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andrey2020 [161]
3 years ago
6

Please help me with my physics! I do not understand and please provide work.

Physics
1 answer:
natali 33 [55]3 years ago
5 0
The first one will be 220N
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The car transports 180 m in 12 s. Calculate its speed and express it in m / s and in km / h
makkiz [27]

Explanation:

Given parameters:

Distance  = 180m

Time  = 12s

Unknown:

Speed  in m/s and km/hr

Solution:

Speed is the distance divided by the time taken,

   Speed  = \frac{distance}{time}  

So;

     Speed  = \frac{180}{12}   = 15m/s

   1000m = 1km

  3600s  = 1hr

Therefore;

   15 x m x \frac{1}{s}  x \frac{1km}{1000m} x \frac{3600s}{1hr}    = 54km/hr

5 0
3 years ago
What formula oan be used to determine the velocity of an object in free fall?​
belka [17]

Answer:

A

Explanation:

8 0
3 years ago
Read 2 more answers
Two identical sticky masses m are moving in the xy-plane, with their momenta at an angle of φ with one another. They are each mo
postnew [5]

Answer:

ucosφ=-v2cosθ2\\\\φ=cos^{-1} (\frac{-v2cosθ2}{cosφ} )

Explanation:

Two identical sticky masses m are moving in the xy-plane, with their momenta at an angle of φ with one another. They are each moving at the same speed v when they collide at the origin of the coordinates and stick together. After the collision, the masses move at an angle −θ2 with respect to the +x axis at speed v2 .1. What was the angle φ?

from the principle of momentum

In a system of colliding bodies,we know that the total momentum before collision will equal to the total momentum after collision.

Take note that momentum is the product of mass and velocity

momentum before collision=momentum after collision

mass, m

u=initial velocity of the identical masses

v2=the common velocity after the collision

Note that the collision is inelastic , since they both moved with the same velocity

umcosφ+umcosφ=(m+m)v2cos−θ2

2mucosφ=2mv2cos−θ2

ucosφ=-v2cosθ2\\\\φ=cos^{-1} (\frac{-v2cosθ2}{cosφ} )

8 0
3 years ago
Based on the thermodynamic functions of enthalpy and entropy, can an unfavorable reaction that has a positive δg at rt be made f
Over [174]
I believe that the answer to the question provided above is yes, an unfavorable reaction that has a positive δg at rt be made favorable by increasing the reaction temperature.
Hope my answer would be a great help for you.    If you have more questions feel free to ask here at Brainly.
7 0
4 years ago
A 1,500 kg car’s speed changes from 30 m/s to 15 m/s after the brakes are applied. Calculate the work done onto the car from the
HACTEHA [7]

The work done onto the car is 506,250 J

The work done on a system implies an increase in the internal energy of the system as a result of some forces acting on the system from the outside.

From the parameters given:

  • The mass of the car = 1500 kg
  • The initial speed = 30 m/s
  • The final speed = 15 m/s

The work done onto the car refers to the change in the kinetic energy (i.e. ΔK.E)

\mathbf{=\dfrac{1}{2} mv_1^2 -\dfrac{1}{2} mv_2^2}

\mathbf{=\dfrac{1}{2} m(v_1^2 - v_2^2)}

\mathbf{=\dfrac{1}{2} \times 1500 \times (30^2 - 15^2)}

= 506,250 J

Therefore, we can conclude that the work done on the car is 506,250 J

Learn more about work done here:

brainly.com/question/18762601

7 0
3 years ago
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