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ikadub [295]
3 years ago
9

When you calculate the SLOPE of a line segment, what does the SLOPE represent? (Choose all that apply) the Distance traveled the

Displacement the Velocity the Acceleration None of the above
Physics
1 answer:
dolphi86 [110]3 years ago
5 0

Answer:

Please find the answer in the explanation

Explanation:

When you calculate the SLOPE of a line segment, what does the SLOPE represent? (Choose all that apply) the Distance traveled the Displacement the Velocity the Acceleration None of the above

The slope of any time graph can not give you distance or displacement except for position - time graph.

When you plot either distance or displacement against time, that is, distance time graph or displacement time graph, you can get speed or velocity as the slope of the line segment.

You can only acceleration as a slope in a line of best fit if velocity is plotted against time. That is, in a velocity time graph.

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Mercury has a radial acceleration of 3.96 × 10−2 m/s2 and its orbital period is T = 88 days. What is the radius of Mercury’s orb
Maslowich

Answer: 58,045,522,878.8 meters

Explanation:

Ok, the data we have is

Period = T = 88 days

Radial acceleration = ar = 3.96x10^-2 m/s^2

And we know that the equation for the radial acceleration is:

ar = v^2/r = r*w^2

Where v is the velocity. r is the radius and w is the angular velocity.

And we know that:

w = 2*pi*f

where f is the frequency, and:

T = 1/f.

Then we can write:

w = 2*pi/T

and our equation becomes:

ar = r*(2*pi/T)^2

Now we solve this for r.

First we need to use the same units in both equations, so we want to write T in seconds.

T = 88 days,

A day has 24 hours, and one hour has 3600 seconds:

T = 88*24*3600 s =7,603,200s

Then:

3.96x10^-2 m/s^2 = r*(2*3.14/7,603,200s)^2

r = (3.96x10^-2 m/s^2) /(2*3.14/7,603,200s)^2 = 58,045,522,878.8 meters

5 0
3 years ago
An electromagnet is a solenoid with a piece of ferromagnetic material within it.
Vlada [557]
ANSWER:
The answer will be OT
3 0
3 years ago
Mercury’s natural state is where the atoms are close to each other but are still free to pass by each other. In which state(s) c
Georgia [21]

Answer:

Mercury's natural state is where the atoms are close to each other but are still free to pass by each other. In which state(s) could mercury naturally exist?

Liquid is the answer

Explanation:

4 0
2 years ago
what is the acceleration of each body of mass 5kg rests on a frictionless table and is connected to a cable that passes over a p
Oksanka [162]

Answer:

6.53 m/s²

Explanation:

Let m₁ = 5 kg and m₂ = 10 kg. The figure is attached and free body diagrams of the objects are also attached.

Both objects (m₁ and m₂) have the same magnitude of acceleration(a). Let g be the acceleration due to gravity = 9.8 m/s². Hence:

T = m₁a               (1)

m₂g - T = m₂a    (2)

substituting T = m₁a in equation 2:

m₂g - m₁a = m₂a

m₂a + m₁a = m₂g

a(m₁ + m₂) = m₂g

a = m₂g / (m₁ + m₂)

a = (10 kg * 9.8 m/s²) / (10 kg + 5 kg) = 6.53 m/s²

Both objects have an acceleration of 6.53 m/s²

8 0
3 years ago
A plank of length L=2.200 m and mass M=4.00 kg is suspended horizontally by a thin cable at one end and to a pivot on a wall at
baherus [9]

Hello!

Let's begin by doing a summation of torques, placing the pivot point at the attachment point of the rod to the wall.

\Sigma \tau = 0

We have two torques acting on the rod:
- Force of gravity at the center of mass (d = 0.700 m)

- VERTICAL component of the tension at a distance of 'L' (L = 2.200 m)

Both of these act in opposite directions. Let's use the equation for torque:
\tau = r \times F

Doing the summation using their respective lever arms:

0 = L Tsin\theta  - dF_g

dF_g = LTsin\theta

Our unknown is 'theta' - the angle the string forms with the rod. Let's use right triangle trig to solve:

tan\theta = \frac{H}{L}\\\\tan^{-1}(\frac{H}{L}) = \theta\\\\tan^{-1}(\frac{1.70}{2.200}) = 37.69^o

Now, let's solve for 'T'.

T = \frac{dMg}{Lsin\theta}

Plugging in the values:
T = \frac{(0.700)(4.00)(9.8)}{(2.200)sin(37.69)} = \boxed{20.399 N}

3 0
1 year ago
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