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skelet666 [1.2K]

2 years ago

15

Which use of a simple machine changes the direction of a force? A. using a ramp to load boxes onto a truck B. using an axe to sp

lit wood C. removing a bottle cap with a bottle opener D. lifting a rock with a crowbar

1 answer:

kiruha [24]2 years ago

8 0

A. <span> using a ramp to load boxes onto a truck </span>

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C2H4O2 is what ??????

alukav5142 [94]

Did you try looking it up ?

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2 years ago

Select the appropriate shape for the given volume form

vredina [299]

Fist one is a cylinder
the second, i believe is a sphere
the third is a rectangular prism
and the last is the same as the first, a cylinder

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2 years ago

Six baseball throws are shown below. In each case the baseball is thrown at the same initial speed and from the same height h ab

Assoli18 [71]

Answer:

EXplained

Explanation:

from conservation of energy

change in potential energy = gain in kinetic energy

so as all he balls are throws from the same height thus the change in potential energy is the same for all the balls thus the gain in kinetic energy is the same for all the balls and as they have the same initial velocity thus the final velocity is the same for all the balls.

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3 years ago

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James rode his bike 0.65 hours and traveled 8.45 km. What was his speed?

maria [59]

Speed = (distance covered) / (time to cover the distance)

= ( 8.45 km) / (0.65 hr)

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= 13 km/hr

5 0

3 years ago

A mass free to vibrate on a level, frictionless surface at the end of a horizontal spring is pulled 35 cm from its equilibrium p

saul85 [17]

Answer:

0.67 s

Explanation:

This is a simple harmonic motion (SHM).

The displacement, $x$ , of an SHM is given by

$x = A\cos(\omega t)$

A is the amplitude and $\omega$ is the angular frequency.

We could use a sine function, in which case we will include a phase angle, to indicate that the oscillation began from a non-equilibrium point. We are using the cosine function for this particular case because the oscillation began from an extreme end, which is one-quarter of a single oscillation, when measured from the equilibrium point. One-quarter of an oscillation corresponds to a phase angle of 90° or $\frac{\pi}{4}$ radian.

From trigonometry, $\sin A =\cos B$ if A and B are complementary.

At $t = 0$ , $x = 3.5$

$3.5 = A\cos(\omega \times0)$

$A =3.5$

So

$x = 3.5\cos(\omega t)$

At $t = 0.12$ , $x = 1.5$

$1.5 = 3.5\cos(0.12\omega)$

$\cos(0.12\omega)=\dfrac{1.5}{3.5}=0.4286$

$0.12\omega =\cos^{-1}0.4286$

$0.12\omega = 1.13$

$\omega = 9.4$

The period, $T$ , is related to $\omega$ by

$T = \dfrac{2\pi}{\omega} = \dfrac{2\times3.14}{9.4}=0.67$

5 0

2 years ago