A shopper pushes a grocery cart 20.0 m at constant speed on level ground, against a 35.0 n frictional force. he pushes in a dire
1 answer:
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To develop this problem it is necessary to apply the concepts given in the balance of forces for the tangential force and the centripetal force. An easy way to detail this problem is through a free body diagram that describes the behavior of the body and the forces to which it is subject.
PART A) Normal Force.
Here,
Normal reaction of the ring is N and velocity of the ring is v
PART B) Acceleration
Negative symbol indicates deceleration.
<em>NOTE: For the problem, the graph in which the turning radius and the angle of suspension was specified was not supplied. A graphic that matches the description given by the problem is attached.</em>
Answer:
98.13m
Explanation:
Complete question
Daniel is 50.0 meters away from a building. Tip of the building makes an angle of 63.0° with the horizontal. What is the height of the building
CHECK THE ATTACHMENT
From the figure, using trigonometry
Tan(θ ) = opposite/adjacent
Where Angle (θ )= 63°
Opposite= X = height of the building
Adjacent= 50 m
Then substitute the values we have
Tan(63)= X/50
1.9626= X/50
X= 1.9626 × 50
X= 98.13m
Hence, the height of the building is 98.13m
Answer:
h=18.05 cm
Explanation:
Given that
m= 25 kg
K= 1300 N/m
x=26.4 cm
θ= 19.5 ∘
When the block just leave the spring then the speed of block = v m/s
From energy conservation
By putting the values
v=1.9 m/s
When block reach at the maximum height(h) position then the final speed of the block will be zero.
We know that
By putting the values
h=0.1805 m
h=18.05 cm