Would it be better to collide head-on with an identical car traveling at the same speed or collide with a stationary brick wall?

Find the force required to do 25 joule work when the force causes a displacement of 0.5 m

Eduardwww [97]

Answer:

Explanation:

The force required can be found by using the formula

$f = \frac{w}{d} \\$

w is the workdone

d is the distance

From the question we have

$f = \frac{25}{0.5} \\$

We have the final answer as

Hope this helps you

6 0

3 years ago

Plants appear green because they do not absorb the green wavelengths of light what happens to those green light waves when they

Arlecino [84]

The green wavelengths are reflected, causing green to be the only color we see in certain parts of plants.

Best of luck, my man.

4 0

4 years ago

Read 2 more answers

(a) a light-rail commuter train accelerates at a rate of 1.15 m/s2. how long does it take it to reach its top speed of 80.0 km/h

snow_lady [41]

A = 1.15m/s2, Vf = 80.0km/h --> we need it in m/s, so:
Vf = 80km/h × 1000m/1km × 1h/3600s
= 22.22m/s
Top speed = Vf, initial speed = Vi
time (t) = V(Vf-Vi) ÷ a
t = (22.22-0)m/s ÷ 1.15m/s2
t = 22.22m/s × s2/1.15m
= 19.32 seconds

6 0

3 years ago

Light of wavelength 600 nm passes though two slits separated by 0.22mm and is observed on a screen 1.1m behind the slits. The lo

Ne4ueva [31]

Answer:

$y(m=1)=\frac{(1)(600*10^{-9}m)(1.1m)}{0.22m}=3*10^{-6}m\\\\y(m=1)=\frac{(-1)(600*10^{-9}m)(1.1m)}{0.22m}=-3*10^{-6}m$

Explanation:

We have to take into account the expression for the position of the fringes

$dsen\theta=m\lambda\\y=\frac{m\lambda D}{d}$

where m is the number of the maximum, d is the separation of the slits, D is the distance to the screen.

(a) By replacing we obtain

$y(m=1)=\frac{(1)(600*10^{-9}m)(1.1m)}{0.22m}=3*10^{-6}m\\\\y(m=1)=\frac{(-1)(600*10^{-9}m)(1.1m)}{0.22m}=-3*10^{-6}m$

(b) more information is required to solve this point. Please complete the information.

HOPE THIS HELPS!

4 0

3 years ago

Read 2 more answers

Consider a lawnmower of mass m which can slide across a horizontal surface with a coefficient of friction μ. In this problem the

inna [77]

Answer:

Fh = u*m*g / (cos(θ) - u*sin(θ))

Explanation:

Given:

- The mass of lawnmower = m

- The angle the handle makes with the horizontal = θ

- The force applied along the handle = Fh

- The coefficient of friction of the lawnmower with ground = u

Find:

Find the magnitude, Fh, of the force required to slide the lawnmower over the ground at constant speed by pushing the handle.

Solution:

- Construct a Free Body Diagram (FBD) for the lawnmower.

- Realize that there is horizontal force applied parallel to ground due to Fh that drives the lawnmower and a friction force that opposes this motion. We will use to Newton's law of motion to express these two forces in x-direction as follows:

F_net,x = m*a

- Since, the lawnmower is to move with constant speed then we have a = 0.

F_net,x = 0

- The forces as follows:

Fh*cos(θ) - Ff = 0

Where, Ff is the frictional force:

Fh = Ff /cos(θ)

Similarly, for vertical direction y the forces are in equilibrium. Using equilibrium equation in y direction we have:

- W - Fh*sin(θ) + Fn = 0

Where, W is the weight of the lawnmower and Fn is the contact force exerted by the ground on the lawnmower. Then we have:

Fn = W + Fh*sin(θ)

Fn = m*g + Fh*sin(θ)

The Frictional force Ff is proportional to the contact force Fn by:

Ff = u*Fn

Ff = u*(m*g + Fh*sin(θ))

Substitute this expression in the form derived for Fh and Ff:

Fh*cos(θ) = u*(m*g + Fh*sin(θ))

Fh*(cos(θ) - u*sin(θ)) = u*m*g

Fh = u*m*g / (cos(θ) - u*sin(θ))

5 0

3 years ago