Ask question

Ask question

All categories

2 years ago

9

An object with an initial velocity of 10 m/s accelerates at a rate of 3.5 m/s2 for 8 seconds. How far will it have traveled duri

ng that time?

1 answer:

ICE Princess25 [194]2 years ago

4 0

Answer:

Distance, d = 192 meters

Explanation:

We have,

Initial velocity of an object is 10 m/s

Acceleration of the object is 3.5 m/s²

Time, t = 8 s

We need to find the distance travelled by the object during that time. Second equation of motion gives the distance travelled by the object. It is given by :

$d=ut+\dfrac{1}{2}at^2$

$d=10\times 8+\dfrac{1}{2}\times 3.5\times 8^2\\\\d=192\ m$

So, the distance travelled by the object is 192 meters.

You might be interested in

an object weighing 15 newtons is lifted from the ground to a height of 0.22 meter what is the increase in the object's gravitati

kicyunya [14]

GPE= weight•height= 15 N• 0.22meter= 3.3 Joules
I hope this helps ~~Charlotte~~

5 0

2 years ago

Read 2 more answers

Which subatomic particle has a negative charge?

ValentinkaMS [17]

Answer:

C. $\displaystyle electron$

Explanation:

0 charge → <em>Neutron</em>

1 charge → <em>Proton</em>

I am joyous to assist you anytime.

4 0

3 years ago

Displacement vectors of 4km north, 2km south, 5km north, 5km south combine to a total displacement of

goldfiish [28.3K]

<u>Answer</u>

The combined displacement is 2km north

<u>Explanation</u>

Since displacement is a vector quantity, we take into account the direction.

Good for us all the displacement vectors are in the same dimension, so we can make north positive and south negative or vice-versa.

We now add to obtain,

$4+-2+5+-5$

This will simplify to

$=4-2+5-5=2$

Therefore the combined displacement is 2km north

5 0

3 years ago

g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart

weeeeeb [17]

Complete Question

A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.

Answer:

The value $v_2 = 4 \sqrt{10} \ m/s$

Explanation:

From the question we are told that

The charge on the first sphere is $q_1 = 2\mu C = 2*10^{-6} \ C$

The charge on the second sphere is $q_2 = 8 \mu C = 8*10^{-6} \ C$

The mass of the second charge is $m = 1.50 \ g = 1.50 *10^{-3} \ kg$

The distance apart is $d = 0.4 \ m$

The speed of the second sphere is $v_1 = 20 \ ms^{-1}$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.8 \ m$ is mathematically represented

$Q = KE + U$

Here KE is the kinetic energy which is mathematically represented as

$KE = \frac{1 }{2} m (v_1)^2$

substituting value

$KE = \frac{1 }{2} * ( 1.50 *10^{-3}) (20 )^2$

$KE = 0.3 \ J$

And U is the potential energy which is mathematically represented as

$U = \frac{k * q_1 * q_2 }{d }$

substituting values

$U = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.8 }$

$U = 0.18 \ J$

So

$Q = 0.3 + 0.18$

$Q = 0.48 \ J$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.4 \ m$ is mathematically represented

$Q_f = KE_f + U_f$

Here $KE_f$ is the kinetic energy which is mathematically represented as

$KE_f = \frac{1 }{2} m (v_2^2$

substituting value

$KE_f = \frac{1 }{2} * ( 1.50 *10^{-3}) (v_2 )^2$

$KE_f = 7.50 *10^{ -4} (v_2 )^2$

And $U_f$ is the potential energy which is mathematically represented as

$U_f = \frac{k * q_1 * q_2 }{d }$

substituting values

$U_f = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.4 }$

$U_f = 0.36 \ J$

From the law of energy conservation

$Q = Q_f$

So

$0.48 = 0.36 +(7.50 *10^{-4} v_2^2)$

$v_2 = 4 \sqrt{10} \ m/s$

6 0

2 years ago

Please help ASAP!

Softa [21]

Answer:

100 newton

Explanation:

newton third law of motion says to every action there is an always an equal and opposite reaction so the magnitude will stay equal but opposite direction

8 0

2 years ago

Read 2 more answers