Ask question

Ask question

All categories

3 years ago

12

While traveling along a highway a driver slows from 34 m/s to 15 m/s in 12 seconds. What is the automobile’s acceleration? (Reme

mber that a negative value indicates a slowing down or deceleration.

1 answer:

Flauer [41]3 years ago

3 0

Answer:

a=-1.58 m/s^2

Explanation:

Vo=34m/s

Vf=15m/s

t=12s

a= dV/dt = (Vf-Vo) /t = (15-34)/12

= - 19/12 = - 1.58m/s2

You might be interested in

Graph are pictorical representations of

inn [45]

Answer:

Graphs are pictorial representations of relationships.

6 0

3 years ago

Read 2 more answers

Can anybody help me with this Physics question! (DUE TOMORROW) (WILL MARK BRAINLIST)

Ymorist [56]

Do you need help with all of them

3 0

3 years ago

Read 2 more answers

What happens when an object with a lower density is placed in a container with an

maksim [4K]

The only thing that definitely happens in every such case is:
The container becomes heavier.

5 0

3 years ago

What type of wave does not need matter to carry energy?

marishachu [46]

Non- mechanical wave does not need matter to carry energy.

e.g:- Light

3 0

3 years ago

Read 2 more answers

Jupiter's semimajor axis is 7.78×1011 m. The mass of the Sun is 1.99×1030 kg. (a) What is the period of Jupiter's orbit in secon

dedylja [7]

Explanation:

It is given that,

Semi major axis of the Jupiter, $a=7.78\times 10^{11}\ m$

Mass of the sun, $M=1.99\times 10^{30}\ kg$

(a) Let T is the period of Jupiter's orbit. It is given by :

$T^2\propto a^3$

$T^2=\dfrac{4\pi^2}{GM}a^3$

$T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.99\times 10^{30}}\times (7.78\times 10^{11})^3$

$T=3.74\times 10^8\ s$

(b) We know that,

$1\ year=3.154\times 10^7\ s$

or

$1\ s=3.171\times 10^{-8}\ year$

$3.74\times 10^8\ s={3.171\times 10^{-8}}\times {3.74\times 10^8}$

T = 11.859 earth years

Hence, this is the required solution.

7 0

4 years ago

Read 2 more answers