Ask question

Ask question

All categories

2 years ago

15

A 1200-kg car accelerates it’s speed from 4 m/s to 10 m/s in 3 seconds. Find the car average speed , the car acceleration , the

distance traveled by , & the net force on car

1 answer:

dsp732 years ago

8 0

Answer:

Given: mass 1200kg

initial velocity: 4m/s

finial velocity: 10 m/s

time 3 sec

then

speed; initial velocity + final velocity/2

4+10/3

: 4.66m/s2

You might be interested in

A 69 kg zebra is traveling 7 m/s east. What is the zebra’s momentum? kg-m/s

Hitman42 [59]

Linear momentum is the product of mass and velocity. In this case, it is simply:
$69\times 7 = 483 kg\cdot m/s$

6 0

2 years ago

At each corner of a square of side l there are point charges of magnitude Q, 2Q, 3Q, and 4Q.What is the magnitude and direction

lbvjy [14]

Answer:

$F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]$

$|F_T|=2\sqrt{34}k\frac{Q^2}{L}$

$\theta=tan^{-1}(\frac{5}{3})=59.03\°$

Explanation:

I attached an image below with the scheme of the system:

The total force on the charge 2Q is the sum of the contribution of the forces between 2Q and the other charges:

$F_T=F_Q+F_{3Q}+F_{4Q}\\\\F_T=k\frac{(Q)(2Q)}{R_1}\hat{i}+k\frac{(3Q)(2Q)}{R_2}\hat{j}+k\frac{(4Q)(2Q)}{R_3}[cos\theta \hat{i}+sin\theta \hat{j}]$

the distances R1, R2 and R3, for a square arrangement is:

R1 = L

R2 = L

R3 = (√2)L

θ = 45°

$F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[cos(45\°)\hat{i}+sin(45\°)\hat{j}]\\\\F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[\frac{\sqrt{2}}{2}\hat{i}+\frac{\sqrt{2}}{2}\hat{j}]\\\\F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]$

and the magnitude is:

$|F_T|=2k\frac{Q^2}{L}\sqrt{3^2+5^2}=2\sqrt{34}k\frac{Q^2}{L}$

the direction is:

$\theta=tan^{-1}(\frac{5}{3})=59.03\°$

4 0

2 years ago

PLEASE HELP ME!!!!!!!

melamori03 [73]

<em><u>This</u></em><em><u> </u></em><em><u>it</u></em><em><u> </u></em><em><u>you</u></em><em><u> </u></em><em><u>can</u></em><em><u> </u></em><em><u>do</u></em><em><u> </u></em><em><u>it</u></em><em><u> </u></em><em><u>boy</u></em><em><u> </u></em><em><u>or</u></em><em><u> </u></em><em><u>girl</u></em><em><u> </u></em>

6 0

2 years ago

If the window is 11 m above the ground, find the time the stone is in flight.

rjkz [21]

d=vi*t+(1/2)gt²

d=11 m
g=9.8 m/s²
vi=0 m/s

11 m=0 m/s*t+(1/2)9.8 m/s²t²
11 m=4.9 m/s²t²
t²=11 m / 4.9 m/s²
t=√(11 m / 4.9 m/s²)=1.489... s≈1.5 s

Answer: the time the sone is in flight is 1.5 s

5 0

2 years ago

A 5.0-kg box has an acceleration of 2.0 m/s2 when it is pulled by a horizontal force across a surface with μK = 0.50. Find the w

Maslowich

Answer:a) 34.5 N; b) 24.5 N; c) 10 N; d) 1J

Explanation: In order to solve this problem we have to used the second Newton law given by:

∑F= m*a

F-f=m*a where f is the friction force (uk*Normal), from this we have

F= m*a+f=5 Kg*2 m/s^2+0.5*5Kg*9.8 m/s^2= 34.5 N

then f=uk*N=0.5*5Kg*9.8 m/s^2= 24.5N

the net Force = (34.5-24.5)N= 10 N

Finally the work done by the net force is equal to kinetic energy change so

W=∫Force net*dr= 10 N* 0.1 m= 1J

7 0

2 years ago

Read 2 more answers