Answer:
Only a backward force is acting, no forward force.
Explanation:
- Once released from the initial push, in absence of friction, the shopping cart would continue moving forward at a constant speed forever.
- As it would move at a constant speed, no net force would be acting on it.
- So, if it is gradually slowing, there must be a net force producing an acceleration in a direction opposite to the movement.
- This force is the kinetic friction force, and is the only force acting on the cart in the horizontal direction.
- As any friction force, opposes to the relative movement between the cart and the horizontal floor, which means that is directed backward.
- This is consistent with the direction of the acceleration of the cart.
Answer:
The direction of the magnetic field at point Z; Into the screen
Explanation:
Answer:
The correct answer is B.
The astronaut will know due to the light from the explosion.
Explanation:
Sound and vibrations require a medium such as air to travel through. Space, there is no air. Only a vacuum. So sound and vibrations are unable to travel. Light requires no medium to travel. It can go through a vacuum.
Therefore the Astronaut will see a bright flash of light as it travels from the explosion to outer space. It is also important to note that light can travel very far because nothing else interacts with its wave particles and as such, it cannot be impeded.
Cheers!
By definition we have that the final speed is:
Vf² = Vo² + 2 * a * d
Where,
Vo: Final speed
a: acceleration
d: distance.
We cleared this expression the acceleration:
a = (Vf²-Vo²) / (2 * d)
Substituting the values:
a = ((0) ^ 2- (60) ^ 2) / ((2) * (123) * (1/5280))
a = -77268 mi / h ^ 2
its stopping distance on a roadway sloping downward at an angle of 17.0 ° is:
First you must make a free body diagram and see the acceleration of the car:
g = 32.2 feet / sec ^ 2
a = -77268 (mi / h ^ 2) * (5280/1) (feet / mi) * (1/3600) ^ 2 (h / s) ^ 2
a = -31.48 feet / sec ^ 2
A = a + g * sin (θ) = -31.48 + 32.2 * sin17.0
A = -22.07 feet / sec ^ 2
Clearing the braking distance:
Vf² = Vo² + 2 * a * d
d = (Vf²-Vo²) / (2 * a)
Substituting the values:
d = ((0) ^ 2- (60 * (5280/3600)) ^ 2) / (2 * (- 22.07))
d = 175.44 feet
answer:
its stopping distance on a roadway sloping downward at an angle of 17.0 ° is 175.44 feet
