Use the motion map to answer the question, describe the position and velocity of the object based moo the motion map?

1. A 3.5 kg object experiences an acceleration of 0.5 m/s2. What net force does the object experience

Leviafan [203]

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 3.5 × 0.5 = 1.75

We have the final answer as

Hope this helps you

7 0

3 years ago

What is the ionosphere

ICE Princess25 [194]

A ionosphere is a layer of the earths atmosphere that is ionized by solar and cosmic radiation <span />

7 0

4 years ago

A 50g ball is released from rest 1.0 above the bottom of thetrack

ludmilkaskok [199]

Answer:

The maximum height of the ball is 2 m.

Explanation:

Given that,

Mass of ball = 50 g

Height = 1.0 m

Angle = 30°

The equation is

$y=\dfrac{1}{4}x^2$

We need to calculate the velocity

Using conservation of energy

$\Delta U_{i}+\Delta K_{i}=\Delta K_{f}+\Delta U_{f}$

Here, ball at rest so initial kinetic energy is zero and at the bottom the potential energy is zero

$\Delta U_{i}=\Delta K_{f}$

Put the value into the formula

$mgh=\dfrac{1}{2}mv^2$

Put the value into the formula

$50\times10^{-3}\times9.8\times1.0=\dfrac{1}{2}\times50\times10^{-3}\times v^2$

$v^2=\dfrac{2\times50\times10^{-3}\times9.8\times1.0}{50\times10^{-3}}$

$v=\sqrt{19.6}$

$v=4.42\ m/s$

We need to calculate the maximum height of the ball

Using again conservation of energy

$\dfrac{1}{2}mv^2=mgh$

Here, h = y highest point

Put the value into the formula

$\dfrac{1}{2}\times50\times10^{-3}\times(4.42)^2=50\times10^{-3}\times9.8\times h$

$y=\dfrac{0.5\times(4.42)^2}{9.8}$

$y=0.996\ m$

Put the value of y in the given equation

$y=\dfrac{1}{4}x^2$

$x^2=4\times0.996$

$x=\sqrt{4\times0.996}$

$x=1.99\ m\ \approx 2 m$

Hence, The maximum height of the ball is 2 m.

4 0

3 years ago

When are scalars used?

Shkiper50 [21]

When you have to wait some

7 0

3 years ago

The Sun's declination is 0° at the _________. A. summer and winter solstice B. summer and winter equinox C. vernal and autumnal

Olegator [25]

-- "Declination zero" means the object is in the sky at some point directly over the Earth's equator.

-- If it's the sun and it appears to be over the equator, then that tells us that the Earth's axis is not tilted toward or away from it.

-- That in turn tells us that the Earth is at one of the two equinoxes in its orbit, either the Spring one or the Autumn one. <em> (D)</em>

-- (The first days of Summer and Winter coincide with solstices, not equinoxes.)

5 0

3 years ago

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