All categories

4 years ago

A car traveling at 37 m/s starts to decelerate steadily. It comes to a complete stop in 6 seconds.what is its acceleration? m/s

1 answer:

7 0

Magnitude of acceleration = (change in speed) / (time for the change)

= (end speed - start speed) / (time for the change)

= (0 - 37 m/s) / (6 sec)

= ( - 37 / 6) (m/s²)

= - 6-1/6 m/s²

= -6.1667 m/s² (rounded)

You might be interested in

A pressure cylinder has a diameter of 150-mm and has a 6-mm wall thickness. What pressure can this vessel carry if the maximum s

myrzilka [38]

Answer:

p = 8N/mm2

Explanation:

given data ;

diameter of cylinder = 150 mm

thickness of cylinder = 6 mm

maximum shear stress = 25 MPa

we know that

hoop stress is given as = $\frac{pd}{2t}$

axial stress is given as = $\frac{pd}{4t}$

maximum shear stress = (hoop stress - axial stress)/2

putting both stress value to get required pressure

$25 = \frac{ \frac{pd}{2t} -\frac{pd}{4t}}{2}$

$25 = \frac{pd}{8t}$

t = 6 mm

d = 150 mm

therefore we have pressure

p = 8N/mm2

7 0

3 years ago

Which statement best compares an asteroid with Earth?

IgorLugansk [536]

Answer:C) Both have rocky composition

6 0

3 years ago

What is the name of the particle that compounds are made of?

VashaNatasha [74]

Answer:

Explanation:

atoms

8 0

3 years ago

What is a work out time setting? (Gym)

tigry1 [53]

Answer:

It determines how long you do a certain workout.

3 0

3 years ago

A railroad track and a road cross at right angles. An observer stands on the road and watches an eastbound train traveling at 60

mamaluj [8]

Answer:

After 4 s of passing through the intersection, the train travels with 57.6 m/s

Solution:

As per the question:

Suppose the distance to the south of the crossing watching the east bound train be x = 70 m

Also, the east bound travels as a function of time and can be given as:

y(t) = 60t

Now,

To calculate the speed, z(t) of the train as it passes through the intersection:

Since, the road cross at right angles, thus by Pythagoras theorem:

$z(t) = \sqrt{x^{2} + y(t)^{2}}$

$z(t) = \sqrt{70^{2} + 60t^{2}}$

Now, differentiate the above eqn w.r.t 't':

$\frac{dz(t)}{dt} = \frac{1}{2}.\frac{1}{sqrt{3600t^{2} + 4900}}\times 2t\times 3600$

$\frac{dz(t)}{dt} = \frac{1}{sqrt{3600t^{2} + 4900}}\times 3600t$

For t = 4 s:

$\frac{dz(4)}{dt} = \frac{1}{sqrt{3600\times 4^{2} + 4900}}\times 3600\times 4 = 57.6\ m/s$

4 0

3 years ago