Ask question

Ask question

All categories

Elena-2011 [213]

3 years ago

5

A truck on the freeway originally moving at 6.6 meters/second accelerates uniformly with acceleration a = 2.8 meters/second2 for

7.9 seconds. How far did the truck travel during that time interval?

1 answer:

Sonja [21]3 years ago

8 0

Answer:

139.514 metres

Explanation:

Initial velocity of the truck = 6.6 m/s

Acceleration of the truck = 2.8 m/s^2

Time interval = 7.9 s

Therefore we use the formula,

s = ut + 1/2 at^2

*where s(the distance travelled)...u(the initial velocity)...t(the time period)

; s = 6.6(7.9) + 1/2 (2.8)(7.9)^2

; s = 52.14 + 87.374

The distance moved by the truck = 139.514m

You might be interested in

What is the maximum speed with which a 1200-kg car can round a turn of radius 94.0 m on a flat road if the coefficient of static

JulijaS [17]

Answer:

<h2> v= 21.47m/s </h2>

Explanation:

For the car to turn at the about the centripetal force must not be greater than the static friction between the tires and the road

we will use the expression relating centripetal force and static friction below

let U represent the coefficient of static friction

Given that

U= 0.50

mass m= 1200-kg

radius r= 94.0 m

Assuming g= 9.81 m/s^2

$U*m*g=\frac{mv^2}{r}$

$U*g=\frac{v^2}{r}$

substituting our given data in to expression we can solve for the speed V

$0.5*9.81=\frac{v^2}{94}$

making v the subject of formula we have

$0.5*9.81=\frac{v^2}{94}\\\v= \sqrt{0.5*9.81*94} \\\\v= \sqrt{461.07} \\\\v= 21.47$

v= 21.47m/s

<em><u>hence the maximum velocity of the car is 21.47m/s</u></em>

7 0

3 years ago

A 0.0400-g positive charged ball with charge q = 6.40 μC is resting on a flat, frictionless horizontal surface. For a time of t

Leto [7]

Answer:

The height is 0.1014 m

Explanation:

Given that,

Mass = 0.0400 g

Charge $q= 6.40\ \mu C$

Time t = 0.0420 s

Electric field $E=7.80\times10^{2}\ N/C$

We need to calculate the electric force on the particle

Using formula of electric force

$F=qE$

Put the value into the formula

$F_{e}=6.40\times10^{-6}\times7.80\times10^{2}$

$F_{e}=0.004992= 0.499\times10^{-2}\ N$

We need to calculate the gravitational force

Using formula of force

$F=mg$

Put the value into the formula

$F_{g}=0.0400\times10^{-3}\times9.8$

$F_{g}=0.000392 = 0.392\times10^{-3}\ N$

We need to calculate the net force

$F_{net}=F_{e}-F_{g}$

$F_{net}=0.499\times10^{-2}-0.392\times10^{-3}$

$F_{net}=0.004598= 0.4598\times10^{-2}\ N$

We need to calculate the acceleration

Using newton's law

$F = ma$

$a = \dfrac{F}{m}$

$a=\dfrac{0.4598\times10^{-2}}{0.0400\times10^{-3}}$

$a =114.95\ m/s^2$

We need to calculate the height

Using equation of motion

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

$s=0+\dfrac{1}{2}\times114.95\times(0.0420)^2$

$s=0.1014\ m$

Hence, The height is 0.1014 m

3 0

2 years ago

A 12 V battery is connected to a 1200 Ω resistor. How much current is flowing through the resistor?

Oduvanchick [21]

Explanation:

Use Ohm's law.

V = IR

12 V = I (1200 Ω)

I = 0.01 A

7 0

3 years ago

Match the microscopes with their useable

umka21 [38]

Microscopes with life science and physics

6 0

3 years ago

Read 2 more answers

What is the importance of measurement? <br>

Kruka [31]

Answer:

To know the time ,length area

8 0

2 years ago