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sveta [45]
3 years ago
11

which of this is not example of simple machine? A) a crowbar,B)a knife,C)a bicycle,D)a wheelchair ramp

Physics
1 answer:
DaniilM [7]3 years ago
3 0
B. Knife HOpe this helps :3
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Ulleksa [173]
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The average distance between earth and moon would be - 384,000 kms. 

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3 years ago
A person kicks a ball, giving it an initial velocity of 20.0 m/s up a wooden ramp. When the ball reaches the top, it becomes air
Alex Ar [27]

Answer:

(a) Height is 4.47 m

(b) Height is 4.37 m

Solution:

As per the question:

Initial velocity of teh ball, v_{o} = 20.0 m/s

Angle made by the ramp, \theta = 22.0^{\circ}

Distance traveled by the ball on the ramp, d = 5.00 m

Now,

(a) At any point on the projectile before attaining maximum height, the velocity can be given by the eqn-3 of motion:

v^{2} = v_{o}^{2} - 2gH

where

H = dsin22^{\circ} = 5sin22^{\circ}

g = 9.8 m/s^{2}

v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ}

v = \sqrt{400 - 19.6\times 5sin22^{\circ}} = 19.06 m/s

Now, maximum height attained is given by:

h = \frac{(vsin\theta)^{2}}{2g}

h = \frac{(19sin(22^{\circ}))^{2}}{2\times 9.8} = 2.60 m

Height from the ground = 5sin22^{circ} + 2.86 = 1.87 + 2.60 = 4.47m

(b) now, considering the coefficient of friction bhetween ramp and the ball, \mu = 0.150:

velocity can be given by the eqn-3 of motion:

v^{2} = v_{o}^{2} - 2gH - \mu gd

v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ} - 0.150\times 9.8\times 5

v = \sqrt{400 - 19.6\times 5sin22^{\circ} - 0.150\times 9.8\times 5} = 18.7 m/s

Now, maximum height attained is given by:

h = \frac{(vsin\theta)^{2}}{2g}

h = \frac{(18.7sin(22^{\circ}))^{2}}{2\times 9.8} = 2.50 m

Height from the ground = 5sin22^{circ} + 2.86 = 1.87 + 2.50 = 4.37 m

6 0
3 years ago
Water drips from the nozzle of a shower onto the floor 190 cm below. The drops fall at regular (equal) intervals of time, the fi
VARVARA [1.3K]

Answer:

Second drop: 1.04 m

First drop: 1.66 m

Explanation:

Assuming the droplets are not affected by aerodynamic drag.

They are in free fall, affected only by gravity.

I set a frame of reference with the origin at the nozzle and the positive X axis pointing down.

We can use the equation for position under constant acceleration.

X(t) = x0 + v0 * t + 1/2 * a *t^2

x0 = 0

a = 9.81 m/s^2

v0 = 0

Then:

X(t) = 4.9 * t^2

The drop will hit the floor when X(t) = 1.9

1.9 = 4.9 * t^2

t^2 = 1.9 / 4.9

t = \sqrt{0.388} = 0.62 s

That is the moment when the 4th drop begins falling.

Assuming they fall at constant interval,

Δt = 0.62 / 3 = 0.2 s (approximately)

The second drop will be at:

X2(0.62) = 4.9 * (0.62 - 1*0.2)^2 = 0.86 m

And the third at:

X3(0.62) = 4.9 * (0.62 - 2*0.2)^2 = 0.24 m

The positions are:

1.9 - 0.86 = 1.04 m

1.9 - 0.24 = 1.66 m

above the floor

8 0
3 years ago
What tool is used to determine the mass of an object?
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- if this answer is what you were looking for, don’t forget to heart and maybe if you really liked it brainliest :)
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Landon attends an early childhood program that is located at a community center which also runs an adult care program.
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Good for Landon. What’s the question?
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