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3 years ago

What is the relationship between the speed of impact and the dropping height?

Physics

1 answer:

SVETLANKA909090 [29]3 years ago

8 0

The difference is that speed means how fast u go and height is how tall you are..

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4. A spring with a mass of 400.0 g is set into simple harmonic motion. The graph of the force of the spring vs. displacement is

Fynjy0 [20]

4 percent i think ok because maybe i think is e

5 0

4 years ago

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A river flows at 2m/s the velociy of ferry relative to the shore is 4m/s

Alex_Xolod [135]

Answer:

Explanation:

.....

7 0

3 years ago

Hunter aims directly at a target (on the same level) 75.0 m away. (a) If the bullet leaves the gun at a speed of 180 m/sby how m

tankabanditka [31]

Answer:

Part a)

$\delta y = 0.85 m$

Part b)

$\theta = 0.65 degree$

Explanation:

Part a)

As we know that the target is at distance 75 m from the hunter position

so here we will have

$x = v_x t$

here we know that

$v_x = 180 m/s$

so we have

$75.0 = 180 (t)$

$t = 0.42 s$

now in the same time bullet will go vertically downwards by distance

$\delta y = \frac{1}{2}gt^2$

$\delta y = \frac{1}{2}(9.81)(0.42^2)$

$\delta y = 0.85 m$

Part b)

In order to hit the target at same level we need to shot at such angle that the range will be 75 m

so here we have

$R = \frac{v^2 sin(2\theta)}{g}$

$75 = \frac{180^2 sin(2\theta)}{9.81}$

$\theta = 0.65 degree$

4 0

4 years ago

A bike is traveling to the left with the speed of 27 m/s when the rider slams on the brakes. The bike skids for 41.5 m with the

Darina [25.2K]

The acceleration of the bike is equal to $8.78\frac{m}{s^{2}}$ opposing the movement (to the right).

Why?

Since after the rider slams the brakes, the bike came to a stop, we know that the acceleration is opposite to the velocity/movement.

We can use the following equation to calculate the acceleration:

$v_{f}^{2}=v_{o}^{2}+2*a*d$

So, substituting, we have: (let's consider negative to the left and positive to the right)

$0=(-27\frac{m}{s} )^{2}+2*a*41.5m\\\\0=729\frac{m^{2} }{s^{2} }+83*a\\\\a=\frac{-729\frac{m^{2}}{s^{2}}}{83m}=-8.78\frac{m}{s^{2}}$

The negative sign means that the acceleration's direction is opposite to the movement.

Hence, we have that the acceleration is equal to $8.78\frac{m}{s^{2}}$ opposing the movement (to the right).

Have a nice day!

5 0

4 years ago

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You are working for a manufacturing company. Your supervisor has an idea for controlling the position of a small bead by using e

Mashutka [201]

You are working for a manufacturing company, which is mathematically given as

$m=3\sqrt{2}$
$m=\frac{15\sqrt{5}}{16}$
x=0.747a
$m/n=\frac{(x^2+a^2)3/2}{x^3}$

<h3>What is the value of m that will place the movable bead in equilibrium at x-a a ....?</h3>

Generally, the equation for the force of equilibrium is mathematically given as

F=2fcos\theta

Therefore

$K(npq^2/a^2)=2\frac{kmpq^2}{a\sqrt{2}^2}0.5\\\\np=mP/ \sqrt{2}\\\\where n=3$

$m=3\sqrt{2}$

By force equilibrium

$K(npq^2/(2a^2))=2*\frac{kmpq^2}{a\sqrt{5a}^2}* \frac{2a}{\sart{5a}}$

Therefore

$n/4=2/5*m*2/\sqrt{5}\\\\m= \frac{5\sqrt{5}}{16}\\\\\\\\$

$m=\frac{15\sqrt{5}}{16}$

$K(npq^2/x^2)=2\frac{kmpq^2}{a\sqrt{x^2+a^2}^2}0.5*x/\sqrt{x^2+a^2}$

x^2+a^2=(14/3)^{2/3}x^2

x=a/1.338

x=0.747a

By force equilibrium

$K(npq^2/x^2)=2\frac{kmpq^2n}{\sqrt{x^2+a^{3/2}}^2}\\\\n/x^2=\frac{2mx}{(x^2+a^2)^{3/2}}$

$m/n=\frac{(x^2+a^2)3/2}{x^3}$