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3 years ago

The variable plotted on the horizontal or x-axis is called the

Physics

1 answer:

IRISSAK [1]3 years ago

7 0

Answer:

Independent Variable

Explanation:

The variables that are plotted on X-axis, the horizontal line of the graph are termed as independent variables. The variables plotted on Y-axis which is vertical side of the graph are dependent variables.

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When a mass increases, its gravitational force_____.

Reil [10]

The correct answer is A

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3 years ago

Which of these is an example of a mechanical wave? sunlight light from a flashlight microwaves used to heat food police siren

harina [27]

Police siren is the correct answer. Hope this helps.

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3 years ago

Read 2 more answers

A toy train rolls around a horizontal 1.0-m-diameter track.The coefficient of rolling friction is 0.10.a) What is the magnitude

valina [46]

Answer:

1.962 rad / s², 1.6 s

Explanation:

Radius of the part = 1.0m / 2 = 0.5 m

angular speed = 30 rpm = 30 rpm × (2πrad / rev) × 1 minutes / 60 seconds = 3.142 rads⁻¹

μk = frictional force / normal ( mg )

normal is the force acting upward against the force of gravity

frictional force = - μk mg

since the body came to rest then

Fnet + Ff = 0

Fnet = - Ff

Fnet = ma

ma = - μk mg

a = - μkg where g = 9.81

a = - 0.1 × 9.81 = 0.981 m/s²

magnitude of angular acceleration = tangential acceleration / radius = 0.981 / 0.5 = 1.962 rad / s²

b) time for the train to come to rest = angular velocity / angular angular acceleration = 3.142/ 1.962 = 1.6 s

The equation earlier derived answer this question

Fnet + Ff = 0 since the body came to a rest

Fnet = - Ff and Ff = - μk mg, Fnet = ma

ma = - μk mg

m cancel m on both side

a = - μkg since it magnitude

a = μkg

5 0

4 years ago

A bird can fly 25 km/h. How long does it take to fly 3.5km?

monitta

The equation of motion that we can use in this case is:

t = d / v

where t is time, d is distance, v is velocity

Therefore calculating for t:

t = 3.5 km / (25 km / h)

<span>t = 0.14 h = 8.4 minutes</span>

6 0

3 years ago

What is the maximum distance we can shoot a dart, provided our toy dart gun gives a maximum initial velocity of 5.84 m/s?

poizon [28]

1) The maximum distance can be reached when the dart is shot with an angle of $\theta=45^{\circ}$ above the horizontal (see demonstration of this fact at the end).

2) The motion of the dart is an uniform motion on the x-axis (horizontal direction) with constant velocity $v_i = 5.84~m/s$ and it is an uniformly accelerated motion on the y-axis (vertical direction), with the gravitational acceleration $g=9.81~m/s^2$ acting downwards. So we can write the laws of motion on both directions:
$S_x(t) =( v_i \cos{\theta}) t$
$S_y(t) = (v_i \sin{\theta})t - \frac{1}{2} gt^2$
where the negative sign means that g points downwards.

3) First of all we can find the time at which the dart reaches the ground. This can be found by requiring $S_y(t)=0$ :
$(v_i \sin{\theta})t - \frac{1}{2} gt^2=0$
From this we find two solutions: $t=0~s$ , corresponding to the beginning of the motion (so we are not interested in this one), and
$t= \frac{2 v_i \sin{\theta}}{g} =0.84~s$

4) Now that we now when the dart reaches the ground, we can use this information to find the distance covered on the x-axis, by using $t=0.84~s$ inside the equation of $S_x(t)$ written at point 2:
$S_x(0.84~s)= 5.84~m/s \cdot \cos{45^{\circ}} \cdot 0.84~s }=3.47~m$

--------------------------
DEMONSTRATION OF POINT 1:

Calling v the initial velocity of the dart, and using a coordinate system where the x-axis coincides with the horizontal direction and the y-axis with the vertical direction, we can write the law of motion on both directions:
$S_x(t) = v cos{\theta} t$
$S_y(t) = v sin{\theta} t + \frac{1}{2} g t^2$
where t is the time, $\theta$ is the initial angle of the dart, and $g= -9.81 m/s^2$ is the gravitational acceleration.
The maximum horizontal distance can be found by requiring that $S_y=0$ . This condition occurs twice: when the motion starts ( $t=0$ ) and when the dart falls to the ground (let’s call $t_s$ the time at which this happens). Therefore we can find $t_s$ by requiring $S_y(t_s)=0$ , i.e.:
$v sin{\theta} t_s + \frac{1}{2} g t_s^2 =0$
which has two solutions: $t_s=0$ (beginning of the motion) and $t_s = - \frac{2 v sin{\theta}}{g}$ .So we can find the maximum horizontal distance covered by the dart by substituting this $t_s$ into the law of motion for $S_x(t)$ :
$S_x(t_s) = v cos\theta t_s = -\frac{2 v^2 cos\theta sin\theta}{g}$
Since
$sin2\theta = 2 cos\theta sin\theta$ ,
we can write
$S_x(t_s) = - \frac{v^2 sin2\theta}{g}$
Since g is negative, the maximum of this function occurs for $sin2\theta=1$ , and this happens when $\theta=45^{\circ}$ .

4 0

4 years ago