A car starts from rest and accelerates at 2.0 m/s2 for 10 seconds. What is the car's displacement at the end of 10 seconds?

If a freely falling rock were equipped with a speedometer, by how much would itd speed readings increase with each second if it

Len [333]

" 20 m/s² " means that if gravity is the ONLY force on the object
(the object is in 'free fall'), then its speed increases by 20 m/s
every second.

That's the answer to your question. Now, let me ask you
another one:

How does a speedometer tied to a falling rock work ?
How can it measure the rock's speed ?

Maybe one way would be to have a little tiny propeller on
the front of the speedometer, and it could measure how fast
the propeller is spinning as the rock falls through the air ?
Great idea. But we already said the rock is in free-fall,
so there's no air resistance, we can't have any air, and
there's nothing to spin the propeller.

How would you do it ? How can you measure the rock's speed ?

3 0

3 years ago

A uniform electric field of magnitude 6.8 × 10 5 N/C points in the positive x direction. (a) Find the electric potential differe

Naya [18.7K]

Answer:

i)4080000V

ii)4080000V

iii)5766400v

Explanation:

Hello!

To solve this problem we will use the following steps

1. We will use the equation that defines the distance between two points in order to find the distance from the origin to each of the points that the problem asks for.

$d=\sqrt{x^2+y^2}$

x= horizonalt component

y=vertical component

d=distance

i)

x=0

y=6

$d=\sqrt{0^2+6^2}=6m$

ii)

x=6

y=0

$d=\sqrt{6^2+0^2}=6m$

iii).

x=6

y=6

$d=\sqrt{6^2+6^2}=8.48m$

2.we calculate the voltage at each point using the equation that relates the distance the voltage and the electric field

V=Ed

where

V=EPD= the electric potential difference

E=electric field magnitude=6.8 × 10^5 N/C.

d=distance

i)

V=Ed=(6.8 × 105 N/C.)(6)=4080000V

ii)

i)

V=Ed=(6.8 × 105 N/C.)(6)=4080000V

iii)

i)

V=Ed=(6.8 × 105 N/C.)(8.48m)=5766400v

8 0

3 years ago

Can someone pls help me. WILL MARK U BRAINLIEST!!!

Dmitry [639]

You can look up the answer key to that , that’s what I always do and it always pops up

8 0

3 years ago

The half-life of a certain element is 100 days. How many half-lives will it be before only one-eighth of this element remains?

Natalka [10]

Answer:

3

Explanation:

The half-life is the time it takes for the amount of radioactive isotope to halve. Therefore, we have:

- After 1 half-life, only 1/2 of the element will be left

- After 2 half-lives, only 1/4 of the element will be left

- After 3 half-lives, only 1/8 of the element will be left

So, it will take 3 half-lives for the element to become 1/8 of its original amount.

Mathematically, this can be also verified by using the equation

$\frac{N(t)}{N_0}=(\frac{1}{2})^\frac{t}{\tau_{1/2}}$

where

N(t) is the amount of the element left at time t

N0 is the initial amount of the element

$\tau_{1/2}$ is the half-life

Substituting $t=3\tau_{1/2}$ (3 half-lives), we find

$\frac{N(t)}{N_0}=(\frac{1}{2})^3=\frac{1}{8}$

7 0

3 years ago

-In the Bohr model, as it is known today, the electron is imagined to move in a circular orbit about a stationary proton. The fo

vampirchik [111]

Answer:

$4.3859007196\times 10^{-9}\ m$

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

v = Velocity of electron = $2.4\times 10^5\ m/s$

q = Charge of electron = $1.6\times 10^{-19}\ C$

m = Mass of electron = $9.11\times 10^{-31}\ kg$

r = Radius

The electrical and centripetal force will balance each other

$\dfrac{kq^2}{r^2}=\dfrac{mv^2}{r}\\\Rightarrow r=\dfrac{kq^2}{mv^2}\\\Rightarrow r=\dfrac{8.99\times 10^9\times (1.6\times 10^{-19})^2}{9.11\times 10^{-31}\times (2.4\times 10^5)^2}\\\Rightarrow r=4.3859007196\times 10^{-9}\ m$

The radius of the orbital is $4.3859007196\times 10^{-9}\ m$

4 0

3 years ago