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Naddik [55]
3 years ago
7

it takes 1.4 minutes for Jonny to get to class which is a distance of 82 meters away s. what is Johnny’s average speed in second

s?
Physics
1 answer:
Marat540 [252]3 years ago
4 0
We have to use the equation speed=distance/time.
We want the average speed in second so we have to change the minutes into seconds. We can do this by multiplying the minutes by 60 (60seconds in 1 minute). 60x1.4= 84 seconds.
Speed=distance/time, distance is 82 and time is 84 so speed=82/84
Average speed = 0.98ms^-1 (2dp)
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Three cars (car F, car G, and car H) are moving with the same speed and slam on their brakes. The most massive car is car F, and
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To solve this problem it is necessary to apply the concepts related to Normal Force, frictional force, kinematic equations of motion and Newton's second law.

From the kinematic equations of motion we know that the relationship of acceleration, velocity and distance is given by

v_f^2=v_i^2+2ax

Where,

v_f = Final velocity

v_i = Initial Velocity

a = Acceleration

x = Displacement

Acceleration can be expressed in terms of the drag coefficient by means of

F_f = \mu_k (mg)  \rightarrowFrictional Force

F = ma \rightarrow Force by Newton's second Law

Where,

m = mass

a= acceleration

\mu_k = Kinetic frictional coefficient

g = Gravity

Equating both equation we have that

F_f = F

\mu_k mg=ma

a = \mu_k g

Therefore,

v_f^2=v_i^2+2ax

0=v_i^2+2(\mu_k g)x

Re-arrange to find x,

x = \frac{v_i^2}{2(-\mu_k g)}

The distance traveled by the car depends on the coefficient of kinetic friction, acceleration due to gravity and initial velocity, therefore the three cars will stop at the same distance.

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3 years ago
Some people will realize that they are dreaming while the dream is happening this is called
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Answer:

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Read 2 more answers
At the same instant that a 0.50-kg ball is dropped from 25 m above Earth, a second ball, with a mass of 0.25 kg, is thrown strai
zhannawk [14.2K]

Answer:

The center of mass of the two-ball system is 7.05 m above ground.

Explanation:

<u>Motion of 0.50 kg ball:</u>

Initial speed, u = 0 m/s

Time = 2 s

Acceleration = 9.81 m/s²

Initial height = 25 m

Substituting in equation s = ut + 0.5 at²

                 s = 0 x 2 + 0.5 x 9.81 x 2² = 19.62 m

Height above ground = 25 - 19.62 = 5.38 m

<u>Motion of 0.25 kg ball:</u>

Initial speed, u = 15 m/s

Time = 2 s

Acceleration = -9.81 m/s²

Substituting in equation s = ut + 0.5 at²

                 s = 15 x 2 - 0.5 x 9.81 x 2² = 10.38 m

Height above ground = 10.38 m

We have equation for center of gravity

          \bar{x}=\frac{m_1x_1+m_2x_2}{m_1+m_2}

          m₁ = 0.50 kg

          x₁ = 5.38 m        

          m₂ = 0.25 kg

          x₂ = 10.38 m    

Substituting

         \bar{x}=\frac{0.50\times 5.38+0.25\times 10.38}{0.50+0.25}=7.05m

The center of mass of the two-ball system is 7.05 m above ground.  

8 0
3 years ago
An object of mass m travels along the parabola yequalsx squared with a constant speed of 5 ​units/sec. What is the force on the
Dmitriy789 [7]

Explanation:

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First, we need to find the radius of curvature.  This is given by the equation:

R = [1 + (y')²]^(³/₂) / |y"|

y' = 2x and y" = 2:

R = [1 + (2x)²]^(³/₂) / |2|

R = (1 + 4x²)^(³/₂) / 2

At x = √2:

R = (1 + 4(√2)²)^(³/₂) / 2

R = (9)^(³/₂) / 2

R = 27 / 2

R = 13.5

So the centripetal force is:

F = m v² / r

F = m (5)² / 13.5

F = 1.85 m

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