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3 years ago

12

shelly starts from rest on her bicycle at the top of a hill. After 6.0s she has reached a final velocity of 14m/s. what is shell

ys acceleration.

2 answers:

earnstyle [38]3 years ago

8 0

Divide 14 by 6 and there is your answer with the unit of m

Anna007 [38]3 years ago

7 0

Answer:

Acceleration, $a=2.33\ m/s^2$

Explanation:

It is given that,

Initial speed of Shelly, u = 0

After 6 seconds, she has reached a final velocity of 14 m/s, v = 14 m/s

We need to find the acceleration of Shelly. The formula to find the acceleration is given by :

$a=\dfrac{v-u}{t}$

$a=\dfrac{14\ m/s-0}{6\ s}$

$a=2.33\ m/s^2$

So, the acceleration of Shelly's is $2.33\ m/s^2$ . Hence, this is the required solution.

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In an attempt to impress its friends, an acrobatic beetle runs and jumps off the bottom step of a flight of stairs. The step is

Aleks04 [339]

Answer:

0.3677181864 m

Explanation:

u = Velocity = 1.5 m/s

$\theta$ = Angle = 20°

y = -20 cm

Velocity components

$u_x=ucos\theta\\\Rightarrow u_x=1.5cos20\\\Rightarrow u_x=1.40953\ m/s$

$u_y=usin\theta\\\Rightarrow u_y=1.5sin20\\\Rightarrow u_y=0.51303\ m/s$

Acceleration components

$a_x=0$

$a_y=-9.81\ m/s^2$

$y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow -0.2=0.51303\times t+\dfrac{1}{2}\times -9.81t^2\\\Rightarrow 4.905t^2-0.51303t-0.2=0$

$t=\frac{-\left(-0.51303\right)+\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}, \frac{-\left(-0.51303\right)-\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}\\\Rightarrow t=0.26088, -0.15629$

Time taken is 0.26088 seconds

$x=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow x=1.40953\times 0.26088\\\Rightarrow x=0.3677181864\ m$

The distance the beetle travels on the ground is 0.3677181864 m

6 0

3 years ago

Three +3.0-μC point charges are at the three corners of a square of side 0.50 m. The last corner is occupied by a −3.0-μC charge

kramer

Answer:

E = 440816.32 N/C

Explanation:

Given data:

Three point charge of charge equal to +3.0 micro coulomb

fourth point charge = - 3.0 micro coulomb

side of square = 0.50 m

$K =1/4 \pi \epsilon_0 = 8.99 \times 10^9$ N.m^2/c^2

Due to having equal charge on center of square, 2 charge produce equal electric field at center and other two also produce electric field at center of same value

So we have

$E_1 + E_3 = 0$

$E =E_2 + E_4$

$E = 2 E_2$

[ $E_2 =\frac{2\times k \times q}{r^2}$

[ $r= \frac{(0.5^2 + 0.5^2)^2}{2} = 0.35 m$ ]

plugging all value

$E = 2 E_2$

$E = 2 E_2 =\frac{2\times k \times q}{r^2}$

$E = \frac{2 \times 8.99 \times 10^93\times 10^{-6}}{0.35^2}$

E = 440816.32 N/C

3 0

3 years ago

Read 2 more answers

Write a hypothesis about the use of an object's physical

KengaRu [80]

If the mass of the object and the volume of the object is determined;

Then, the density of the object is determined by taking the ratio of the mass and volume.

<h3>What is density of an object?</h3>

The density of an object is the ratio of the mass and volume of that object.

Mathematically;

Density = mass/volume

To determine the density of an object therefore, the physical characteristics of mass and the volume of the object are measured.

The mass of the object is obtained using a scale or a balance.

The volume of the object if a solid is obtained using a displacement bottle. If it is a liquid, a measuring cylinder is used.

The density of the object is then obtained by taking the ratio of the mass and the volume of the object.

In conclusion, the density of an object is determined from the volume and mass ratio.

Learn more about density at: brainly.com/question/1354972

#SPJ1

3 0

1 year ago

Astronaut mark uri is space-traveling from planet x to planet y at a speed of relative to the planets, which are at rest relativ

lyudmila [28]

<span>From the point of view of the astronaut, he travels between planets with a speed of 0.6c. His distance between the planets is less than the other bodies around him and so by applying Lorentz factor, we have 2*</span>√1-0.6² = 1.6 light hours. On the other hand, from the point of view of the other bodies, time for them is slower. For the bodies, they have to wait for about 1/0.6 = 1.67 light hours while for him it is 1/(0.8) = 1.25 light hours. The remaining distance for the astronaut would be 1.67 - 1.25 = 0.42 light hours. And then, light travels in all frames and so the astronaut will see that the flash from the second planet after 0.42 light hours and from the 1.25 light hours is, 1.25 - 0.42 = 0.83 light hours or 49.8 minutes.

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3 years ago

Please help me out i'm so depressed and such a failure

aksik [14]

Answer:

I reckon towards b. Let me know if im right

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3 years ago