of the steel wire is 0.3 mm.

* The length of the wire is 4 m.

* The area of cross section of wire is,

$A=2\times10^{-6}m^2$

* The young modulus of the steel is,

$Y=2.1\times10^{11}\text{ Pa}$

Solution:

The young modulus of the steel in terms of the force and extension is,

$Y=\frac{F\times l}{A\times dl}$

where F is the force acting on the steel wire,, l is the original length of the wire, dl is the extension of the wire, and A is the area,

Substituting the known values,

$\begin{gathered} 2.1\times10^{11}=\frac{F\times4}{2\times10^{-6}\times0.3\times10^{-3}} \\ F=0.315\times10^2\text{ N} \\ F=31.5\text{ N} \end{gathered}$

Thus, the force which produce the extension of 0.3 mm of the steel wire is 31.5 N.

7 0

4 months ago