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A car accelerates uniformly from rest to speed 6.6 m/s in 6.5 s .Find the distance the car travel during this time .

kirill [66]

Answer:

<em>The distance the car traveled is 21.45 m</em>

Explanation:

<u>Motion With Constant Acceleration </u>

It occurs when an object changes its velocity at the same rate thus the acceleration is constant.

The relation between the initial and final speeds is:

$v_f=v_o+at\qquad\qquad [1]$

Where:

a = acceleration

vo = initial speed

vf = final speed

t = time

The distance traveled by the object is given by:

$\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]$

Solving [1] for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

Substituting the given data vo=0, vf=6.6 m/s, t=6.5 s:

$\displaystyle a=\frac{6.6-0}{6.5}$

$a = 1.015\ m/s^2$

The distance is now calculated with [2]:

$\displaystyle x=0*6.5+\frac{1.015*6.5^2}{2}$

x = 21.45 m

The distance the car traveled is 21.45 m

6 0

3 years ago

A jet is travelling at a speed of 1200 km/h and drops cargo from a height of 2.5 km above the ground Calculate the time it takes

OLEGan [10]

a) Time of flight: 22.6 s

To calculate the time it takes for the cargo to reach the ground, we just consider the vertical motion of the cargo.

The vertical position at time t is given by

$y(t) = h +u_y t - \frac{1}{2}gt^2$

where

h = 2.5 km = 2500 m is the initial height

$u_y = 0$ is the initial vertical velocity of the cargo

g = 9.8 m/s^2 is the acceleration of gravity

The cargo reaches the ground when

$y(t) = 0$

So substituting it into the equation and solving for t, we find the time of flight of the cargo:

$0 = h - \frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(2500)}{9.8}}=22.6 s$

b) 7.5 km

The range travelled by the cargo can be calculated by considering its horizontal motion only. In fact, the horizontal motion is a uniform motion, with constant velocity equal to the initial velocity of the jet:

$v_x = 1200 km/h \cdot \frac{1000 m/km}{3600 s/h}=333.3 m/s$

So the horizontal distance travelled is

$d=v_x t$

And if we substitute the time of flight,

t = 22.6 s

We find the range of the cargo:

$d=(333.3)(22.6)=7533 m = 7.5 km$

7 0

3 years ago

Find the energy u of the capacitor in terms of c and q by using the definition of capacitance and the formula for the energy in

gtnhenbr [62]

The formula for the energy in a capacitor , u in terms of q and c is q²/2c

<h3>What is the energy of a capacitor?</h3>

The energy of a capacitor u = 1/2qv where

q = charge on capacitor and
v = voltage across capacitor.

<h3>What is the capacitance of a capacitor?</h3>

Also, the capacitance of a capacitor c = q/v where

q = charge on capacitor and
v = voltage across capacitor.

So, v = q/c

<h3>The formula for energy of the capacitor in terms of q and c</h3>

Substituting v into u, we have

u = 1/2qv

= 1/2q(q/c)

= q²/2c

So, the formula for the energy in a capacitor , u in terms of q and c is q²/2c

Learn more about energy in a capacitor here:

brainly.com/question/10705986

#SPJ12

3 0

2 years ago

Wo siblings are arguing over what way to pull a 10 kg wagon. Boris wants to pull it to the right. Boris puts 220 N of force on t

Mrrafil [7]

Answer:

a) Then net force acting on the wagon 30 [N] in a negative direction to the left

b)Acceleration of the wagon 3 m/s² in a negative direction to the left

Explanation:

The coordinates x and y show the positive direction in x-axis ( to the right) and negative direction to the left. In north-south direction positive y values are to the north and negative values to the south

The free-body diagram shows 4 forces acting

In y-axis:

The weight ( P = m*g = 10 [kg]*9,8 [m/s²] = 98 [N] negative

Normal reaction force Fn = P = 98 [N] positive ( surface is frictionless)

∑Fy = 0 P - Fn = 0 P = Fn = 98 [N]

In x-axis:

Boris pulling force to the right (positive ) 220 [N]

Natasha pulling force to the left ( negative) 250 [N]

∑Fx = m*a

220 [N] - 250 [N] = 10 Kg * a [N] = Kg * m /s²

-30 [kg*m/s²] = 10 * Kg * a

-30/10 = - 3 m/s² = a

Sign (-) means the direction of acceleration vector is to the left (the same direction of the movement )

Then net force acting on the wagon 30 [N] in negative direction to the left

Acceleration of the wagon 3 m/s² in negative direction to the left

5 0

3 years ago

A physicist is investigating a beam of laser light of wavelength 550 nm. The light strikes a target that is 189 meters away from

Lena [83]

Answer: here you go I was looking for this answer everywhere,I have it now so it’s 6.30 x 10^-7 s

Explanation:

I hope this helps☺️

8 0

3 years ago