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Sauron [17]
2 years ago
9

What is another word for a change in velocity/change in time

Physics
2 answers:
Leona [35]2 years ago
7 0
Changed is another word
inna [77]2 years ago
5 0

Answer:

acceleration

Explanation:

acceleration =velocity final-velocity initial /time

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A nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the
zalisa [80]

Answer:

a) -2.516 × 10⁻⁴ V

b) -1.33 × 10⁻³ V

Explanation:

The electric field inside the sphere can be expressed as:

E= \frac{kqr}{R^3}

The potential at a distance can be represented as:

V(r) - V(0) = -\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) = [\frac{qr^2}{8 \pi E_0R^3 }]₀

V(r) =   -[\frac{qr^2}{8 \pi E_0R^3 }]₀

Given that:

q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

 = 0.56 × 10⁻² m

R = 1.29 cm

  =  1.29 × 10⁻² m

E₀ = 8.85 × 10⁻¹² F/m

Substituting our values; we have:

V(r) = -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}

V(r) = -2.15  × 10⁻⁴ V

The difference between the radial distance  and center can be expressed as:

V(r) - V(0) = -\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) =  [\frac{qr^2}{8 \pi E_0R^3 }]^R

V(r) = -\frac{qR^2}{8 \pi E_0R^3 }

V(r) = -\frac{q}{8 \pi E_0R }

V(r) = -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}

V(r) = -0.00133

V(r) = - 1.33 × 10⁻³ V

8 0
2 years ago
When looking at liquids in a container, that have layered on one another;
Mashcka [7]

Answer: C. At the bottom

4 0
2 years ago
Whats the difference between an inner planet and an outer planet
Kryger [21]
The outer planets<span> are further away, larger and made up mostly of gas. The </span>inner planets<span> (in order of distance from the sun, closest to furthest) are Mercury, Venus, Earth and Mars. After an asteroid belt comes the </span>outer planets<span>, Jupiter, Saturn, Uranus and Neptune.</span>
8 0
2 years ago
Read 2 more answers
A student throws a rock horizontally from the edge of a cliff that is 20 m high. The rock has an initial speed on 10 m/s. If air
fiasKO [112]

The distance of the rock from the base of the cliff is C) 20 m

Explanation:

The motion of the rock in this problem is a projectile motion, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

We start by analyzing the vertical motion to find the time of flight of the rock (the time it takes to reach the ground). We can do it by using the suvat equation:

s=u_y t+\frac{1}{2}at^2

where, taking downward as positive direction,

s = 20 m is the vertical displacement of the rock

u_y=0 is the initial vertical velocity

t is the time of flight

a=g=9.8 m/s^2 is the acceleration of gravity

Solving for t,

t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(20)}{9.8}}=2.02 s

Now we can analzye the horizontal motion: the rock moves horizontally with a constant velocity of

v_x = 10 m/s

Therefore, the horizontal distance covered after a time t is

d=v_x t

and substituting t = 2.02 s, we find the final distance of the rock from the base of the cliff:

d=(10)(2.02)=20 m

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

6 0
2 years ago
An object is dropped from a bridge. A second object is thrown downward 1.48 s later. They both reach the water 48.1 m below at t
pashok25 [27]

To solve this problem we will apply the linear motion kinematic equations. With the data provided we will calculate the time of the first object to fall. Later we will get the time difference between the two. This difference will allow us to find the free fall distance. Through the distance we will find the initial velocity, that is,

x = v_0 t +\frac{1}{2}at^2

48.1 = 0*t + \frac{1}{2} (9.8)t^2

t = 3.13s

The second object is thrown downward at one second later and it meets the first object at the water is

t' = 3.13 -1.48

t' = 1.65s

The distance of the object will travel due to free fall acceleration is

x = v_0 t+\frac{1}{2} at^2

x = 0*(1.65) +\frac{1}{2}(9.8)(1.65)^2

x = 13.34m

The distance of the object will travel due to its initial velocity is

v_0 = \frac{d_0}{t}

d_0 = v_0 t

48.1-13.34 = v_0 (1.65)

v_0 = 21.06m/s

Therefore the initial speed of the second object is 21.06m/s

8 0
3 years ago
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