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4 years ago

What two forces act on falling bodies?

Physics

1 answer:

Alex Ar [27]4 years ago

7 0

The two forces are: Air resistance and gravity.

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Calculate the instantaneous speed of<br> an apple 8 seconds after being dropped<br> from rest.

Lynna [10]

Vfinal = Vinitial + at
Vinitial = 0
a = 9.8 m/s^2
t = 8s
Vfinal = 0 + 9.8 * 8
ANSWER: 78.4 m/s

8 0

3 years ago

What information does the atomic mass of an element provide?

andriy [413]

C
The atomic mass is the sum of the number of neutrons and protons.

6 0

3 years ago

How much work does the electric field do in moving a proton from a point with a potential of +145 v to a point where it is -55 v

Aleks04 [339]

The work done by the electric field in moving a charge is the negative of the potential energy difference between the two locations, which is the product between the magnitude of the charge q and the potential difference $\Delta V$ :
$W=-\Delta U=-q\Delta V=-q (V_f -V_i)$
The proton charge is $q=1.6 \cdot 10^{-19}C$ , and the two locations have potential of $V_i = +145 V$ and $V_f=-55 V$ , therefore the work is
$W=-(1.6 \cdot 10^{-19}C)(-55 V-(+145 V))=3.2 \cdot 10^{-17}J$

5 0

3 years ago

What would be your estimate of the age of the universe if you measured a value for Hubble's constant of H0 = 30 km/s/Mly ? You c

Maksim231197 [3]

Answer:

The age of the universe would be 9.9 billion years

Explanation:

We can calculate an estimate for the age of the Universe from Hubble's Law. Let's suppose the distance between two galaxies is D and the apparent velocity with which they are separating from each other is v. At some point, the galaxies were touching, and we can consider that time the moment of the Big Bang.

Thus, the time it has taken for the galaxies to reach their current separations is:

$\displaystyle{t=D/v}$

and from Hubble's Law:

$v =H_0D$

Therefore:

$\displaystyle{t=D/v=D/(H_0\times D)=1/H_0}$

With the given value for the Hubble's constant we have:

$H_0=(30\ km/s/Mly) \times (1 Mly/ 9.461 \times 10^{18} km) = 3.17\times 10^{-18}\ 1/s$

and thus,

$t=1/H_0 = 1/(3.17\times 10^{-18} 1/s) = 0.315 \times 10^{18}\ s \approx 9988584474.8858\ years \approx 9.9\ billion\ years$

6 0

3 years ago

3 a A motorcyclist starts from rest and reaches

andrew-mc [135]

The acceleration of the body is 2 m/s^2 while the deceleration is - 1.2 m/s^2.

<h3>What is the acceleration?</h3>

Let us recall that the acceleration is the change in the speed of a body with time. We have been told that the body accelerates for 3s and then decelerates to 2s. This implies that the total time that the object spent in motion is 5 s.

Thus;

v = u + at

v = final velocity

u = initial velocity

a = acceleration

t = time taken

v - u/t = a

a = 6 - 0/3

= 2 m/s^2

Again;

v - u/t = a

a = 0 - 6/5

a = - 1.2m/s^2

Learn more about acceleration:brainly.com/question/12550364

#SPJ1

<h3 />

7 0

2 years ago