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densk [106]
3 years ago
5

A motorcycle skids to a stop on a road.

Physics
1 answer:
Svetlanka [38]3 years ago
5 0

Answer:

Friction of the road on the motorcycle in the opposite direction

Explanation:

Khanacademy

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To choose particular advancements to help the data security outline. A SecSDLC will display an extensive, repeatable, and predictable methodology for coordinating security into your application advancement process. It will include characterizing necessities, architecting the application, building and testing the application, conveying to creation, and keeping up the application underway.
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PLZZ HELP AND ANSWER AND THE FIRST ONE THAT THOSE THAT WILL BE MARKED BRAINIEST!
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4 years ago
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A bar 4.0m long weights 400 N. Its center of gravity is 1.5m from on end. A weight of 300N is attached at the light end. What ar
Debora [2.8K]

Answer:

The resulting, needed force for equilibrium is a reaction from a support, located at 2.57 meters from the heavy end. It is vertical, possitive (upwards) and 700 N.

Explanation:

This is a horizontal bar.

For transitional equilibrium, we just need a force opposed to its weight, thus vertical and possitive (ascendent). Its magnitude is the sum of the two weights, 400+300 = 700 N, since weight, as gravity is vertical and negative.

Now, the tricky part is the point of application, which involves rotational equilibrium. But this is quite simple if we write down an equation for dynamic momentum with respect to the heavy end (not the light end where the additional weight is placed). The condition is that the sum of momenta with respect to this (any) point of the solid bar is zero:

0=\Sigma_{i}M=400\cdot1.5+300\cdot4-d\cdot700

Where momenta from weights are possitive and the opposed force creates an oppossed momentum, then a negative term. Solving our unknown d:

d=\frac{400\cdot1.5+300\cdot4}{700} =2.57 m

So, the resulting force is a reaction from a support, located at 2.57 meters from the heavy end (the one opposed to the added weight end).

8 0
4 years ago
Suppose you designed a spacecraft to work by photon pressure. The sail was a completely absorbing fabric of area 1.0 km2 and you
Alekssandra [29.7K]

Answer:

(a) F = 6.14 *10⁻⁴ N

(b) P = 6.14* 10⁻¹⁰ Pa

(c) t = 27.2 min

Explanation:

Area of sail A = 1.0 km² = 1.0 * 10⁶m²

Wavelength of light  λ = 650 nm = 650 * 10⁻⁹ m

Rate of impact of photons R = 1 mol/s = 6.022 * 10²³ photons/s

(a)

Momentum of each photon is Ρ = h/λ = 6.625 * 10⁻³⁴ / 650 * 10⁻⁹

      = 1.0192 * 10⁻²⁷ kg.m/s

Since the photons are absorbed completely, in every collision the above momentum is transferred to the sail.  

Momentum transferred to the sail per second is product of rate of impact of photons and momentum transferred by each photon.

dp/dt = R * h/ λ

This is the force acting on the sail.

F = R * h/λ = 6.022 * 10²³ * 1.0192 * 10⁻²⁷ = 6.14 *10⁻⁴ N

F = 6.14 *10⁻⁴ N

b)

Pressure exerted by the radiation on the sail = Force acting on the sail / Area of the sail

P = F/A =  6.14 * 10⁻⁴ / 1.0 * 10⁶ =  6.14* 10⁻¹⁰ Pa

P = 6.14* 10⁻¹⁰ Pa

c)  

Acceleration of spacecraft a = F/m = 6.14 * 10⁻⁴ /1.0 = 6.14 * 10⁻⁴m/s²

As the spacecraft starts from rest, initial speed u=0,m/s ,

final speed is u = 1.0 m/s after time t  

v = u+at  

t = 1.0 - 0/ 6.14 * 10⁻⁴ =  1629s = 27.2 min

t = 27.2 min

4 0
4 years ago
Calculate the speed of a person who walked 100 m in 25 seconds. Show all of your work. Step 1: Using a light color, highlight th
satela [25.4K]

Answer:

4 m/s

Explanation:

to find m/s, divide the meters term by the seconds term. 100m/25s = 4m/s

4 0
3 years ago
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