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Marina CMI [18]
3 years ago
8

An object moving at 30 m/s takes 5 sec. to come to a stop.  What is the object’s acceleration​

Physics
1 answer:
Klio2033 [76]3 years ago
4 0

Answer:

6 m/s²

Explanation:

a=v/t

a= 30 m/s÷ 5 sec= 6 m/s²

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I need to find 1).a,b,c
Aleksandr [31]
Let's cut through the weeds and the trash
and get down to the real situation:

                  A stone is tossed straight up at  5.89 m/s .
                  Ignore air resistance.


Gravity slows down the speed of any rising object by  9.8 m/s every second.
So the stone (aka Billy-Bob-Joe) continues to rise for

                     (5.89 m/s / 9.8 m/s²)  =  0.6 seconds.

At that timer, he has run out of upward gas.  He is at the top
of his rise, he stops rising, and begins to fall.

His average speed on the way up is  (1/2) (5.89 + 0) = 2.945 m/s .

Moving for 0.6 seconds at an average speed of  2.945 m/s,
he topped out at

                    (2.945 m/s) (0.6 s) =  1.767 meters above the trampoline.

With no other forces other than gravity acting on him, it takes him
the same time to come down from the peak as it took to rise to it.

   (0.6 sec up) + (0.6 sec down)  =  1.2 seconds until he hits rubber again.



 
5 0
4 years ago
What are main causes by tornadoes
brilliants [131]
Updrafts and downdrafts in a thunderstorm interacting with wind shear
7 0
4 years ago
An object has a kinetic energy of 275 j and the momentum of magnitude of 25.0 kg m/s. find the speed and mass of the object
lyudmila [28]
<span>Recall formula for Kinetic energy is: KE = 1/2mv^2, where KE = 275J and momentum (which is 25.0 kg m/s) = m*v Therefore substitute for KE and mv in the equation above to get speed => 275 = 0.5 * 25 * v v = 275/12.5 v = 22m/s to get mass m, recall momentum = m*v => 25= 22*m m= 25/22 = 1.3663kg</span>
6 0
4 years ago
Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150
Sever21 [200]

Answer:2.89\approx 2.9^{\circ}C/s

Explanation:

Given

Power\left ( P\right )=150 MW

mass of core\left ( m\right )=1.60\times 10^5 kg

Average specific heat \left ( C\right )=0.3349 KJ/kg^{\circ}C

And rate of increase of temperature =\frac{\mathrm{d}T}{\mathrm{d} t}

Now

P=mc\frac{\mathrm{d}T}{\mathrm{d} t}

150\times 10^6=1.60\times 10^5\times 0.3349\times \frac{\mathrm{d}T}{\mathrm{d} t}

Thus \frac{\mathrm{d}T}{\mathrm{d} t}=[tex]\frac{1.60\times 10^5\times 0.3349}{150\times 10^6}

\frac{\mathrm{d}T}{\mathrm{d} t}=2.89\approx 2.9^{\circ}C/s

6 0
4 years ago
Hank and Harry are two ice skaters whiling away time by playing 'tug of war' between practice sessions. They hold on to opposite
Alex73 [517]

Answer:

the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1

Explanation:

Given the data in the question;

Hank and Harry are two ice skaters, since both are on top of ice, we assume that friction is negligible.

We know that from Newton's Second Law;

Force = mass × Acceleration

F = ma

Since they hold on to opposite ends of the same rope. They have the same magnitude of force |F|, which is the same as the tension in the rope.

Now,

Mass_{Hank × Acceleration_{Hank = Mass_{Henry × Acceleration_{Henry

so

Mass_{Hank /  Mass_{Henry = Acceleration_{Henry / Acceleration_{Hank

given that; magnitude of Hank's acceleration is 1.26 times greater than the magnitude of Harry's acceleration,

Mass_{Hank /  Mass_{Henry = 1 / 1.26

Mass_{Hank /  Mass_{Henry = 0.7937 or [ 0.7937 : 1 ]

Therefore, the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1 ]

8 0
3 years ago
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