All categories

2 years ago

A ball is pushed into a spring-loaded launcher with a force of 20 N, which compresses the spring 0.08 m.

Physics

1 answer:

damaskus [11]2 years ago

3 0

Answer:

0.8J

Explanation:

Given parameters:

Force = 20N

Compression = 0.08m

Unknown:

Spring constant = ?

Elastic potential energy = ?

Solution:

To solve this problem, we use the expression below:

F = k e

F is the force

k is the spring constant

e is the compression

20 = k x 0.08

k = 250N/m

Elastic potential energy;

EPE = $\frac{1}{2}$ k e² = $\frac{1}{2}$ x 250 x 0.08²

Elastic potential energy = 0.8J

You might be interested in

Transparent

marusya05 [52]

Answer:

A..................

4 0

3 years ago

A man with a mass of 65.0 kg skis down a frictionless hill that is 5.00 m high. At the bottom of the hill the terrain levels out

anzhelika [568]

Answer:

The horizontal distance is 4.823 m

Solution:

As per the question:

Mass of man, m = 65.0 kg

Height of the hill, H = 5.00 m

Mass of the backpack, m' = 20.0 kg

Height of ledge, h = 2 m

Now,

To calculate the horizontal distance from the edge of the ledge:

Making use of the principle of conservation of energy both at the top and bottom of the hill (frictionless), the total mechanical energy will remain conserved.

Now,

$KE_{initial} + PE_{initial} = KE_{final} + PE_{final}$

where

KE = Kinetic energy

PE = Potential energy

Initially, the man starts, form rest thus the velocity at start will be zero and hence the initial Kinetic energy will also be zero.

Also, the initial potential energy will be converted into the kinetic energy thus the final potential energy will be zero.

Therefore,

$0 + mgH = \frac{1}{2}mv^{2} + 0$

$2gH = v^{2}$

$v = \sqrt{2\times 9.8\times 5} = 9.89\ m/s$

where

v = velocity at the hill's bottom

Now,

Making use of the principle of conservation of momentum in order to calculate the velocity after the inclusion, v' of the backpack:

$mv = (m + m')v'$

$65.0\times 9.89 = (65.0 + 20.0)v'$

$v' = 7.56\ m/s$

Now, time taken for the fall:

$h = \frac{1}{2}gt^{2}$

$t = \sqrt{\frac{2h}{g}}$

$t = \sqrt{\frac{2\times 2}{9.8} = 0.638\ s$

Now, the horizontal distance is given by:

x = v't = $7.56\times 0.638 = 4.823\ m$

7 0

3 years ago

Read 2 more answers

Increase the slit width to 1050 nm. What happens to the width of the central bright fringe? (Always remember to wait a few secon

djyliett [7]

Answer:

Explanation:

In case of Single Slit, diffraction will occur.

Then In Single slit Diffraction, width of central fringe is

$x_c= 2D\lambda/a
$

where D= distance b/w screen and slit

a= slit width

\lambda = wavelength

Thus if Screen width increases keeping other factors same then width of central fringe becomes narrower as

$x_c\propto 1/a$

On increasing the slit width the central bright fringe width The width of the central bright fringe becomes narrower.

3 0

3 years ago

What's the difference between a direct relationship and a positive one?

borishaifa [10]

A positive or direct relationship is one in which the two variables (we will generally call them x and y) move together, that is, they either increase or decrease together. In a negative or indirect relationship, the two variables move in opposite directions, that is, as one increases, the other descremases

5 0

2 years ago

A hawk is flying horizontally at 18.0 m/s in a straight line, 230 m above the ground. A mouse it has been carrying struggles fre

Lisa [10]

Answer:

a) vd = 47.88 m/s

b) θ = 80.9°

c) t = 6.8 s

Explanation:

In the situation of the problem, you can assume that the trajectory of the hawk and the trajectory of the mouse form a rectangle triangle.

One side of the triangle is the horizontal trajectory of the hawk after 2.00s of flight, the other side of the triangle is the distance traveled by the mouse when it is falling down. And the hypotenuse is the trajectory of the hawk when it is trying to recover the mouse.

(a) In order to calculate the diving speed of the hawk, you first calculate the hypotenuse of the triangle.

One side of the triangle is c1 = (18.0m/s)(2.0s) = 36m

The other side of the triangle is c2 = 230m - 3m = 227 m

Then, the hypotenuse is:

$h=\sqrt{(36m)^2+(227m)^2}=229.83m$ (1)

Next, it is necessary to calculate the falling down time of the mouse, this can be done by using the following formula:

$y=y_o+v_ot+\frac{1}{2}gt^2$ (2)

yo: initial height = 230m

vo: initial vertical speed of the mouse = 0m/s

g: gravitational acceleration = -9.8m/s^2

y: final height of the mouse = 3 m

You replace the values of the parameters in (2) and solve for t:

$3=230-4.9t^2\\\\t=\sqrt{\frac{227}{4.9}}=6.8s$

The hawk traveled during 2.00 second in the horizontal trajectory, hence, the hawk needed 6.8s - 2.0s = 4.8 s to travel the distance equivalent to the hypotenuse to catch the mouse.

You use the value of h and 4.8s to find the diving speed of the hawk:

$v_d=\frac{229.83m}{4.8s}=47.88\frac{m}{s}$

The diving speed of the Hawk is 47.88m/s

(b) The angle is given by:

$\theta=cos^{-1}(\frac{c_1}{h})=cos^{-1}(\frac{36m}{229.83m})=80.9 \°$

Then angle between the horizontal and the trajectory of the Hawk when it is descending is 80.9°

(c) The mouse is falling down during 6.8 s

4 0

3 years ago