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Ganezh [65]
2 years ago
15

If something is traveling at 20 m/s constant velocity, is it in equilibrium? If a projectile is launched at a velocity of 20 m/s

, is it in equilibrium?
Physics
1 answer:
Irina18 [472]2 years ago
3 0

If something is traveling at 20 m/s constant speed AND its direction isn't changing, then its velocity is constant.  Another way to say that is: Its acceleration is zero.  Zero acceleration means zero NET force acting on the object, or a group of BALANCED forces acting on it, also called EQUILIBRIUM.  The required answer is: YES.

If a real projectile is launched, the force of gravity acts on it vertically downward.  There's no upward force acting on it to balance gravity.  Therefore, the forces on the projectile are NOT balanced, there IS a net vertical force on it, and it's NOT in equilibrium.  Too bad.


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5 0
2 years ago
The peak intensity of radiation from a star named sigma is 2 x 10^6 mmkay. What is the average surface temperature of Sigma roun
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8 0
3 years ago
Read 2 more answers
1.- Un barco recorre la distancia que separa Gran Canaria de Tenerife (90 km) en 6 horas. ¿Cuál es
bezimeni [28]

Answer:

The speed is 15 km/h or 4.16 m/s.

Explanation:

A boat travels the distance that separates Gran Canaria from Tenerife (90 km) in 6 hours. Which  the speed of the boat in km / h? And in m / s?

Given that,

Distance, d = 90 km = 90000 m

Time, t = 6 hours = 21600 s

Speed = distance/time

v=\dfrac{90\ km}{6\ h}\\\\=15\ km/h

or

v=\dfrac{90000\ m}{21600\ s}\\\\=4.16\ m/s

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7 0
2 years ago
A force of 55N accelerates a 7.5kg wagon at 5.3 m/s^2 along a road. How large is the frictional force?
Varvara68 [4.7K]

Answer:

<h2>15.25 N</h2>

Explanation:

       A force of 55\text{ }N is acting on a wagon along the road. The wagon weights 7.5\text{ }kg. Acceleration of the wagon is given as 5.3\text{ }\frac{m}{s^{2}}.

       Consider the block as the system, the forces acting are Frictional force, Gravitational force, Normal reaction and External force applied by us.

       Gravitational Force and Normal Reaction cancel out each other.

       Net External Force = Mass of system/wagon \times Acceleration of wagon

       F_{ext}-F_{friction}=(7.5\text{ }kg)\times(5.3\text{ }\frac{m}{s^{2}})=39.75\text{ }N\\55\text{ }N-F_{friction}=39.75\text{ }N\\F_{friction}=15.25\text{ }N

F_{friction} has a negative sign because it opposes the motion of the wagon.

∴ Frictional Force = 15.25 N

4 0
2 years ago
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