Prompt

"An object at rest starts accelerating. If it is going 120 m/s after traveling 202 meters, how quickly did it speed up?"

Marizza181 [45]

Answer:

this is just a guess bc i only looked at it for 5 seconds but i think 150 m/s

4 0

3 years ago

Birdman is flying horizontally at a

den301095 [7]

Answer:

68 m

Explanation:

Given that the horizontal velocity of the birdman = 17 m/s and

the height, h= 78 m.

The gravitational force is acting in the downward direction, so it will not change the horizontal speed.

The horizontal speed will remains be constant and will be equal to the initial horizontal speed of the turd.

Initially, the turd was also flying horizontally with the birdman, so the initial velocity of the turd is the same as the horizontal velocity of the birdman, i.e In the horizontal direction, $u_0=17$ m/s.

In the vertical direction, u = 0,

The distance to be traveled, in the direction of application of force, is equals to the height of the turd, i.e

s= 78 m

Let t be the time taken to cover a distance of 78 m.

Now, applying the equation of motion in the vertical direction,

$s=ut+\frac 12 at^2$

where u is the initial velocity and a is the acceleration due to gravity in the direction of displacement,s.

Here, $a=g=9.81 m/s^2$ , so

$78=0\times t +\frac 12 (9.81)t^2$

$\Rightarrow t^2=(78\times2)/9.81$

$\Rightarrow t = 4$ seconds.

Hence, the time taken to reach the ground is 4 seconds.

As the horizontal speed, $u_0=17$ m/s, is constant throughout the journey, so

the horizontal distance covered by turd

$= u\times t$

$= 17 \times 4 = 68$ m.

So, the distance of landing from the start of the field is 68 m as the birdman releases a turd directly above the start of the field.

Hence, the robot must hold the bucket at a distance of 68 m from the start of the field.

5 0

3 years ago

As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a p

monitta

Answer:

force F = 1.66 × $10^{-13}$ N

Explanation:

given data

proton and an electron = 865 nm

solution

we get here force that is express as

force F = k q1 q2 ÷ r² ......................1

put here value and we get

force F = 9 × $10^{9}$ × $\frac{1.6\times (10^{-19})^{2}}{865 \times (10^{-9})^{2}}$

force F = 1.66 × $10^{-13}$ N

4 0

3 years ago

A 0.29 kg particle moves in an xy plane according to x(t) = - 19 + 1 t - 3 t3 and y(t) = 20 + 7 t - 9 t2, with x and y in meters

Artist 52 [7]

Answer:

Part a)

$F = 7.76 N$

Part b)

$\theta = -137.7 degree$

Part c)

$\theta = -127.7 degree$

Explanation:

As we know that acceleration is rate of change in velocity of the object

So here we know that

$x = -19 + t - 3t^3$

$y = 20 + 7t - 9t^2$

Part a)

differentiate x and y two times with respect to time to find the acceleration

$a_x = \frac{d^2}{dt^2}(-19 + t - 3t^3)$

$a_x = \frac{d}{dt}(0 +1 - 9t^2)$

$a_x = -18t$

$a_y = \frac{d^2}{dt^2}(20 + 7t - 9t^2)$

$a_y = \frac{d}{dt}(0 +7 - 18t)$

$a_y = -18$

Now the acceleration of the object is given as

$\vec a = (-18t)\hat i + (-18)\hat j$

at t= 1.1 s we have

$\vec a = -19.8 \hat i - 18 \hat j$

now the net force of the object is given as

$\vec F = m\vec a$

$\vec F = (0.29 kg)(-19.8 \hat i - 18 \hat j)$

$\vec F = -5.74 \hat i - 5.22 \hat j$

now magnitude of the force will be

$F = \sqrt{5.74^2 + 5.22^2} = 7.76 N$

Part b)

Direction of the force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{-5.22}{-5.74}$

$\theta = -137.7 degree$

Part c)

For velocity of the particle we have

$v_x = \frac{dx}[dt}$

$v_x = (0 +1 - 9t^2)$

$v_y = \frac{dy}{dt}$

$v_y = (0 +7 - 18t)$

now at t = 1.1 s

$\vec v = -9.89\hat i - 12.8 \hat j$

now the direction of the velocity is given as

$\theta = tan^{-1}(\frac{v_y}{v_x})$

$\theta = tan^{-1}(\frac{-12.8}{-9.89})$

$\theta = -127.7 degree$

7 0

3 years ago

Which is an example of an exothermic process?

Scorpion4ik [409]

Answer:

an example of an exthermic process is combustion

Explanation:

combustion is like lighting a candle

3 0

3 years ago