A car traveling with constant speed travel 150 km in 7200s.what is the speed of the car?

Physics

1 answer:

scoray [572]3 years ago

6 0

Time = 7200 s. 3600 s = 1hour.

7200s = 7200 / 3600 = 2 hours.

Speed = Distance / time = 150 km / 2 hours = 75 km/h

You might be interested in

A car moves along an x axis through a distance of 900 m, starting at rest (at x = 0) and ending at

OlgaM077 [116]

A car moves along an x axis through a distance of 900 m, starting at rest (at x = 0) and ending at rest (at x = 900 m). Through the first 1/4 of that distance, its acceleration is +6.25 m/s2. Through the next 3/4 of that distance, its acceleration is -2.08 m/s2. What are (a) its travel time through the 900 m and (b) its maximum speed?

Solve for the time at the 1/4 mark. That's 225 m. How? d = (1/2)at^2 ( initial velocity zero). Thus 225 = (1/2) 6.25 t^2. t^2 = ( 225 * 2 ) / 6.25. t = 8.5 sec. 

At the other end t^2 = (675 * 2) / 2.08 -- we reversed the sign and ran time backwards. t = 25.5 sec. 

So total time is 8.5 + 25.5 or 34 sec. 

Since zero initial velocity: v^2 = 2 a d. Here, v^2 = 2 * 6.25 * 225. v = 53 m/s. That's the fastest speed since braking then occurs.

7 0

4 years ago

Imagine you are holding a box in your hand, the force required for you to hold the box with mass 4.54kg and you want to accelera

svetlana [45]

Answer:

(a) Fₓ = 0 N

$F_y$ = 9.08 N

(b)

(a) Fₓ = 0 N

$F_y$ = 9.08 N

Fₓ = 9.08 N

Explanation:

The magnitude of the force will remain the same in each case, which is given as follows:

F = ma (Newton's Second Law)

where,

F = force = ?

m = mass = 4.54 kg

a = acceleration = 2 m/s²

Therefore,

F = (4.54 kg)(2 m/s²)

F = 9.08 N

Now, we come to each scenario:

(a)

Since the motion is in the vertical direction. Therefore the magnitude of the force in x-direction will be zero:

Fₓ = 0 N

For upward direction the force will be positive:

$F_y$ = 9.08 N

(b)

Since the motion is in the vertical direction. Therefore the magnitude of the force in x-direction will be zero:

Fₓ = 0 N

For upward direction the force will be negative:

$F_y$ = - 9.08 N

(c)

Since the motion is in the horizontal direction. Therefore the magnitude of the force in y-direction will be zero:

$F_y$ = 0 N

Fₓ = 9.08 N

3 0

3 years ago

A cheetah is walking at 1.0 m/s when it sees a zebra 25 m away. what acceleration would be required to reach 20.0 m/s in that di

inna [77]

Let
a = acceleration.
v = Speed
r = position
t = time.
To find the answer, you must write the equations of speed, and position known and clear the variables that are needed to find the acceleration.
The answer is a = 7.98m / s ^ 2.
I attach the solution.