Answer:
<u>Matter is a substance that has inertia and occupies physical space. According to modern physics, matter consists of various types of particles, each with mass and size.Matter can exist in several states, also called phases. The three most common states are known as solid, liquid and gas.</u><u>Matter is the Stuff Around You</u><u> </u><u>o</u><u>r</u><u> </u><u>Atoms and compounds are all made of very small parts of matter. Those atoms go on to build the things you see and touch every day. Matter is defined as anything that has mass and takes up space (it has volume).</u><u>Solid ice, water and steam are few examples of matter touched in everyday life. Subatomic particles are also considered as matter.</u>
Answer:
An equipotential surface is circular in the two-dimensional. Since the electric field lines are directed radially away from the charge, hence they are opposite to the equipotential lines. Therefore, the electric field is perpendicular to the equipotential surface.
Answer:
1.84 m
Explanation:
For the small lead ball to be balanced at the tip of the vertical circle just before it is released, the reaction force , N equal the weight of the lead ball W + the centripetal force, F. This normal reaction ,N also equals the tension T in the string.
So, T = mg + mrω² = ma where m = mass of small lead ball, g = acceleration due to gravity = 9.8 m/s², r = length of rope = 1.10 m and ω = angular speed of lead ball = 3 rev/s = 3 × 2π rad/s = 6π rad/s = 18.85 rad/s and a = acceleration of normal force. So,
a = g + rω²
= 9.8 m/s² + 1.10 m × (18.85 rad/s)²
= 9.8 m/s² + 390.85 m/s²
= 400.65 m/s²
Now, using v² = u² + 2a(h₂ - h₁) where u = initial velocity of ball = rω = 1.10 m × 18.85 rad/s = 20.74 m/s, v = final velocity of ball at maximum height = 0 m/s (since the ball is stationary at maximum height), a = acceleration of small lead ball = -400.65 m/s² (negative since it is in the downward direction of the tension), h₁ = initial position of lead ball above the ground = 1.3 m and h₂ = final position of lead ball above the ground = unknown.
v² = u² + 2a(h₂ - h₁)
So, v² - u² = 2a(h₂ - h₁)
h₂ - h₁ = (v² - u²)/2a
h₂ = h₁ + (v² - u²)/2a
substituting the values of the variables into the equation, we have
h₂ = 1.3 m + ((0 m/s)² - (20.74 m/s)²)/2(-400.65 m/s²)
h₂ = 1.3 m + [-430.15 (m/s)²]/-801.3 m/s²
h₂ = 1.3 m + 0.54 m
h₂ = 1.84 m
Answer:
The final position of the ship after the given time period is 42 km West of B.
Explanation:
Given;
average velocity of the ship, v = 35 km/h
time taken for the ship to reach point D, t = 1.2 hours
The position of the ship after the given time period is calculated as follows;
x = v x t
x = (35 km/h) x 1.2 h = 42 km
x = 42 km West of B.
Therefore, the final position of the ship after the given time period is 42 km West of B.
An arrow pointing from the bottom of the ramp to the top, I assume it would be friction.
